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2 years ago

Based on the tug of war pictured below, what is the net force? –190 n 10 n –10 n 190 n

Physics

1 answer:

kirza4 [7]2 years ago

8 0

The net force in the tug of war which is shown in the picture provided will be 0N.

<h3>What is a Force?</h3>

This is defined as the power exerted on a body and the unit of Force is referred to as Newton.

In the tug of war, the net force is gotten by summing up all the forces.

–190N + 10N –10N + 190N = 0N.

Therefore the net force is 0N.

Read more about Force here brainly.com/question/12970081

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A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is

maksim [4K]

D = v^2 / 2ug

d= 3.5^2 / 0,15 x 9.8 m/s^2

the answer should be around 4.2m

hope this helps

8 0

2 years ago

Very large accelerations can injure the body, especially if they last for a considerable length of time. The severity index (SI)

Ludmilka [50]

Answer:

a) The severity index (SI) is 3047.749, b) The injured travels 0.345 meters during the collision.

Explanation:

a) The g-multiple of the acceleration, that is, a ratio of the person's acceleration to gravitational acceleration, is:

$a' = \frac{35\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$a' = 3.569$

The time taken for the injured to accelerate to final speed is given by this formula under the assumption of constant acceleration:

$v_{f} = v_{o} + a \cdot t$

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v_{f}$ - Final speed, measured in meter per second.

$a$ - Acceleration, measured in $\frac{m}{s^{2}}$ .

$t$ - Time, measured in seconds.

$t = \frac{v_{f}-v_{o}}{a}$

$t = \frac{\left(12\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)}{35\,\frac{m}{s^{2}} }$

$t = 0.095\,s$

Lastly, the severity index is now determined:

$SI = \frac{a'^{5}}{2\cdot t}$

$SI = \frac{3.569^{5}}{2\cdot (0.095\,s)}$

$SI = 3047.749$

b) The initial and final speed of the injured are $1.944\,\frac{m}{s}$ and $5.278\,\frac{m}{s}$ , respectively. The travelled distance can be determined from this equation of motion: