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Neko [114]
3 years ago
15

A block is launched up a frictionless 40° slope with an initial speed v and reaches a maximum vertical height h. The same block

is launched up a frictionless 20° slope with the same initial speed v. On this slope, the block reaches a maximum vertical height of _____.
1. h
2. h/2
3. 2h
4. A height greater than h but less than 2h.
5. A height greater than hl2 but less than h.

Physics
1 answer:
Strike441 [17]3 years ago
3 0

Answer: 1. h

Explanation:

The block would reach exactly the same height from the ground. It would travel a greater distance away from the source, but the height away from the earth would remain the same as you are giving it the same energy each time. Therefore, it will reach the same gravitation potential energy.

Another approach to look at it this is seeing it when the Block moves up the slope, its kinetic energy decreases and the potential energy increases. In both cases, the kinetic energy decreases by same amount, therefore the block rises to same height H.

Try to use the formula;

1/2MV2 = mgh

Where V = √(2gh)

I hope this helps

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Explanation:
Kinematics equation for first Object:

but:
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Kinematics equation for the second Object:
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but:

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s


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If kinetic energy of a body is decreased by 10percent then decrease in linear momentum of that body​
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Answer:

E = 1/2 M V^2 = 1/2 P V     since P = M V

E2 / E1 = P2 V2 / (P1 V1)

P2 / P1 = E2 V1 / (E1 V2) = V2^2 V1 / (V1^2 V2) = V2 / V1

E2 / E1 = V2^2 / V1^2

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V2 / V1 = (.9)1/2 = .95

The linear momentum would have to decrease by 5%

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You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As so
grigory [225]

Answer:

Explanation:

Time duration during which acceleration exists in  bicycle =

23 / 12 = 1.91 s

Time duration during which acceleration exists in car

= 47 / 8 = 5.875 s

Distance covered by bicycle during acceleration ( t = 1.91 s )

= 1/2 x 12 x (1.91)²

= 21.88 mi

Distance covered by car during this time ( t = 1.91 s )

= 1/2 x 8 x (1.91)²

7.64 mi ,

velocity of car after 1.91 s

= 8 x 1.91 = 15.28 mi/h

Let after time 1.91 , time taken by them to meet each other be t

Total distance covered by cycle = total distance covered by car

21.88 + 23 t = 7.64 + 15.28t + 4 t²

21.88 = 7.64 - 7.72t +4 t²

4 t² -7.72 t -14.24 = 0

t = 2.83 s

Total time taken

= 2.83 + 1.91

= 4.74 s

So after 4.74 s they will meet each other.

b ) Maximum distance occurs when velocity of both of them becomes equal .

Velocity after 1.91 s of bicycle

12 x 1.91 = 23 mi/h

Velocity after 1.91 s of car

8 x 1.91 = 15.28 mi/h . Let after time t , the velocity of car becomes 23

15.28 + 8t = 23

t = .965 s

So after time .965 s , car has velocity equal to that of bicycle.

The bicycle will travel a distance of

= 21.88 + .965 x 23 = 44.075 mi

car will travel a distance of

7.64 + 15.28 x .965 + .5 x 8 x .965²

= 7.64 + 14.75 + 3.72

= 26.11 mi

Distance between car and bicycle

= 44.075 - 26.11 = 17.965 mi

= 17.965 x 1760

= 31618.4 ft.

5 0
3 years ago
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