A locomotive is accelerating at 1.60 m/s2. It passes through a 20.0-m-wide crossing in a time of 2.40 s. After the locomotive leaves the crossing, how much time is required until its speed reaches 31.6 m/s
Answer:
The time to reach 31.6 m/s is 14.54 s.
Explanation:
Given;
acceleration of the locomotive, a = 1.60 m/s²
distance traveled by the locomotive, s = 20 m
time of motion, t = 2.4 s
final velocity of the locomotive, v = 31.6 m/s
The speed of the of the locomotive as it leaves the 20 m wide crossing;
[tex]v = \frac{d}{t} \\\\v = \frac{20}{2.4} \\\\v = 8.333 \ m/s[/tex]
The time to reach 31.6 m/s is calculated from the equation below;
v = u + at
31.6 = 8.333 + 1.6t
1.6t = 31.6 - 8.333
1.6t = 23.267
t = 23.267 / 1.6
t = 14.54 s
Therefore, the time to reach 31.6 m/s is 14.54 s.
A stone of mass 1kg is thrown at 10m/s upwards making an angle of 37°with the horizontal from a building that is 20m high. Using the law of conservation of energy calculate the speed wjen the stone hits the ground.
Answer:
31.68 m/s
Explanation:
The law of conservation of energy states that energy is not lost or gained it is just converted, in this example, since it is not given any resistance from the wind, you'd have two variables Speed on the Y-axis and on the X-axis, since both of them would result in the same decrease and increase with against gravity, it doesn't matter the value of both.
As the stone continues to go upwards it will continue to lose speed due to de-acceleration from the gravity acting on it, similarly, it will continue to gain Potential energy, instead of kinetic energy, when it reaches its highest point the speed on Y will be "0" and the free fall will start, since the up and down movement will be equal in time the and acceleration would be equal -9.81 m/s and 9.81 m/s because the only acceleration you have is gravity, you only need to calculate how much speed will gain a rock accelerating at 9.81 m/s falling 20 m:
[tex]H=\frac{1}{2}g*t^{2} \\\frac{20}{1/2*9.81} =t^{2} \\\\t^{2} =4.08\\t=\sqrt{4.08} \\t=2.21[/tex]
Now we just add the time accelerating:
[tex]Vf=Vo+at\\Vf=10 m/s+ 2.21*9.81\\Vf=10 m/s+21.68\\Vf=31.68 m/s[/tex]
What is the shortest time that a jet pilot starting from rest can take to reach Mach-3.60 (3.60 times the speed of sound) without graying out? (Use 331 m/s for the speed of sound in cold air.)
Answer:
30.4 s
Explanation:
A pilot , with plane accelerated at 4 g starts greying out . In the problem , the acceleration of jet is 4 g
a = 4 x 9.8 = 39.2 m /s²
initial velocity u = 0
Final velocity = 3.60 times speed of sound
= 3.6 x 331 = 1191.6 m /s
v = u + at
Putting the values
1191.6 = 0 + 39.2 t
t = 30.4 s .
What must the charge (sign and magnitude) of a particle of mass 1.43 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.
Answer:
the sign and magnitude of the charge is - 2 x 10⁻⁵ C.
Explanation:
Given;
mass of the particle, m = 1.43 g = 0.00143 kg
electric field experienced by the particle, E = 700 N/C
The force experienced by the particle is calculated as;
F = mg = EQ
Where;
Q is the magnitude of the charge
[tex]Q = \frac{mg}{E} \\\\Q = \frac{0.00143 \times 9.8}{700} \\\\Q = 2\times 10^{-5} \ C[/tex]
The force must be upward in opposite direction to the electric field. Since the force and the electric field are in opposite direction, the charge must be negative.
Therefore, the sign and magnitude of the charge is - 2 x 10⁻⁵ C.
The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed.
The charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.
What is electric charge?The electric force experienced by the body when placed it into the electromagnetic field is called electric charge.
Given information-
The mass of the particle is 1.43 g or 0.00143 kg.
The magnitude of the downward-directed electric field is 700 N/C.
The magnitude of the free-fall acceleration is 9.80 meter per second squared.
The electric field is defined as the electric force per unit charge. It can be given as,
[tex]E=\dfrac{F}{q}[/tex]
Rewrite the equation for the charge,
[tex]q=\dfrac{F}{E}[/tex]
Force experienced by the particle is equal to the product of mass and free fall acceleration (gravity). Thus,
[tex]q=\dfrac{0.00143\times9.8}{700}\\q=2\times10^{-5}[/tex]
Thus the magnitude of the charge is [tex]2\times10^{-5}[/tex] C.
The particle to remain stationary, when placed in a downward-directed electric the force must be in opposite direction which upward directed. For the opposite direction the sign of the charge should be negative.
Thus the charge of the given particle to remain stationary should be [tex]-2\times10^{-5}[/tex] C.
Learn more about the electric charge here;
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Which of the following statements is true? A. Both warming up and cooling down or important. B. It is more important warm up then it is to cool down. C. Is more important to cool down then is to warm up. D. Both warming up and cooling down are not important
Answer:
No. A is correct because both warming up and cooling down are important
Both warming up and cooling down are important.
This is based on aerobics and human body balance regulation as regards exercising.
Warming up and cooling down in exercising are just based on the level of intensity at which the exercise is carried out.
Now, warming up when exercising involves activities like jogging. Warming up is a very vital part of exercising as it helps to get a person's cardiovascular system ready for the subsequent exercises and physical activities to be engaged. This will help in making sure there is enough blood flowinh to your muscles as well as increasing your body temperature.In another sense, cooling down is also very vital in activity because after the blood pressure and heart rates have been raised after exercising, they will need to be restored to their normal levels at which they were before commencement of the exercise. It also helps to regulate the blood flow.Thus, Both warming up and cooling down are important.
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A scientist reports a measurement of the temperature of the surface of a newly discovered planet as negative 20 Kelvin. What conclusion can you draw from this report
Answer:
The temperature of this newly discovered planet violates the third law of thermodynamic, there is a mistake in this value.
Explanation:
The third law of the Thermodynamic says:
At zero kelvin all molecular movement stops, which means that the entropy will be zero at this temperature.
So we can say there is no thermodynamic system that has temperature values less than 0 K.
The conclusion of the report will be.
The temperature of this new planet violates the third law of thermodynamic, there is a mistake in this value.
I hope it helps you!
Solar flares are disturbances on the Sun’s surface that send large amounts of energy away from the Sun’s surface. These disturbances can often interfere with technology here on Earth. What statement BEST describes the cause of solar flares?
Answer:
with that boom and that boom glare boom that boom glare with that boom the tha ba glare boom that boom glare with that.
Explanation:
A 1 800-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.60 m before coming into contact with the top of the beam, and it drives the beam 13.6 cm farther into the ground before coming to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.
Answer:
F = 614913.88 N
Explanation:
We are given;
Mass of pile driver; m = 1800 kg
Height of fall of pole driver; h = 4.6 m
Depth driven into beam; d = 13.6 cm = 0.136 m
Now, from energy equations and applying to this question, we can write that;
Workdone = Change in potential energy
Formula for workdone is; W = F × d
While the average potential energy here is; W = mg(h + d)
Thus;
Fd = mg(h + d)
Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.
Making F the subject, we have;
F = mg(h + d)/d
F = 1800 × 9.81 × (4.6 + 0.136)/0.136
F = 614913.88 N
A particle has 37.5 J of kinetic energy and 12.5 J of gravitational potential energy at one point during its fall from a tree to the ground. An instant before striking the ground, how much mechanical energy-rounded to the nearest Joule-does the particle have? Ignore air resistance.
Answer: 50J
Explanation:
Mechanical energy follows the same principles of kinetic energy and potential energy, it is conserved. So Ei = Ef.
Mechanical energy is the sum of ALL energy's. There is no friction, so its just kinetic plus potential.
37.5 + 12.5 = 50J
Since the particle has not touched the ground, it has not transferred any energy to the ground yet, therefore the mechanical energy must still be 50J; mostly in kinetic energy with a very small amount of potential because of the low height relative to the ground.
How much horizontal force FFF must a sprinter of mass 48 kgkg exert on the starting blocks to produce this acceleration
Answer:
720N
Explanation:
According to Newtons second law;
Force = Mass* Acceleration
Let the acceleration given be 15m/s²
Mass = 48kg
Substitute
Force = 48 * 15\
Force = 720N
Hence the required horizontal force is 720N
Note that the value of the acceleration was assumed
Two objects are located on an airtrack (you may assume there is no friction). The magnitude of the charge on object A is 5 times higher than on object B, and object A also has a 6 times higher mass than object B (the picture is not scale or necessarily in the correct direction). Each object is accelerating in the direction of the arrows.
a. Draw a system schema.
b. Draw the force body diagrams for each charge and identify all newton 3d law pairs.
c. Write (in symbolic form) expressions for the net force on each object.
d. Find how many times faster/slower the acceleration of object B is compared to object A?
Answer:
Explanation:
From the information given:
The schematic diagram for the system and the force body diagrams for each charge can be seen in the image attached below
(c)
The symbolic expressions for the net force on each object is as follows:
[tex]N_A = m_Ag \\ \\ F_{net}, A= F_A = \dfrac{kQ_AQ_B}{r^2}[/tex]
[tex]N_B = m_Bg \\ \\ F_{net}, B= F_B = \dfrac{kQ_AQ_B}{r^2}[/tex]
(d) From above
[tex]F_A =F_B[/tex]
[tex]m_Aa_A = m_Ba_B \\ \\ (6m_B)a_A = m_B a_B[/tex]
[tex]a_B = 6a_A[/tex]
4.The equation for Kinetic Energy is 1/2 x m x v2. Calculate the KE when the mass is 5 kg and the velocity is 4 m/s.
a.2 joules
b.10 joules
c.20 joules
d.40 joules
Explanation:
KE = ½mv²
= ½ × 5 × 4²
= ½ × 5 × 16
= ½ × 80
= 40 J
Earth has seasons because _____.
it rotates on its axis as it moves around the sun
the temperature of the sun changes
its axis is tilted
the distance between Earth and the sun changes
Answer:
c, its axis is tilted
maybe
As it works its way around the sun, its tilted axis exposes different parts of earth.
C would be it because the roation of Earth on its axis doesn't have anything to do with the exposer of the revolution on its axis
Two masses are suspended by cord that passes over a pulley with negligible mass. The cord also has negligible mass. One of the masses, m1, has a mass of 7.0 kg and the other mass, m2, has a mass of 3.0 kg. The pulley turns on a shaft through the center of the pulley and supports the pulley and all the masses. The vertical force of the shaft on the pulley that supports the whole system is
Answer: F = 98N
Explanation: The shaft have to sustain the pulley, the cord and the two masses. The pulley and the cord have negligible masses, so, they have negligible weight.
The two masses have two vertical forces acting on them: force of traction because of the cord and force due to gravitational force, also known as weight.
So, the vertical force the shaft has to support is the sum of the weight of each mass:
[tex]F_{net}=F_{g}_{1}+F_{g}_{2}[/tex]
[tex]F_{net}=m_{1}.g+m_{2}.g[/tex]
[tex]F_{net}=g(m_{1}+m_{2})[/tex]
[tex]F_{net}=9.8(7+3)[/tex]
[tex]F_{net}=[/tex] 98
The vertical force that supports the whole system is 98 N.
if you make a sound by tapping on a glass of water what is the order of vibration
A. glass - water - air
C. air - water - glass
B. water - glass - air
Answer:
A.) glass water air
Explanation:
hope this helps :) have a great day!!
Answer:
A.) glass - water - air
good luck, i hope this helps :)
The equilibrant is the equal to the resultant magnitude but opposite in direction.
True
False
Answer and I will give you brainiliest
Answer:
The answer is False......
Answer:
true
it is equal but opposite
5. Friction has two components __________ and ___________.
Answer:
static friction and kinetic friction.
Explanation:
Four point charges each having charge Q are located at the corners of a square having sides of length a. Find symbolic expressions for the work required to bring a fifth charge q from infinity to the center of the square.
Answer:
Explanation:
We shall first calculate electric potential at the centre of square due to four charges of Q .
potential = k Q / r , k = 9 x 10⁹ , Q is charge which is placed at distance r from the point where potential is measured .
Distance of corner of square to centre = √ 2 a / 2 = a / √2 = .707 a
potential due to one charge
= k x Q / .707 a
= 9 x 10⁹ x Q / .707a
= 12.73 x 10⁹ Q/a
Potential due to 4 charges
= 4 x 12.73 x 10⁹ Q/a
= 50.92 x 10⁹ Q/a .
Potential at infinity = 0
Work done in carrying charge q from infinity to centre of square
= potential difference x charge
= ( 50.92 x 10⁹ Q/a - 0 ) x q
= 50.92 x 10⁹ x Qq / a J .
A 12 V battery is connected across a device with variable resistance. As the resistance of the device increases, determine whether the following quantities increase, decrease, or remain unchanged. HINT (a) The current through the device. increases decreases remains unchanged (b) The voltage across the device. increases decreases remains unchanged (c) The power consumed by the device. increases decreases remains unchanged
Answer:
a) DECREASE, b) INCREASE, c) power remains constant
Explanation:
The resistance and the battery are connected in series.
a) How the current changes when the resistance changes
V = I R
I = V / R
I = 12 / R
if the resinification increases, the current must DECREASE
b) when you check the expression
V= I R
, if R increases the voltage of the INCREASE
c) the power in an electric circuit is
P = I V
Let's analyze this expression, the voltage increases linearly with the increase in resistance and the current decreases linearly with the increase in r, for which the two effects are compensated and the dissipated power remains constant
The nucleus of 8Be, which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons).
(a) What is the force between the two alpha particles when they are 6.60 ✕ 10−15 m apart? N
(b) What is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 4.0026 u. m/s2
Answer:
a) F = 21.16 N, b) a = 3.17 10²⁸ m / s
Explanation:
a) The outside between the alpha particles is the electric force, given by Coulomb's law
F = [tex]k \frac{ q_1 q_2}{r^2}[/tex]
in that case the two charges are of equal magnitude
q₁ = q₂ = 2q
let's calculate
F = [tex]9 \ 10^9 \ \frac{ (2 \ 1.6 \ 10^{-19} )^2 }{ (6.60 \ 10^{-15} )^2 }[/tex]
F = 21.16 N
this force is repulsive because the charges are of the same sign
b) what is the initial acceleration
F = ma
a = F / m
a = [tex]\frac{21.16}{4.0026 \ 1.67 \ 10^{-27} }[/tex]21.16 / 4.0025 1.67 10-27
a = 3.17 10²⁸ m / s
this acceleration is in the direction of moving away the alpha particles
An aircraft flies 800km due East and then 600km due north. Determine the magnitude of it's displacement. please it's urgent
Answer:
Magnitude of it's displacement = 1,000 km
Explanation:
Given:
Distance towards east = 800 km
Distance towards north = 600 km
Find:
Magnitude of it's displacement
Computation:
Magnitude of it's displacement = √800² + 600²
Magnitude of it's displacement = √640000 + 360000
Magnitude of it's displacement = √10000000
Magnitude of it's displacement = 1,000 km
There are two metallic spheres and a positive-charge generator. Throughout the entire problem, one sphere is always closest to the generator (call it C), and one is always farther away (call it F). Both spheres start electrically neutral and are in contact with each other. These spheres are now brought near the positivecharge generator (call it G), but do not touch it. The two metallic spheres C and F are then separated from each other, and then individually moved away from the positive-charge generator.
Immediately after the two spheres are moved away from the positive-charge generator, the sphere farther away from the charge generator is charged_______
a. positive
b. negative
c. neutral
d. cannot be determined
Answer:
The correct answer is C
Explanation:
In this problem we must remember that charges of the same sign repel and of the same sign attract.
Generator G induces a charge in the closest spheres C, otherwise the charge is negative, therefore in the furthest sphere F, with the ac that is in contact, a positive charge is formed, they do not indicate that the spheres It is in contact with Earth, so the charges remain on them.
As the spheres move away from the generator, the load is redistributed, they return to their initial load, that is, they remain Neutral since no load left or entered the spheres, there was only a redistribution due to the proximity of the generator.
The correct answer is C
what is the frequency of a wave having a period equal to 18 seconds
Explanation:
The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.
The relation between wave period and frequency is as follows.
T = \frac{1}{f}T=
f
1
where, T = time period
f = frequency
It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.
T = \frac{1}{f}T=
f
1
or, f = \frac{1}{T}f=
T
1
= \frac{1}{18 sec}
18sec
1
= 0.055 per second (1cycle per second = 1 Hertz)
or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz
Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .A cylinder of gas is at room temperature (20°C). The air conditioner breaks down, and the temperature rises
to 46°C. What is the new pressure of the gas relative to its initial pressure (Pi)?
Answer: we divide this equation by pV and use {C}_{p}={C}_{V}+ . ... The temperature, pressure, and volume of the resulting gas-air mixture
Explanation:
The new pressure of the gas relative to its initial pressure P₁ is P₁ × 1.09
From the question given above, the following data were obtained:
Initial temperature (T₁) = 20 °C = 20 + 273 = 293 K
Initial pressure (P₁) = P₁
New temperature (T₂) = 46 °C = 46 + 273 = 319 K
New pressure (P₂) =?The new pressure of the gas can be obtained as follow:
P₁ / T₁ = P₂ / T₂
P₁ / 293 = P₂ / 319
Cross multiply
293 × P₂ = P₁ × 319
Divide both side by 293
P₂ = (P₁ × 319) / 293
P₂ = P₁ × 1.09Thus, the new pressure relative to it's initial pressure is P₁ × 1.09
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A lake has a surface area of 410 m2 and a volume of 1140 m3 . Suppose that during the day, sunlight with a power averaging 820 W/m2 shines on the lake, and that about 10% of this power is absorbed in the lake, producing heat. Assume that the temperature of the lake water stays constant because the absorbed solar power is exactly balanced by heat lost due to evaporation of water from the lake surface. What is the evaporation rate, in g/s
Answer:
Explanation:
Total solar energy falling on total surface per second
= 410 x 820 W
= 336200 W
10 % of 336200 = 33620 J is converted into heat which is absorbed by lake water . But its temperature does not rise because heat is used up in evaporating water in the form of vapor .
Total heat released during evaporation = 33620 J
Let evaporation rate be m gram /s
heat absorbed by m gram water = m x latent heat of evaporation
= m x 2260 J .
Given ,
m x 2260 = 33620
m = 14.87 g /s .
You are working in a shoe test laboratory measuring the coefficients of friction for running shoes on a variety of surfaces. The shoes are pushed against the surface with a downward force of 400 N, and a sample of the surface material is then pulled out from under the shoe by a machine. The machine pulls with a force of 300 N before the material begins to slide. When the material is sliding, the machine only has to pull with a force of 200 N to keep the material moving. What is the coefficient of static friction between the shoe and the material
Answer:
0.75
Explanation:
Since the static frictional force is the maximum force applied just before sliding, our frictional force, F is 300 N.
Since F = μN where μ = coefficient of static friction and N = normal force = 400 N (which is the downward force applied against the surface).
So, μ = F/N
= 300 N/400 N
= 3/4
= 0.75
So, the coefficient of static friction μ = 0.75
PLEASEEEE HELP MEEEEE!!!!
Answer:
D. Yes
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Using Figure 2, what is the momentum of Train Car A before the collision?
A
180,000 kg*m/s
B
0 kg*m/s
C
11,250 kg*m/s
D
4 kg*m/s
Answer:
Option A. 180000 Kgm/s.
Explanation:
From the question given above, the following data were obtained:
For Train Car A:
Mass of train car A = 45000 Kg
Velocity of train car A = 4 m/s
Momentum of train car A =?
For Train Car B:
Mass of train car B = 45000 Kg
Velocity of train car B = 0 m/s
Momentum is simply defined as the product of mass and velocity. Mathematically, it can be expressed as:
Momentum = mass × velocity
With the above formula, the momentum of train car A before collision can be obtained as follow:
Mass of train car A = 45000 Kg
Velocity of train car A = 4 m/s
Momentum of train car A =?
Momentum = mass × velocity
Momentum = 45000 × 4
Momentum of train car A = 180000 Kgm/s
dose contact or noncontact force weaken with distance
Answer:
The more massive an object is, the greater the gravitational force. Since gravitational force is inversely proportional to the distance between two interacting objects, more separation distance will result in weaker gravitational forces.
Explanation:
hope this help:)
The all-digital touch-tone phones use the summation of two sine waves for signaling. Frequencies of these sine waves are defined as 697, 770, 852, 941, 1209, 1336, 1477, and 1633 Hz. Since the sampling rate used by the telecommunications is 8000 Hz, convert those eight analog frequencies into digital frequencies of radians and cycles.