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amm1812
1 year ago
13

Two motorcycles travel along a straight road heading due north. At t = 0 motorcycle 1 is at x = 50 m and moves with a constant s

peed of 6.50 m/s; motorcycle 2 starts from rest at x = 0 and moves with constant acceleration. Motorcycle 2 passes motorcycle 1 at the time t = 10.08.
Part B What is the position of the two motorcycles when motorcycle 2 passes motorcycle 1? Express your answer to three significant figures and include the appropriate units.

Part C What is the acceleration of motorcycle 2? Express your answer to three significant figures and include the appropriate units.​
Physics
1 answer:
Oduvanchick [21]1 year ago
3 0

Answer:

Vf = 23 m/s

Explanation:

First we need to find the distance covered by the motorcycle 2 when it passes motorcycle 1. Using the uniform speed equation for motorcycle 1:

s₁ = v₁t₁

where,

s₁ = distance covered by motorcycle 1 = ?

v₁ = speed of motorcycle 1 = 6.5 m/s

t₁ = time = 10 s

Therefore,

s₁ = (6.5 m/s)(10 s)

s₁ = 65 m

Now, for the distance covered by motorcycle 2 at the meeting point. Since, the motorcycle started 50 m ahead of motorcycle 2. Therefore,

s₂ = s₁ + 50 m

s₂ = 65 m + 50 m

s₂ = 115 m

Now, using second equation of motion for motorcycle 2:

s₂ = Vi t + (1/2)at²

where,

Vi = initial velocity of motorcycle 2 = 0 m/s

Therefore,

115 m = (0 m/s)(10 s) + (1/2)(a)(10 s)²

a = 230 m/100 s²

a = 2.3 m/s²

Now, using 1st equation of motion:

Vf = Vi + at

Vf = 0 m/s + (2.3 m/s²)(10 s)

Vf = 23 m/s

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A planar loop consisting of four turns of wire, each of which encloses 200 cm2, is oriented perpendicularly to a magnetic field
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0.6A

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7 0
2 years ago
A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init
Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

h=\frac{gt^2}{2}

For second stone

h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}---2

Equating 1 &2 we get

\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}

\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0

26.46t-35.721-52.92t+142.884=0

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4 0
2 years ago
A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when
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Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

F_t - mg = ma

here since speed is constant or it is at rest

so we will have

a = 0

F_t = mg

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

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so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

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F_t = mg - ma

so tension force is less than the weight of crate if it is accelerating downwards

4 0
2 years ago
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