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3 years ago

A 50.0 kg driver is riding at 35.0 m/s in her red sports car when she must suddenly slam on the brakes to avoid

Physics

1 answer:

egoroff_w [7]3 years ago

8 0

Answer:

3500N

Explanation:

Given parameters:

Mass of driver = 50kg

Speed = 35m/s

Time = 0.5s

Unknown:

Average force the seat belt exerts on her = ?

Solution:

The average force the seat belt exerts on her can be deduced from Newton's second law of motion.

F = mass x acceleration

So;

F = mass x $\frac{change in velocity }{time}$

F = 50 x $\frac{35}{0.5}$ = 3500N

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Discuss whether any work is being done by each of the following agents and, if so, whether the work is positive or negative: (a)

Aleks04 [339]

Answer:

a) As the chicken is still, the displacement is zero, which implies that the work is zero.

b) as the person is still there is no displacement therefore the work is zero

c) Lagraua applies a vertical force and the displacement is vertical, therefore the Work is positive

d) the force of gravity is directed downwards and the displacement is upwards, therefore the angle between it is 180º and the 180º fly is -1. Consequently the lock is negative

e) when the person meticulously feels the upward force and the displacement is downward, therefore the work is negative

Explanation:

Work is defined by the expression

W = F. r

bold letters indicate vectors, we can write this expression as a module

W= F r cos θ

where is at the angle between force and displacement.

Let's apply this expression to the different cases

a) As the chicken is still, the displacement is zero, which implies that the work is zero.

b) as the person is still there is no displacement therefore the work is zero

c) Lagraua applies a vertical force and the displacement is vertical, therefore the Work is positive

d) the force of gravity is directed downwards and the displacement is upwards, therefore the angle between it is 180º and the 180º fly is -1. Consequently the lock is negative

e) when the person meticulously feels the upward force and the displacement is downward, therefore the work is negative

7 0

3 years ago

If a 80kg diver jumps off of a 5 m high dive into a regulation diving pool, how much should the temperature of the pool go up?

Agata [3.3K]

Answer:

The answer cannot be determined.

Explanation:

The energy of the diver when he hits the pool will be equal to its potential energy $mgh$ , and for the temperature of the pool to rise up, this energy has to be converted into the heat energy of the pool.

The change in temperature ${\Delta}T$ then will be

${\Delta}T=\frac{{\Delta}Q}{mc} .$

Where m is the mass of water in the pool, c is the specific heat capacity of water, and ${\Delta}Q$ is the added heat which in this case is the energy of the diver.