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1 year ago

Lucy and her bike together have a mass of 120kg. She slows down from 4.5m/s to 3.5m/s. How much kinetic energy does she lose?

1 answer:

7 0

The kinetic energy of a moving object is given by
$K= \frac{1}{2}mv^2$
where m is the object's mass and v its velocity.

In our problem, the initial kinetic energy is:
$K_i = \frac{1}{2} m v_i^2 = \frac{1}{2}(120 kg) (4.5 m/s)^2=1215 J$

while the final kinetic energy is:
$K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(120 kg)(3.5 m/s)^2= 735 J$

So, the kinetic energy lost by Lucy and her bike is
$\Delta K = K_i - K_f = 1215 J - 735 J = 480 J$

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Natasha_Volkova [10]

Answer:

Explanation:

Given

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so time taken is $t_1=\frac{100}{1.8}=55.55\ s$

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Where d is the distance

V0 is the initial velocity

A is the acceleration

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From the graph a = 4/3 m/s2

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1 year 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

choli [55]

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

$F_t - mg = ma$

$F_t = mg + ma$

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

$F_t - mg = -ma$

$F_t = mg - ma$

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

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Read 2 more answers

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Answer:

Part A. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull.

Part B. The magnitude of normal force acting on the suitcase is equal to the sum of the weight of the suitcase and the man.

Explanation:

Part A. This is because when the man pulls on the suit upwards, he exerts a force in the upward direction. This takes part of the force of weight of the suitcase and decreases the force the suitcase is exerting on the ground. Thus, the normal force (force exerted by suitcase on the ground) also decreases by the same force as the pull.

Part B. The statements for this part were not given in the question, but the answer reflects what is going to happen in that scenario. Since the man sits on the suitcase, the total weight acting on the ground through the suitcase is that of the suitcase plus the man. Since this force (acting on the ground) is normal force, the statement given in the answer is correct.

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