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3 years ago

Playing with a stress ball you squeeze it as

Physics

1 answer:

shutvik [7]3 years ago

8 0

Answer:

it dies

Explanation:

MWUAHHAHAHHA

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Apilot of mass 70 kg rides a fighter jet The fighter jet moves in a vertical circle of radius 100 m at a constant

Cerrena [4.2K]

Answer:

the force exerted by the seat on the pilot is 10766.7 N

Explanation:

The computation of the force exerted by the seat on the pilot is as follows:

$F = Mg + \frac{MV^2}{R}\\\\= 70 \times 9.81 + \frac{70 \times 120^2}{100}\\\\= 10766.7 N$

Hence, the force exerted by the seat on the pilot is 10766.7 N

4 0

3 years ago

You are operating a pwc in an area where swimmers are in the water. When must you slow your pwc to "slow, no wake speed"?.

Lostsunrise [7]

You should slow your pwc to "slow, no wake speed" when within 100 feet of anchored vessels or non-motorized craft.

This is the process of operating a personal watercraft at the slowest possible speed.

This helps to maintain steerage which prevents different forms of accident or risks when in motion in the water.

Read more about Slow-no-wake here brainly.com/question/10410716

#SPJ1

4 0

2 years ago

A firefighter of mass 81 kg slides down a vertical pole with an acceleration of 3 m/s 2 . The acceleration of gravity is 10 m/s

natima [27]

Answer:

The force of friction that acts on him is

$F_k=567N$

Explanation:

The firefighter with an acceleration of 3m/s^2 take the gravity acceleration as 10m/s^2 isn't necessary to know the coefficient of friction just to know the force of friction:

$F=m*a$

$F=F_w-F_k$

$m*a=F_w-F_k$

$F_w=81kg*10m/s^2=810N$

Sole to Fk

$81kg*3m/s^2=810N-F_k$

$F_k=810N-243N$

$F_k=567N$

4 0

3 years ago

6) what is ararge of speed

irina [24]

Answer:

in everyday use and in kinematic the speed of an object is the magnitude of the change of its position it is thus a scalar quantity

5 0

3 years ago

A 12.0 kg rock slides off the edge of a bridge and falls into the water 25.0 meters below. what is the kinetic energy of the roc

Pavel [41]

During the fall, all the initial potential energy of the rock
$U=mgh$
has converted into kinetic energy of motion
$K= \frac{1}{2}mv^2$
where h is the initial height of the rock, m its mass, and v its velocity just before hitting the water. So, for energy conservation, we have
$U=K$
and so we can find the value of K, the kinetic energy of the rock just before hitting the ground:
$K=U=mgh = (12.0 kg)(9.81 m/s^2)(25.0 m)=2943 J$

7 0

3 years ago