Answer:
4ac (quadruples)
Explanation:
The general formula for centripetal acceleration is given as follows:
[tex]a_{c} = \frac{v^2}{r}\\\\[/tex] -------------- equation (1)
where,
ac = centripetal acceleration
v = tangential speed
r = radius of circular path
Now, if we double tangential speed, v' = 2v, then the acceleration will become:
[tex]a_{c}' = \frac{(2v)^2}{r}\\\\a_{c}' = \frac{4v^2}{r}[/tex]
using equation (1):
[tex]a_{c}' = 4a_{c}[/tex]
therefore, the correct answer is:
4ac (quadruples)
what is gravity ?short ans
Answer:
Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward one another. On Earth, gravity gives weight to physical objects, and the Moon's gravity causes the ocean tides.
In the year 2000, a company made $4.7 million in profit. For each consecutive year after that, their profit increased by 15%. How much would the company's profit be in the year 2004, to the nearest tenth of a million dollars?
Answer: $8.2 million.
Explanation:
If we have a quantity A, and we have an increase of x%, this can be written as:
A + (x%/100%)*A
Now, for this particular case we have:
In year 2000 (we can define this year as the year zero, y = 0) the initial value is $4.7 million.
The next year, y = 1, there is an increase of 15%, then we will have a profit of:
P = $4.7 million + (15%/100%)*$4.7 million = $4.7 million + 0.15*$4.7 million
P = $4.7 million*(1 + 0.15) = $4.7 million*(1.15)
in the next year, y = 2, the profit will be:
P = $4.7 million*(1.15) + (15%/100%)* $4.7 million*(1.15)
= $4.7 million*(1.15) + 0.15* $4.7 million*(1.15)
= $4.7 million*(1.15)^2
We already can see the pattern, the profit in the year y will be:
P(y) = $4.7 million*(1.15)^y
In particular, in the year 2004 we have y = 4, then the profit that year will be:
P(4) = $4.7 million*(1.15)^4 = $8.2 million.
What is the force of gravity acting on a 2000 kg spacecraft when it is about the Earth at a in orbit distance of 8.375 x 106 m?
Answer:
W = 13.44 KN
Explanation:
First, we need to find the value of acceleration due to gravity at the specified height:
[tex]g' = g(1-2\frac{h}{R})[/tex]
where,
g' = value of acceleration due to gravity at given height = ?
g = value of acceleration due to gravity at surface of earth = 9.81 m/s²
h = height of space craft = 8.375 x 10⁶ m - 6.37 x 10⁶ m = 2.005 x 10⁶ m
R = Radius of Earth = 6.37 x 10⁶ m
Therefore,
[tex]g' = (9.81\ m/s^2)(1-2\frac{2.005\ x\ 10^6\ m}{6.37\ x\ 10^6\ m})\\\\g' = (9.81\ m/s^2)(1 - 0.315)\\g' = 6.72\ m/s^2[/tex]
Now, we can find the weight or force of gravity on space craft:
[tex]W = mg'[/tex]
where,
W = Force of gravity on space craft = ?
m = mass of space craft = 2000 kg
Therefore,
[tex]W = (2000 kg)(6.72\ m/s^2)[/tex]
W = 13.44 KN
I’ll give brainliest if it’s correct ;-;z
Someone help!!!
3 different earthquakes that were measured with broadband seismographs, 3 earthquakes long period, and 3 earthquakes measured with short period instruments.
Explanation:
what is the question? could you pls provide it
Force is measured in what unit? What is this unit equal to in other units?
Force is measured in Newton ( kg.m. /s²)
if your guitar pick is producing a harsh sound, you should use your fingernail instead relax your wrist, arm, and shoulder make sure your wrist and arm are at a ninety degree angle press the pick more firmly onto the strings
If your guitar pick is producing a harsh sound, you should make sure your wrist and arm are at a ninety degree angle.
Making of harsh sounds due to the pick from a guitar is seen often during guitar practices from beginners. There are numerous reasons for this. One of the reasons that can be considered as the most often mistake is the way a guitarist holds the pick. In order to avoid the sound produced by the pick, make sure that your wrist and arm are in a ninety degree angle.
This will provide a comfortable position to use the picks. You should not hold the pick firmly either. Try to hold the pick gently. Remember lesser the pick gets in contact with the string, lesser the sound produced. Also try to use a proper pick for guitar.
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help me please!!!!!!!!!!!!!!!
Answer:
10
Explanation:
would
Supposing d(t) is known to have value D,
what procedure will find the time t at which
this happens?
2. Set d(t) equal to v^2/a
6. Set a to zero
Phase conductors 6 awg and smaller ____ permitted to be re-identified as equipment ground conductors using green paint, tape, or other effective means
Phase conductors 6 awg and smaller are not permitted to be re-identified as equipment ground conductors using green paint, tape, or other effective means.
Depending on the type of system offered, the grounded conductor of a service is typically a neutral conductor, though it can also be a phase conductor. A grounded phase conductor and no grounded neutral conductor are present in a corner-grounded delta system, for instance. The unbalanced load is returned to the source by the neutral. The conductor that has been purposefully grounded is the grounded conductor. The neutral is the conductor that is purposefully grounded in the majority of wiring systems used in industrial facilities, businesses, and homes. For safety reasons, the grounding conductor is employed. The grounding wire does not carry current under typical circumstances. The grounding wire offers a low resistance channel for the current in a fault situation.
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A 0.5-kg ball accelerated at 50 m/s 2 . What force was applied?
Answer:
25N
Explanation:
soln:
given,
mass(m)=0.5kg
acceleration(a)=50m/s^2
force(F)=?
we know,
F=m*a
or,F=0.5*50
•°•F=25N
Look at the picture
Answer:
Your answer is d
Explanation:
hello help me pls! i need serious help
Answer:c
Explanation:
A certain superconducting magnet in the form of a solenoid of length 0.500m can generate a magnetic field of 9.00T in its core when its coils carry a current of 75.0A . Find the number of turns in the solenoid.
The number of turns in the solenoid is [tex]4.78\times 10^4 \ turns[/tex].
Solenoids are essentially coils of cord. These generate a magnetic subject which strives a pressure over a steel element.A solenoid is a fundamental time period for a coil of cord that we use as an electromagnet.We additionally consult with the tool which could convert electric power into mechanical power as a solenoid.Actually it generates a magnetic subject for developing linear movement from the electrical cutting-edge. With using a magnetic subject.Magnetic field at the centre of a solenoid of length 'L' having 'N' number of turns with a current 'I' is given by [tex]B=\frac{u_0NI}{L}[/tex] ...(1) where [tex]u_0[/tex] is the permittivity of free space whose value is [tex]4\pi \times10^{-7}Tm/A[/tex] .It is given that superconducting magnet in the form of a solenoid of length 0.500m can generate a magnetic field of 9.00T in its core when its coils carry a current of 75.0A .
Putting above values in equation (1) , we get
[tex]9.00=\frac{4\pi \times10^{-7}\times N \times75.0}{0.500} \\\\N=\frac{9.00\times0.500}{4\pi \times10^{-7}\times75.0} \\\\N=4.78\times10^4 \ turns[/tex]
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Question 10 While looking in the yard, Jill noticed that the areas directly under the trees had less grass than the other areas in the yard. Which of the following is the BEST explanation for this observation? A The trees are blocking the sunlight. B The grass is receiving too much fertilizer. 2 C The areas under the trees are too cool for grass to grow. D The areas under the trees are receiving less water.
Answer:
The trees are blocking the sunlight.
Explanation:
B is obviously not the answer because fertilizer wouldn't have that effect. For C, shade doesn't change the temperature that much that it would inhibit plant growth. For D, even if the tree blocks the water, the surrounding ground will still be wet and therefore the grass should still have a sufficient amount of water to grow. Sunlight can't be gotten from somewhere else, unlike water, so this would inhibit plant growth.
A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is the direction of its final velocity? Answer in direction (deg)
Answer:
(3504 m/s, 66°)
The final velocity of the missile is approximately 3504 m/s at an angle of 66 degrees.
Explanation:
We are going to use the horizontal and vertical components of this object since it is in projectile motion.
We are given the initial velocity of the missile; 1350 m/s at an angle of 25 degrees. We are also given the displacement in the x-direction: 23,500 meters at an angle of 55 degrees. The total time of this projectile in motion is 10.20 seconds.
Using this information, we can break up the motion of the projectile into its horizontal and vertical components and solve for the unknown variables [tex]a[/tex] and [tex]v_f[/tex].
Horizontal direction (x):List out the known variables:
[tex]v_i=1350[/tex] [tex]v_f=?[/tex] [tex]\triangle x=23500[/tex] [tex]a=?[/tex] [tex]t=10.20[/tex]Since we have initial velocity, displacement, and time, we can use one of the kinematic equations for constant acceleration that contains these variables, including acceleration, and solve for a:
[tex]\triangle x=v_it + \frac{1}{2}at^2[/tex]We are solving for acceleration in the x-direction, so this equation should be in terms of the x-direction:
[tex]\triangle x_x=(v_i)_x t +\frac{1}{2}a_x t^2[/tex]Let's break up the displacement into its horizontal (x) component:
[tex]23500cos(55)[/tex]Let's break up the initial velocity into its horizontal component:
[tex]1350cos(25)[/tex]Plug these values into the equation and 10.20 seconds for t.
[tex]23500cos(55)=1350cos(25) \cdot t + \frac{1}{2}a_x (10.20)^2[/tex]Solve for [tex]a_x[/tex].
[tex]23500cos(55)=12479.85823+52.02a_x[/tex] [tex]999.1880243=52.02a_x[/tex] [tex]a_x=19.20776671[/tex]The acceleration in the x-direction is about 19.21 m/s².
Now, we can use this acceleration and solve for the final velocity in the x-direction using this constant acceleration kinematic equation:
[tex]v_f=v_i+at[/tex]Use this equation in terms of the x-direction:
[tex](v_f)_x=(v_i)_x+a_xt[/tex]Plug the known values into the equation and solve for [tex](v_f)_x[/tex].
[tex](v_f)_x=1350cos(55)+19.20776671 \cdot 10.20[/tex] [tex](v_f)_x=1419.434733[/tex]The final velocity in the x-direction is about 1419.43 m/s.
Vertical direction (y):This process will be the same as solving for acceleration and final velocity in the x-direction, except this time we will be using the vertical (y) components.
[tex]\triangle x_y=(v_i)_y t +\frac{1}{2}a_y t^2[/tex]Vertical component of initial velocity:
[tex]1350sin(25)[/tex]Vertical component of displacement:
[tex]23500sin(55)[/tex]Plug known values into the equation and solve for [tex]a_y[/tex].
[tex]23500sin(55)=1350sin(25) \cdot (10.20) + \frac{1}{2}a_y (10.20)^2[/tex] [tex]23500sin(55)=5819.453464 +52.02a_y[/tex] [tex]13430.61958=52.02a_y[/tex] [tex]a_y=258.181845[/tex]The acceleration in the y-direction is about 258.18 m/s².
Now let's use the same equation we used previously to solve for the final velocity in the y-direction.
[tex]v_f=v_i+at[/tex]Use this equation in terms of the y-direction:
[tex](v_f)_y=(v_i)_y+a_y t[/tex]Plug known values into the equation and solve for [tex](v_f)_y[/tex].
[tex](v_f)_y=1350sin(25)+258.181845 \cdot 10.20[/tex] [tex](v_f)_y=3203.989472[/tex]The final velocity in the y-direction is about 3203.99 m/s.
Final velocity and direction:The magnitude of the final velocity is the square root of the final velocity in the horizontal and vertical directions squared and added together.
[tex]|v|=\sqrt{(v_f)^2_x + (v_f)^2_y}[/tex]Plug the final velocity in the x and y-directions into the equation.
[tex]|v|=\sqrt{(1419.434733)^2+(3203.989472)^2}[/tex] [tex]|v|=\sqrt{2014794.961+10265548.54}[/tex] [tex]|v|=\sqrt{12280343.5}[/tex] [tex]|v|=3504.332105[/tex]The magnitude of the final velocity is about 3504 m/s. We can solve for the direction of the final velocity by using arccos to find the angle that is formed with the x-axis.
[tex]\displaystyle\theta = cos^-^1 \big{(} \frac{v_x}{|v|}\big{)} }[/tex]Plug in the x-component of the initial velocity and the magnitude of the final velocity of the projectile into the equation.
[tex]\displaystyle \theta =cos^-^1 \big{(}\frac{1419.434733}{3504.332105}\big{)}}[/tex] [tex]\displaystyle \theta = cos^-^1 \big{(} .4050514308 \big{)}}[/tex] [tex]\displaystyle \theta = 66.1056499[/tex]The direction of the final velocity is about 66 degrees.
A wire carrying a current I is bent into the shape of an equilateral triangle of side L. (a) Find the magnitude of the magnetic field at the center of the triangle.
The magnetic field at the center of the triangle is [tex]\frac{4.50 \mu_{0} I}{\pi L}[/tex].
An equilateral triangle with side L is formed by the bend of a wire carrying a current I.
The magnetic field created by a straight wire is given as:
[tex]B = \frac{ \mu_{0} I}{2 \pi a}[/tex]
The cosines of the complementary angles are identical to the sines of the angles that appear in that equation.
The magnetic field from a distance "a" from the wire,
We obtain:
[tex]tan{~}30^{\circ} = \frac{a}{L/2}\\a = 0.2887L[/tex]
Therefore, the magnetic field at P which is the center of the triangle will be:
[tex]B = \frac{\mu_{0} I}{4 \pi a}(cos{~}\theta_{1} - cos{~}\theta_{2})[/tex]
[tex]B = \frac{\mu_{0} I}{4 \pi \times 0.2887L}(cos{~}30^{\circ} - cos{~}150^{\circ})[/tex]
[tex]B = \frac{\mu_{0} I \times 1.732}{\pi \times L}[/tex]
[tex]B = \frac{1.50 \mu_{0} I}{ \pi L}[/tex]
Every side produces the same amount of field in the same direction, which in the illustration is perpendicular to the paper.
Therefore, the total field is:
[tex]\frac{1.50 \mu_{0} I}{ \pi L} = \frac{4.50 \mu_{0} I}{ \pi L}[/tex]
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7. A student is rotating an object on a rope that is 4.5m long. If we increase the length
of the rope so that the student rotates in uniform circular motion with the mass
rotating 9.0 m away from the center of the rotation (2x farther), what will be the
effect on the tangential velocity for the athlete if we keep the time for 1 revolution
the same?
It will be 16 times greater
0
O
It will be 2 times greater
0
о
It will be 4 times greater
0
It will be the same
CLEAR ALL
7 08 09 10 0 11
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The tangential velocity is proportional to the length of the rope hence when the length of the rope is doubled, the tangential velocity becomes two times greater.
We know that the tangential velocity of an object moving along a circular path depends on its radius and it's angular velocity.
In this case, we have been told that the motion of the object is uniform meaning that the angular velocity is held constant and the length of the rope which is the radius was doubled.
Since the tangential velocity is directly proportional to the length of the rope, the tangential velocity will be two times greater when the length of the rope is doubled.
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Which modes of heat transfer require a medium? Which does not? Explain why a medium is/is not
needed for each response.
Answer:
Conduction and convection requires Modes of transfer of heat ! Radiation doesn't require any medium !!
In terms of heat transfer, radiation is the emission of thermal energy in the form of infrared waves. ... Since heat is carried by electromagnetic waves, it does not need a physical medium to transfer it. Instead it radiates through space - this is how the Earth is heated by the Sun despite space being a vacuum
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a ping-pong ball is held submerged in a bucket of water by a string attached to the bucket's bottom. part a salt is now added to the water in the bucket, increasing the density of the liquid. what happens to the tension in the string ?
The tension force in the string will increase because the difference between the densities became increased.
What is the free body diagram?Free-body diagrams are utilized to display the relative direction and strength of all forces that are being applied to an item in a certain scenario. A unique illustration of the geometric diagrams that were covered in a previous lesson is the free-body diagram. We will make use of these graphics throughout the entire study of physics.
A thread fastened to the bucket's base is holding a ping-pong ball immersed in water. The water in the bucket is now supplemented with salt, making the liquid denser.
Before adding the salt let T be the tension in the string.
Due to the growing disparity in densities, the force applied to the rope will increase.
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a sound has a frequency of 230 hz. what is the velocity of this sound if it has a wavelength of 46 cm?
Answer:
V=f×λ
Where,
V is the velocity of the wave measure using m/s.
f is the frequency of the wave measured using Hz.
λ is the wavelength of the wave measured using m.
transform 46 cm in m = 0.46m and there u have it
Explanation:
can light cause plastic to melt? If yes does it matter what kind of light?
Answer:
plastics are susceptible to degradation due to ultraviolet (UV) light from the sun. With enough exposure, UV light can cause a chemical reaction in the plastic, which results in scission, or severing, of those big polymer molecules.
Explanation:
Answer:
the short answer is yes and sun or strong uv lights
Explanation:
LEDs cannot melt plastic fixtures as they just do not get that hot, not even at the base These new bulbs are based on field-induced polymer electroluminescent (FIPEL) technology, with a twist. Without heat as a continual stressor, the polymer should remain stable for years. Any kind of light bulbs, from fluorescent to incandescent to halogen, can cause fires if they are not used correctly. As more heat is put into the plastic, more bonds are broken and the material becomes more like a liquid and less like a solid. Semi-crystalline polymers have a melting temperature which is the point at which the secondary bonds in the crystalline regions break. At this point the mateiral is completely liquid. There is melt and there is soften. Certainly they soften and become weak in sunlight both by the heat and by the UV degredation of the plastic. Unfortunately, the plastics used for garbage bags have a low melting point but by design, they should not melt in strong sunlight. the underlined sentences are the most important
S A nonuniform electric field is given by the expression→E = ay i^ + bz j^ + cxk^where a, b , and c are constants. Determine the electric flux through a rectangular surface in the x y plane, extending from x=0 to x=w and from y=0 to y=h .
The electric flux is 2chω² where the k^ term was eliminated since k. k^ =0
The electric force per unit charge is referred to as the electric field. It is assumed that the field's direction corresponds to the force it would apply to a positive test charge. From a positive point charge, the electric field radiates outward, and from a negative point charge, it radiates in.
We are given an electric field in the general form
E→ =ay i^+bz j^+cx k^
In the xy plane z=0 so that the electric field reduces to
E→ =ay i^+cx k^
To obtain the flux, we integrate
ϕ E=∫ E→⋅d A→ =∫ (ay i^ +cx k^) ⋅k^
dAϕE=ch∫
x=0w
xdx=2chω²
where the k^ term was eliminated since k. k^ =0
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courtney dauwalter leaves her home at 7 a.m. and takes his usual path to the top of mount elbert, arriving at 7 p.m. the following morning, she starts at 7 a.m. at the top of mount elbert and takes the same path back, arriving at her home at 7 p.m. determine if there exists a point on the path the courtney will cross at exactly the same time of day on both days. justify your answer.
There exists a point on the path the Courtney will cross at exactly the same time of day on both days.
Given that Courtney first takes his usual path and travel between 7.a.m. and 7.p.m.
Next morning, she travel backs between the time 7 a.m. and 7 p.m.
The time taken by Courtney on each day = 12 hours
12/2 = 6 hours.
6 hours from 7 a.m. is 1 pm. So this 6 hours will divide the total journey into two equal parts.
At 1 p.m. on first day, she will be in the middle of her way from home to Mount Elbert.
At 1.p.m. of the second day also, she will be in the middle of her way from Mount Elbert to her home.
So she cross the exact middle point of the path at exactly 1.pm on both days.
Intermediate Value theorem states this result. Since the distance travelled is a continuous function, there is a particular value of time corresponding to the exact same point crossed by Courtney on both days.
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Tessa uses a toy slingshot to launch a tennis ball across the park for her dog to fetch. For her first launch, she
uses 100 N of force. Her second launch uses 200 N of force, and her third launch uses 300 N. Which launch had
the greatest acceleration of the tennis ball?
Answer:
See the explanation below.
Explanation:
To solve this problem we must apply Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration and this force can be calculated by means of the following equation.
F = m*a
where:
F = force [N] (units of Newtons)
m = mass [kg]
a = acceleration [m/s²]
The mass of the tennis ball will always be the same therefore it will never change.
Now clearing a:
[tex]a=\frac{F}{m}[/tex]
If the mass of the ball remains the same:
[tex]a = \frac{100}{m} ; a = \frac{200}{m};a =\frac{300}{m}[/tex]
We see that for a force of 300 [N], the acceleration exerted on the ball must be greater. Therefore with the force of 300 [N] the greatest acceleration is achieved.
Hi im really confused:
How much energy would be needed to raise the temperature of the air in a room by 5.0°C if the room measures 4.0m x 4.0m x 3.0m? (Density of air = 1.0kg/m³)
Assume that the room has no furniture and that the walls gain no thermal energy.
The energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ
U = [tex]c_{p}[/tex] m ΔT
U = Energy
[tex]c_{p}[/tex] = Specific heat
m = Mass
ΔT = Change in temperature
ρ = Density
V = Volume
ρ = 1000 g / m³ (Dry air )
= 1 J / g K
ΔT = 5 °C
V = 4 * 4 * 3
V = 48 m³
m = ρ V
m = 1000 * 48
m = 48000 g
U = 1 * 48000 * 7
U = 336000 J
U = 336 kJ
Therefore, the energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ
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What is the relationship between magnetism and the movement of electrical charges in generators ( the device).
Answer:
Electricity and magnetism are two related phenomena produced by the electromagnetic force. Together, they form electromagnetism. A moving electric charge generates a magnetic field. A magnetic field induces electric charge movement, producing an electric current.
Explanation:
What does this tell you about the direction and magnitude of the acceleration acting on the cannonball throughout its duration of flight?
Answer:
Direction remains the same but velocity changes.
Explanation:
This tell us about the direction and magnitude of the acceleration acting on the cannonball throughout its duration of flight that its direction remains the same but its magnitude of the acceleration is continuously changing. The cannonball moves in the direction in which the cannon was fired while the velocity is highest after the fire but decreases when goes higher and when it comes back to the ground so its velocity increases against so we can say that both positive and negative acceleration occurs. Positive acceleration means increase in the magnitude of velocity whereas negative acceleration means decrease in velocity.
Aluminium is a non magnetic metal. However, it can become magnetised under special conditions. Explain how. (4 marks)
Answer:
Exposing it to a very strong magnetic field
Explanation:
Aluminium is a non magnetic metal thus it is not magnetic under normal conditions. This is because it interacts with other magnets. The only condition that magnetism can be induced in aluminium is exposing it to a very strong magnetic field.
A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpendicular to the proton's velocity. It leaves the field-filled region with velocity -20.0j Mm/s. Determine(a) the direction of the magnetic field.
The the direction of magnetic field is perpendicular to the velocity of proton and is directed towards the + z axis.
We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.
We have to determine the direction of magnetic field.
What is the magnitude of force on the charged particle moving in a uniform magnetic field?The magnitude of force on the charged particle moving in a uniform magnetic field is given by -
F = qvB sinθ
According to the question, we have -
Entering Velocity (v) = 20 i m/s.
Magnetic field intensity (B) = 0.3 T.
Leaving velocity (u) = - 20 j m/s.
Now -
F = q |v| |B| sinθ
F = 1.6 x [tex]10^{-19}[/tex] x 20 x 0.3 x sin(90)
F = 9.6 x [tex]10^{-19}[/tex] Newton.
Now, using the Fleming's left hand rule, the the direction of magnetic field is perpendicular to the velocity of proton and is directed towards the + z axis. The force is towards the '- y' axis. Therefore -
B = + 0.3 k
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A ship has a constant velocity of 8.33 m/sec. How far does it travel in done day?
Distanced travelled by a ship with a constant velocity in a day is 719712m
VELOCITY is defined as rate of change of displacement in a given interval of time.
it is a vector quantity.
its unit is m/s
to calculate the distance of a ship travelled
distance = speed x time
d = s x t ----1.
velocity of ship = 8.33m/s
time taken = 1 day =86400 sec
now using the above values in equation 1 we get
d =8.33m/s x 86400 sec
d = 719712m
thus a ship travels 719712m in a day.
Distanced travelled by a ship with a constant velocity in a day is 719712m
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