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Svetlanka [38]
2 years ago
12

A student practicing for a cross country meet runs 250 m in 30 s. What is the average speed

Physics
2 answers:
den301095 [7]2 years ago
6 0

Total distance traveled = 250m

Total time taken = 30 seconds

Now,

Average speed = total distance

__________

time taken

= 250 m ÷30 second

= 8.3m/s

Kaylis [27]2 years ago
5 0

Answer:8.3m/sec 30 sec,

Explanation:

A student practicing for a track meet, ran 250 m in 30 sec. a. What was her average speed? 250 m = 8.3 m/sec 30 sec.

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A plane is landing at an airport. The plane has a massive amount of kinetic energy due to it's motion. When the plane lands, it
marshall27 [118]

Answer:

A. The brakes used a coil system to convert the kinetic energy into potential energy stored in the brakes

Explanation:

Based on the law of conservation of energy, the brakes used a coil system to convert the kinetic energy into potential energy stored in the brakes.

The law of conservation of energy states that energy is neither created nor destroyed in a system but it is transformed from one form to another.

As the airplane slows down, the kinetic energy which is presented in the motion of the plane is gradually converted to potential energy.

The potential energy is the energy due to the position of a body.

8 0
2 years ago
A person looking out the window of a stationary train notices that raindrops are falling vertically down at a speed of 3.84 m/s
JulijaS [17]

Answer:

the train is moving at the speed of v = 1.79 m/s

Explanation:

given,

rain drop is falling vertically down with the speed of  = 3.84 m/s

angle of the rain drop = 25°                      

tan θ = \dfrac{v}{u}                      

tan 25° = \dfrac{v}{3.84}                      

v =3.84 × tan 25°                      

v = 1.79 m/s                  

hence, the train is moving at the speed of v = 1.79 m/s

3 0
3 years ago
A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box
forsale [732]
We need to see what forces act on the box:

In the x direction:

Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

In the y direction:

N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force. 

From N-Gcosα=0 we get: 

N=Gcosα, we will need this for the force of friction.

Now to solve for Fh:

Fh=ma + Ff + Gsinα,

Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²

Fh=ma + μmgcosα+mgsinα

Now we plug in the numbers and get:

Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
8 0
3 years ago
The electric field everywhere on the surface of a thin, spherical shell of radius 0.770 m is of magnitude 860 N/C and points rad
11111nata11111 [884]

Answer:

(a) Q = 7.28\times 10^{14}

(b) The charge inside the shell is placed at the center of the sphere and negatively charged.

Explanation:

Gauss’ Law can be used to determine the system.

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The negative charge at the center attracts the same amount of positive charge at the surface of the shell.

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3 years ago
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