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arlik [135]
3 years ago
14

A roller coaster goes from 10 m/s to 30 m/s in 5 seconds. What is it's acceleration? Vf-Vi/T

Physics
1 answer:
Olin [163]3 years ago
6 0
Acceleration = (final velocity-initial velocity)/time taken
=(30-10)/5
= 4m/s2
Hope this helps
Please mark me brainliest
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Compare the de Broglie wavelength of a golf ball moving at 70.0 miles per hour (31.3 m/s) to that of an alpha particle moving at
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3.bullet is in xray range of wavelength

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Since when was the light we see now emanating from the quasar? Note that the distance between the Earth and the quasar is 598 Mp
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If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular
erastova [34]

Answer;

D. The car would begin to move in the direction it was headed in a straight line.

Explanation;

-Centripetal force is any net force causing uniform circular motion. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration.

-The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

-Therefore,If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced then the car would begin to move in the direction it was headed in a straight line.

6 0
3 years ago
Read 2 more answers
The Coulomb force between two charges q1 and q2 at separation r in air is F. If half of the separation is filled with medium of
vodomira [7]

Answer:

The new Coulomb force is q₁q₂/9πε₀r²

Explanation

The coulomb force between the two charges q₁ and q₂ at a distance r in air is given by F = q₁q₂/4πε₀r².

Now, let us assume the material of dielectric constant κ = 9 is placed between them on the side of the q₁ charge. The value of its effective charge is now q₃ = q₁/κ at a distance of d = r/2 from the q₂ charge.

Since we have air between q₂ and q₃, the coulomb force between them is

F' = q₂q₃/4πε₀d²

= q₂(q₁/κ)/4πε₀(r/2)²

=  4q₂q₁/κ4πε₀r²

= 4/κ(q₂q₁/4πε₀r²)

= 4/9 × (q₂q₁/4πε₀r²)

= q₁q₂/9πε₀r²

So, the new Coulomb force is q₁q₂/9πε₀r²

3 0
3 years ago
A car is moving at 25.5 m/s when it accelerates at 1.94 m/s^2 for 2.3 s. What is the car's final speed? (Keep in mind direction
Stolb23 [73]

Answer:

29.96m/s

Explanation:

Given parameters:

Initial speed  = 25.5m/s

Acceleration  = 1.94m/s²

Time  = 2.3s

Unknown:

Final speed of the car  = ?

Solution:

To solve this problem, we are going to apply the right motion equation:

    v = u  + at

v is the final speed

u is the initial speed

a is the acceleration

t is the time taken

 Now insert the parameters and solve;

      v  = 25.5 + (1.94 x 2.3)  = 29.96m/s

3 0
3 years ago
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