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castortr0y [4]
2 years ago
7

A4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects encounter the same const

ant braking force, and are brought to rest. Which object travels the greater distance before stopping? a) the 4.0 kg object b) the 1.0 kg object c) Both objects travel the same distance d) It cannot be determined from the information given.
Physics
1 answer:
Evgesh-ka [11]2 years ago
6 0

Answer:

Both objects travel the same distance.

(c) is correct option

Explanation:

Given that,

Mass of first object = 4.0 kg

Speed of first object = 2.0 m/s

Mass of second object = 1.0 kg

Speed of second object = 4.0 m/s

We need to calculate the stopping distance

For first particle

Using equation of motion

v^2=u^2+2as

Where, v = final velocity

u = initial velocity

s = distance

Put the value in the equation

0= u^2-2as_{1}

s_{1}=\dfrac{u^2}{2a}....(I)

Using newton law

a=\dfrac{F}{m}

Now, put the value of a in equation (I)

s_{1}=\dfrac{8}{F}

Now, For second object

Using equation of motion

v^2=u^2+2as

Put the value in the equation

0= u^2-2as_{2}

s_{2}=\dfrac{u^2}{2a}....(I)

Using newton law

F = ma

a=\dfrac{F}{m}

Now, put the value of a in equation (I)

s_{2}=\dfrac{8}{F}

Hence, Both objects travel the same distance.

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Part b)

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Explanation:

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Part b)

Horizontal speed of the canister is given as

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x = 24.5 (1.65)

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Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

v_x = 24.5 m/s

final velocity in Y direction

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now magnitude of velocity is given as

v = \sqrt{v_x^2 + v_y^2}

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v = 27.3 m/s

direction of velocity is given as

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[tex]\theta = 26.35 degree

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On a distant planet, a rock falls in 48.4 s from the top of a 1.10e+02 m cliff to the planet surface below. What is the accelera
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