9. A bullet of mass 8g is stationary in the barrel of a gun of mass 4kg. The trigger is pulled and a force of 2.8kN is exerted on the bullet for 1.7ms. Calculate or find: a) Velocity of the bullet as it leaves the barrel of the gun. b) Recoil velocity of the gun. c) Show by calculation which type of collision is taking place.
Answers
After the trigger is pulled, the momentum of the system is:
p = mv
where p is the momentum, m is the mass, and v is the velocity.
Before the bullet is fired:
p = 0
After the bullet is fired:
p = (0.008 kg)(v) + (4 kg)(-u)
where u is the recoil velocity of the gun.
Since there are no external forces acting on the system, the initial momentum is equal to the final momentum:
0 = (0.008 kg)(v) + (4 kg)(-u)
Solving for u, we get:
u = -(0.008 kg)(v)/(4 kg) = -0.002 v
Substituting this into the equation for the momentum, we get:
p = (0.008 kg)(v) + (4 kg)(0.002 v) = 0.016 v
The force applied to the bullet is given by:
F = ma
where F is the force, m is the mass, and a is the acceleration.
Rearranging this equation to solve for acceleration, we get:
a = F/m = (2.8 kN)/(0.008 kg) = 350,000 m/s^2
Using the kinematic equation:
v = u + at
where u is the initial velocity (which is zero), t is the time, and a is the acceleration, we can solve for the final velocity:
v = at = (350,000 m/s^2)(1.7 × 10^-3 s) = 595 m/s
Therefore, the velocity of the bullet as it leaves the barrel of the gun is 595 m/s.
b) The recoil velocity of the gun is given by:
u = -(0.008 kg)(v)/(4 kg) = -0.002 v = -1.19 m/s
c) In an elastic collision, kinetic energy is conserved, while in an inelastic collision, kinetic energy is not conserved.
The initial kinetic energy of the system is zero, since both the bullet and the gun are stationary. After the collision, the kinetic energy of the system is:
K = (1/2)mv^2 + (1/2)Mu^2
where K is the kinetic energy, m is the mass of the bullet, v is its velocity, M is the mass of the gun, and u is its recoil velocity.
Substituting the values we found earlier, we get:
K = (1/2)(0.008 kg)(595 m/s)^2 + (1/2)(4 kg)(1.19 m/s)^2 = 1340 J
The work done on the bullet is:
W = Fs = (2.8 kN)(2 × 10^-3 m) = 5.6 J
Since the work done on the bullet is less than the change in kinetic energy of the system, the collision is inelastic.
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Related Questions
Sonu is observing his image in a plane mirror. The distance between the mirror and his image is 4 m. If he moves 1 m towards the mirror, the distance between Sonu and his image will be: (a) 3 m (b) 5 m (c) 6 m (d) 8 m
Answers
Answer:
Originally Sonu and image is 8 m (4 m to mirror and 4 m to image)
If he moves 1 m towards the mirror the image distance will be reduced to (c) 6 m
A train is moving to the right. The force of air resistance and the surface friction from the tracks totals 259,800 N . In order for the train to keep moving at a constant velocity, how much forward force is the engine providing?
Answers
The engine is providing a forward force of 259,800 N to keep the train moving at a constant velocity.
What is forward force?
Forward force is the force applied in the direction of motion, which causes an object to move or maintain a constant velocity.
Constant velocity refers to the motion of an object moving in a straight line at a steady speed, without changing its direction. The velocity of the object is constant when its speed and direction remain unchanged over time, regardless of whether the object is moving or stationary.
Since the train is moving at a constant velocity, the net force acting on the train is zero. Therefore, the forward force provided by the engine must be equal in magnitude and opposite in direction to the force of air resistance and surface friction.
So, the magnitude of the forward force provided by the engine is:
Magnitude of forward force = Magnitude of force of air resistance and surface friction
Magnitude of forward force = 259,800 N
Therefore, the engine is providing a forward force of 259,800 N to keep the train moving at a constant velocity.
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How far away is the epicenter if P waves takes 12 minutes to get there
Answers
Answer:
4000 km
Explanation:
Answer: 1,6 KM
Explanation:
To determine the distance from the epicenter of an earthquake if P waves take 12 minutes to arrive, we need to use the formula:
distance = (time * 8) / 60
where "time" is the time it takes for the P waves to arrive in minutes, and "distance" is the distance from the epicenter in kilometers.
Substituting the given value of time, we get:
distance = (12 * 8) / 60
distance = 1.6 kilometers
Therefore, the distance from the epicenter of the earthquake is approximately 1.6 kilometers.
A 50g
block is attached to a
horizontal spring with spring
constant k = 3600 N/m
The spring is compressed by 5 cm,
as shown in the figure.
When the spring is released, will the
block be able to cross the top of the hill with
h =10m?
Answers
The spring can not cross the hill.
What is the elastic potential energy?Elastic potential energy is the potential energy stored in an object when it is deformed, or stretched or compressed, by a force. When an elastic object is stretched or compressed, it has the potential to return to its original shape and size, and the work done to stretch or compress it is stored as elastic potential energy.
We know that;
E = 1/2Ke^2
E = 0.5 * 3600 * (5 * 10^-2)^2
E = 4.5 J
Now the GPE of the hill is;
mgh = 50 * 10^-3 * 10 * 9.8
= 4.9 J
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Why do ionizing smoke detectors contain an isotope that undergoes alpha decay rather than beta or gamma decay?
Answers
Ionizing smoke detectors contain an isotope that undergoes alpha decay (such as americium-241) because alpha particles are more ionizing than beta or gamma particles.
What is Isotopes?
Isotopes are variants of an element that have the same number of protons in their atomic nuclei but differ in the number of neutrons. This means that isotopes of the same element have the same atomic number (which determines the element's chemical properties) but different atomic masses.
In an ionizing smoke detector, the alpha particles emitted by the radioactive source ionize the air in the detector's chamber, creating a current that is detected by a circuit. When smoke particles enter the chamber, they attach to the ions and reduce the current, triggering the detector's alarm.
Alpha particles are heavier and have a higher charge than beta or gamma particles, which means they interact more strongly with matter and create more ionization as they travel through air. This makes them more effective at ionizing the air in the detector's chamber, and therefore more sensitive to the presence of smoke particles.
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a skateboarder who weighs 68kg is going down a ramp at the very bottom of the ramp they have 2400 J of kinetic energy calculate the speed in m/s to one decimal place
Answers
Answer:
The skateboarder has 2400 J of kinetic energy. We can use the formula for kinetic energy to find the speed:
KE = 1/2 mv^2
KE = 1/2 (68 kg) (v^2)
2400 J = 1/2 (68 kg) (v^2)
2400 J = 34 kg v^2
2400 J / 34 kg = v^2
70.59 m/s = v^2
What is the resistance of resistor R1?
(1 point)
3.00
2.00
7.50
5.00
Answers
Answer:
Current in first junction R1, R2 = 5 amps
Since current thru R2 = 3 amps the current thru R1 = 2 amps
V1 = V = I1 R1 = 15 volts
R1 = 15 volts / 2 amps = 7.5 ohms
A pendulum swings along path WXYZ and back again.
Resistive forces can be ignored.
which statement describes the Total energy of the bob?
A. it has a maximum value at X
B. it has a maximum value at Y
C. it has a maximum value at Z
D. it has the same value at W,X,Y and Z
- give answer with explanation please :) -
Answers
a puck moves 2.35 m/s in a -22° direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50.0° direction what is Δy
Answers
A puck moves at 2.35 m/s in a -22° angle.It is pushed for 0.215 seconds by a hockey stick, changing its velocity to 6.42 m/s in a 50.0° direction. The negative sign indicates that the puck has fallen below its initial height.
How to solve the given problem with the help of trigonometry?
To solve the problem, we need to use the equations of motion and trigonometry.
First, let's find the initial velocity components:
[tex]\mathrm V_{i,x} = v_i cos(\theta_i) = 2.35 cos(-22^\circ) \approx 2.216\text{ m/s}[/tex]
[tex]\mathrm V_{i,y} = v_i sin(\theta_i) = 2.35 sin(-22^\circ) \approx -0.891\text{ m/s}[/tex]
where [tex]v_i[/tex] is the initial velocity and [tex]\theta_i[/tex] is the initial angle in degrees.
Next, we need to find the final velocity components:
[tex]\mathrm V_{f,x} = v_f cos(\theta_f) = 6.42 cos(50^\circ) \approx 4.121\text{ m/s}[/tex]
[tex]\mathrm V_{f,y} = v_f sin(\theta_f) = 6.42 sin(50^\circ) \approx 4.936\text{ m/s}[/tex]
where [tex]v_f[/tex] is the final velocity and [tex]\theta_f[/tex] is the final angle in degrees.
The change in velocity components are:
[tex]\Mathrm\Delta V_x = V_{f,x} - V_{i,x} \approx 1.905\text{ m/s}[/tex]
[tex]$\Delta V_y = V_{f,y} - V_{i,y} \approx 5.827\text{ m/s}[/tex]
The time of the push is t = 0.215 s.
Using the equation of motion, we can find the change in position in the y-direction:
[tex]$\Delta y = V_{i,y} t + \frac{1}{2} a_y t^2[/tex]
where a_y is the acceleration in the y-direction.
Since the puck is in flight, the only force acting on it is gravity, which causes an acceleration of [tex]-9.8 m/s^2[/tex] in the y-direction.
Plugging in the values, we get:
[tex]\Delta y = (-0.891\text{ m/s})(0.215\text{ s}) + \frac{1}{2} (-9.8\text{ m/s}^2)(0.215\text{ s})^2 \approx -0.139\text{ m}[/tex]
Note that the negative sign indicates that the puck has fallen below its initial height.
Therefore, the change in position in the y-direction is approximately 0.139 m.
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Answer:
0.434
Explanation:
Similarities between thermonic emissions and liquid vaporisation
Answers
Both processes involve a change in the state of matter. In thermionic emissions, electrons are emitted from a solid metal surface to form a gas or plasma, while in liquid vaporization, a liquid is converted into a gas or vapor.
Liquid vaporisation, also known as evaporation, is the process by which a liquid transforms into its gaseous state at a temperature below its boiling point. This occurs due to the random motion of molecules in a liquid, where some of the molecules near the surface gain enough kinetic energy to overcome the intermolecular forces holding them together and escape into the air as a gas.
During the process of liquid vaporization, the liquid absorbs energy from its surroundings in the form of heat, which increases the average kinetic energy of the molecules and causes them to evaporate. The rate of evaporation depends on several factors such as temperature, humidity, and surface area of the liquid.
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5. Deltas are built up by ____.
A. Leaching
B. Desposition
C. Erosion
D. Abrasion
Answers
Answer:
B. Disposition
Explanation:
How are deltas formed?In order for a delta to form, water must flow slowly so that the sediment can be deposited to create new landforms.
The higher up a ramp a marble is released, the further from A it lands. Why is this, from an energy perspective?
Answers
This is because the higher the marble is released, the more potential energy it has.
What is potential energy?Potential energy is the energy that is stored in an object due to its position, shape, or composition. This energy is released when the object is disturbed from its current state. Potential energy can be found in many different forms, such as gravitational, elastic, chemical, electrical, magnetic, and nuclear energy. Gravitational potential energy is the energy of an object due to its vertical position, and is calculated by multiplying its mass by the gravitational acceleration and its height. Elastic potential energy is the energy stored in a deformed material, such as a spring, that is released when the material returns to its original shape.
This potential energy is converted to kinetic energy as the marble rolls down the ramp, and this kinetic energy determines how far the marble will travel away from point A. The higher the initial potential energy, the further the marble will travel.
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An electron moves in a circular path with a radius of 0. 20 m at a constant speed of 1. 5 X 105 m/s. What is the period of its motion?
Answers
The period of motion of the electron is 2.09 × 10^-6 seconds.
What is period of motion ?
Period of motion can be defined as the time it takes for an object to complete one full cycle of its motion. In the context of circular motion, the period is the time it takes for an object to travel around a full circle and return to its starting position.
The period of motion, T, of an object moving in a circle can be calculated using the following formula:
T = 2πr/v
Where
r is the radius of the circle v is the speed of the objectIn this case, the electron moves in a circular path with a radius of 0.20 m at a constant speed of 1.5 × 10^5 m/s. Plugging these values into the formula gives:
T = 2π(0.20 m) / (1.5 × 10^5 m/s)
T = 2.09 × 10^-6 s
Therefore, the period of motion of the electron is 2.09 × 10^-6 seconds.
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One mole of an ideal gas expands isothermally at 270 K from 1 litreto 3. 5 litres. The
temperature is then increased to 300 K at constant volume. Next the gas is compressed
isothermally at 300 K to its initial volume. The gas is then returned to its initial state. Calculate the net work done during this process
Answers
The net work done during this process is -992 J. The negative sign indicates that work was done on the system, i.e. the gas lost energy to the surroundings.
W1 = (1 mol)(8.314 J/mol·K)(270 K) ln(3.5 L / 1 L) ≈ 5978 J
Next, the temperature is increased at constant volume, so no work is done:
W2 = 0
Then, the gas is compressed isothermally at 300 K to its initial volume, so we can use the same equation as before to calculate the work done:
W3 = (1 mol)(8.314 J/mol·K)(300 K) ln(1 L / 3.5 L) ≈ -6969 J
Finally, the gas is returned to its initial state, so the work done is:
W4 = 0
Therefore, the net work done is:
Wnet = W1 + W2 + W3 + W4 = 5978 J + 0 J - 6969 J + 0 J = -992 J
work done is defined as the product of force and displacement, where force is the amount of energy required to move an object and displacement is the distance over which the object is moved. Mathematically, work done is expressed as W = F x d, where W is the work done, F is the force applied and d is the displacement.
Work done can be positive, negative, or zero depending on the direction of force and displacement. When the force is in the same direction as displacement, work done is positive, when the force is in the opposite direction to displacement, work done is negative, and when the force and displacement are perpendicular, work done is zero.
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Problem: Calculate the length and orientation of a road of length 5 meters in a frame of reference which is moving with a velocity 0. 6c in a direction making an angle of 30 deg with the rod. Solution: e = 35 ^ 0 * 45
Answers
The length and orientation of the road in a frame of reference that is moving with the given velocity in a direction making an angle of 30 degrees with the road is calculated to be 2.641 meters and 16.83 degrees respectively.
Let's first calculate the velocity of the road in the frame of reference that is moving with a velocity of 0.6c. We can use the velocity addition formula to calculate this:
v' = (v - u) / (1 - uv/c²)
where
v = velocity of the road
u = velocity of the frame of reference = 0.6c
c = speed of light
Since the road is at rest in its own frame of reference, v = 0. Substituting the values, we get:
v' = (-0.6c) / (1 - 0.6c × 0 / c²)
v' = -0.8824c
The negative sign indicates that the road appears to be moving in the opposite direction in the moving frame of reference.
Now, let's calculate the length of the road in the moving frame of reference. We can use the length contraction formula to calculate this:
L' = L × √(1 - v'²/c²)
where
L = length of the road
v' = velocity of the road in the moving frame of reference
Substituting the values, we get:
L' = 5 × √(1 - (-0.8824c)²/c²)
L' = 2.641 meters
Therefore, the length of the road in the moving frame of reference is 2.641 meters.
Finally, let's calculate the orientation of the road in the moving frame of reference. We are given that the direction of the velocity of the frame of reference makes an angle of 30 degrees with the road. We can use the following formula to calculate the angle between the road and the velocity of the frame of reference in the moving frame of reference:
tan(e') = (tan(e) - u/c) / (1 - u×tan(e)/c)
where
e = angle between the road and the velocity of the frame of reference
u = velocity of the frame of reference = 0.6c
e' = angle between the road and the velocity of the frame of reference in the moving frame of reference
Substituting the values, we get:
tan(e') = (tan(30) - 0.6c/c) / (1 - 0.6c×tan(30)/c)
tan(e') = 0.294
Taking the inverse tangent, we get:
e' = 16.83 degrees
Therefore, the orientation of the road in the moving frame of reference is 16.83 degrees and the length of the road in the moving frame of reference is 2.641 meters.
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The complete question is :
Given a road of length 5 meters, calculate the length and orientation of the road in a frame of reference that is moving with a velocity of 0.6c in a direction making an angle of 30 degrees with the road.
A 700 N (70 kg) skydiver falls towards the Earth. If the force due to air
resistance is 322 N, what is the acceleration of the skydiver?
Label your answer as m/s/s and use +/- to show direction.
Answers
The acceleration of the skydiver is approximately +5.4 m/s².
What is the acceleration?We can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass times its acceleration:
Net force = mass x acceleration
In this case, the net force is the force due to gravity minus the force due to air resistance:
Net force = force due to gravity - force due to air resistance
So we have:
(700 N) - (322 N) = (70 kg) x acceleration
Simplifying this equation, we get:
378 N = (70 kg) x acceleration
To find the acceleration, we can divide both sides by the mass:
acceleration = 378 N / (70 kg)
acceleration = 5.4 m/s²
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If 400 grams of water is to be heated from 24.0°C to 100.0°C to make a cup of tea, how much heat must be added? The specific heat of water is 4180 J/g∙C
Answers
Taking into account the definition of calorimetry, if 400 grams of water is to be heated from 24.0°C to 100.0°C to make a cup of tea, the heat added is 127254.4 J.
Definition of sensible heatWhen heat absorbs or releases from a substance causes a temperature change in it without affecting its physical state (phase change), it is called sensible heat.
The expression that allows to calculate heat exchanges is:
Q = c× m× ΔT
where:
Q is the heat exchanged by a body of mass m.c is the specific heat substance.ΔT is the temperature variation.Heat addedIn this case, you know:
Q= ?c= 4.180 J/g∙Cm= 400 gΔT= Tfinal - Tinitial= 100 C - 24 C= 76 CReplacing in the definition of sensible heat:
Q = 4.180 J/g∙C× 400 g× 76 C
Solving:
Q= 127254.4 J
Finally, the heat added is 127254.4 J.
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If a 1800 kg car starts at rest and speeds up to 5. 0 m/s then the work done is 22,500 J
Answers
According to the solving the statement is correct. The work done on the car as it speeds up to 5.0 m/s is 22,500 J.
What constitutes work completed?The work performed by a force is the sum of the movement and the component of something like the applied pressure of the object in the displacement direction. Work is accomplished when we push a block with a certain amount of force, "F," which causes the body to move with a certain amount of acceleration. Work completed is represented by the formula W = F.
According to the given information:KEf = (1/2)mv²
= (1/2)(1800 kg)(5.0 m/s)²
= 22,500 J
The work done on the car is equal to the change in kinetic energy:
W = KEf - KEi
= 22,500 J - 0 J
= 22,500 J
Therefore, the statement is correct. The work done on the car as it speeds up to 5.0 m/s is 22,500 J.
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I understand the the question you are looking for is:
If a 1800 kg car starts at rest and speeds up to 5. 0 m/s then the work done is?
State whether the following statements are true or false. Write the question number and T or F next to the number. For example, 4. 3. 1. T, if the answer for question 4. 3. 1 is true. 4. 3. 1 Magnetomotive force is the production of magnetic field by current in a conductor. (1) 4. 3. 2 25 mA is necessary for the coil with 4000 turns to produce 1000 A. T of mmf. (1) 4. 3. 3 When unlike poles, that is, south and north poles, are placed close together, an attraction force is produced by the magnetic fields. (1) 4. 3. 4 Magnetic lines of force never cross one another. (1) 4. 3. 5 When a magnetic field is moved past a stationery wire, there is relation motion. -1
Answers
The first two statements are false and the remaining three statements are true.
1. Magnetomotive force is the production of magnetic field by current in a conductor.
This statement is false. Magnetomotive force (MMF) is a measure of the strength of a magnetic field produced by a magnetic circuit, not by current in a conductor.
2. 25 mA is necessary for the coil with 4000 turns to produce 1000 A-T of MMF.
This statement is false. To produce 1000 A-T of MMF in a coil with 4000 turns, the current required would be:
MMF = N * I
1000 A-T = 4000 turns * I
I = 0.25 A or 250 mA
3. When unlike poles, that is, south and north poles, are placed close together, an attraction force is produced by the magnetic fields.
This statement is true. According to the laws of magnetism, opposite poles (north and south) attract each other, while like poles (north and north, or south and south) repel each other.
4. Magnetic lines of force never cross one another.
This statement is true. Magnetic lines of force (also known as magnetic field lines) are imaginary lines used to represent the direction and strength of a magnetic field. These lines are always continuous and form closed loops, never crossing one another.
5. When a magnetic field is moved past a stationary wire, there is relative motion.
This statement is true. According to Faraday's law of electromagnetic induction, a changing magnetic field induces an electromotive force (EMF) in a nearby conductor. This EMF causes current to flow in the conductor, creating a voltage.
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2. A huge snow storm dropped 10 inches of snow on Joe's driveway. His driveway is 126 meters long and Joe must shovel to get his car out of the garage. Joe pushes the shovel with a force of 10 N. How much work (in Newton meters) does Joe do to clear one path down his driveway?
Answers
The amount of work that Joe does to clear one path down his driveway is 1260 Nm.
How much work does Joe do to clear one path down his driveway?
The amount of work that Joe does to clear one path down his driveway is given by the product of the force he applies and the distance he moves the shovel:
Work = Force x Distance
In this case, the force is 10 N and the distance is the length of the driveway, which is 126 meters. Therefore:
Work = 10 N x 126 m = 1260 Nm
So Joe does 1260 Nm of work to clear one path down his driveway.
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In which type of relationship do both organisms benefit?
Answers
Answer: I think the answer is symbiotic tell me if I am right
Explanation:
Answer:
Symbiotic
Explanation:
Pathogenic relationships refer to things like microbes or viruses, think along the lines of how diseases spread throughout the body. One party is benefitted but the other is not.
'Microorganism' and 'macroorganism' refer to types of organisms rather than types of relationships.
Make up your own average speed question and provide the answer
Answers
For instance, someone travelling at an average speed of 40 miles divided by 40 minutes, or 1 mile per minute, is travelling 20 miles north and then 20 miles south (60 mph).
What is the typical speed?Calculating average speed involves dividing the whole distance travelled by the total time it took to cover that distance. Speed is the rate of movement of something at a specific time. Average speed refers to the average speed during the course of a journey.
What does the speed formula mean?Speed equals distance x time is the speed equation. The meter per second, abbreviated as m/s or ms-1, is the SI unit of speed.
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A 53. 0 kg stunt pilot who has been diving her airplane vertically pulls out of the dive by changing her course to a circle in a vertical plane. What is the apparent weight of the pilot at the lowest point of the pullout?
Answers
The apparent weight of the pilot at the lowest point of the pullout 1040 N.
The centripetal force, F = mv²/r, where m is the mass of the pilot, v is her speed, and r is the radius of the circle. At the lowest point of the pullout, the radius of the circle is equal to the distance from the center of the circle to the point where the plane changes direction (i.e., the height above the ground).
Since the plane is diving vertically, the initial speed is, v₀ = √(2gh), where h is the initial height of the plane above the ground, and g is the acceleration due to gravity. At the lowest point of the pullout, all of the potential energy that the plane had at the start of the dive is converted into kinetic energy. Therefore, the speed, v = √(2gh).
Substituting the expressions for v and r into the expression for the centripetal force,
F = m * (2gh) / (2h) = mg
Therefore, the apparent weight of the pilot at the lowest point of the pullout is twice her actual weight,
2 * (53.0 kg * 9.81 m/s²) = 1040 N
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How much work w, in joules, would you have to do to bring the third charge, q3, from very far away to the point p?
Answers
The work required to bring the third charge from very far away to the point P is approximately 36 millijoules (to two significant figures). The units are joules (J).
To calculate the work required to bring the third charge q3 from very far away to the point P, we need to know the electric potential at point P due to the other two charges q1 and q2. The electric potential at point P is given by:
V = kq1/r1 + kq2/r2
U = q3 V
where q3 is the charge of the third charge.
r1 = 0.1 m
r2 = 0.2 m
q1 = -2 μC = -2 × 10^-6 C
q2 = 3 μC = 3 × 10^-6 C
Substituting these values into the electric potential formula, we get:
V = kq1/r1 + kq2/r2
V = (9 × 10^9 N m^2/C^2) × [(-2 × 10^-6 C)/0.1 m + (3 × 10^-6 C)/0.2 m]
V ≈ 9000 V
final = q3 V
final = (4 μC) × (9000 V)
final = 36000 μJ
final = 36 mJ
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PHYSICAl SCIENCE : How do jewelers detectives antique dealer or other scientist could use electromagnetic waves to determine the composition of unknown materials in the course of their jobs?
Answers
Answer: It detects any waves or any substance.
Explanation: Don't really know how to explain it but it detects specific stuff on every material, like a demand for Example, if it is real then it could scratch glass but glass couldn't scratch it.
Compute the ratio between the luminosity from part b to the mass from part a. How does that ratio compare with the sun's ratio of luminosity to mass?
Answers
L/M = 1000 L Sun/MSun Based on the spectral category and luminosity group of a star, a classification is made. Cepheids are an illustration of a variable star.
A)M≈1MSun
b)L≈1000LSun
c)1000
D) Wrong; every main-sequence star upon that HR diagram turns helium into hydrogen. T 10 7 years
SUPER HUGE AND RED GIANT:
The H-R diagram's upper right corner may be located there. It is incredibly cool but very bright.
THE SUN, FOR EXAMPLE, IS IN THE Major, AS ARE THE HEATEST AND LIGHTEST STARS IN The GALAXY.
Very hot and faint White Dwarfs with a radius that isn't significantly greater then Earth.
Stars on the lower left of both the H-R diagram have smaller radii and higher temperatures than low-luminosity main-sequence stars.
Main-sequence stars with a lengthy lifetime are much less brilliant, colder, smaller, and far less massive than main-sequence stars having short lifetimes.
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''Complete question"
What is the approximate mass of the most massive stars left on the main sequence of this star cluster?
a) M≈0.1M Sun
b) M≈1 M Sun
c) M≈10M Sun
d) M≈100M Sun
Part B
What is the luminosity of the most luminous stars in the cluster?
a) L≈ 1LSun
b) L≈ 10LSun
c) L≈ 100LSun
d) L≈ 1000LSun
Part C
Compute the ratio between the luminosity from part B to the mass from part A. How does that ratio compare with the Sun's ratio of luminosity to mass?
L / M = __ LSun /MSun
Part D
Estimate the maximum amount of time these very luminous stars can last as red giants from your answer to part C.
a) t≈ 10^5 years
b) t≈ 10^7 years
c) t≈ 10^9 years
d) t≈ 10^11 years
Mars's moon Phobos is only about 12 km wide. Your weight on Phobos would be
0.001 what it is on Earth. Its escape velocity is 12 m/s. This velocity results in enough
kinetic energy to produce zero PE at infinite distance. You could launch yourself into
orbit around Phobos with a strong, very fast jump. Compare this jumping speed to
escape velocity, and describe what would happen if you jumped at an angle.
Answers
If you jumped at an angle, it would depend on the angle you jumped at and the speed you jumped at. If you jumped at an angle that was greater than the escape velocity of 12 m/s, you would be able to escape.
What is angle ?An angle is a figure formed by two rays, or line segments, that extend from a common endpoint. It is a measure of the amount of turn between the two rays. The angle is measured in degrees, with a full circle being 360 degrees. Angles can be acute, obtuse, right, reflex, or straight. Acute angles measure less than 90 degrees, whereas obtuse angles measure more than 90 degrees. Right angles measure exactly 90 degrees, reflex angles are greater than 180 degrees, and straight angles measure exactly 180 degrees.
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Different types of waves on the electromagnetic spectrum share the same _____?
Answers
Answer:
Different types of waves on the electromagnetic spectrum share the same speed, which is the speed of light in a vacuum, approximately 299,792,458 meters per second (or about 186,282 miles per second). All types of electromagnetic waves, including radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays, travel at the same speed in a vacuum. This is one of the fundamental properties of the electromagnetic spectrum.
If a 50g object is moving initially with 40m/s, after 2 min. it covered 670 m with 90 m/s.
a- Calculate the force applied causing it to speed up.
b- Calculate the work done by this force.
c- If the coefficient of friction for the ground is 0.34, What is the work done by friction?
d- What is the net work done?
Answers
Answer:
a) We can use the following formula to calculate the acceleration of the object:
a = (v_f - v_i) / t
where a is the acceleration, v_f is the final velocity, v_i is the initial velocity, and t is the time interval. Substituting the given values, we get:
a = (90 m/s - 40 m/s) / (2 min * 60 s/min) = 0.83 m/s^2
The force applied causing it to speed up can be found using Newton's second law of motion:
F = m * a
where F is the net force, m is the mass of the object, and a is the acceleration. Substituting the given values, we get:
F = 0.05 kg * 0.83 m/s^2 = 0.042 N
Therefore, the force applied causing the object to speed up is 0.042 N.
b) The work done by this force can be calculated using the following formula:
W = F * d
where W is the work done, F is the net force, and d is the displacement of the object. The displacement of the object is given by:
d = 670 m
Substituting the given values, we get:
W = 0.042 N * 670 m = 28.14 J
Therefore, the work done by the force is 28.14 J.
c) The work done by friction can be calculated using the formula:
W_friction = F_friction * d
where W_friction is the work done by friction, F_friction is the force of friction, and d is the displacement of the object. The force of friction can be calculated using:
F_friction = μ * F_norm
where μ is the coefficient of friction and F_norm is the normal force. The normal force is equal to the weight of the object, which is given by:
F_weight = m * g
where m is the mass of the object and g is the acceleration due to gravity. Substituting the given values, we get:
F_weight = 0.05 kg * 9.81 m/s^2 = 0.49 N
The normal force is equal in magnitude to the weight of the object, so we have:
F_norm = F_weight = 0.49 N
Substituting the given coefficient of friction, we get:
F_friction = 0.34 * 0.49 N = 0.17 N
The work done by friction can now be calculated by substituting the values we have found:
W_friction = 0.17 N * 670 m = 113.9 J
Therefore, the work done by friction is 113.9 J.
d) The net work done can be calculated as the sum of the work done by the applied force and the work done by friction:
W_net = W_applied + W_friction
Substituting the values we have found, we get:
W_net = 28.14 J + 113.9 J = 142.0 J
Therefore, the net work done is 142.0 J.
what is the qinghai earthquake compressional wave
Answers
The earth quakes are produced due to the sudden movements of tectonic plates in the earth's crust. The earthquake is one of the major natural disasters.
What is earthquake?An earthquake is defined as the shaking of the surface of the earth because of the sudden release of energy in the earth's crust. The waves created during the earthquake are called seismic waves.
The Qinghai earthquake was one of the severe earthquake which occurred on April 14, 2010 in the isolated southern Yushu Tibetan area in Qinghai province, which is in china on the northeastern portion of the plateau of Tibet.
Many peoples were killed and damaged many belongings also.
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