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-Dominant- [34]

3 years ago

15

Two football players are attempting to tackle each other. If one football player has a mass of 100 kg and pushes with a force of

100N onto another football player whose mass is 80kg, how much force does the other player push back with?

1 answer:

Vaselesa [24]3 years ago

8 0

Answer:

is it 20kg. Two opposing forces pushing onto each other

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a child goes down the slide, starting from rest. if the length of the slide is 2m and it takes the child 3 seconds to go down th

Doss [256]

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6 0

4 years ago

A bungee cord has a spring constant of 82.5 N/m. After jumping, a person oscillates with a period of 7.50 s. What is the mass of

Arada [10]

Answer:

Explanation:

T = 2π √ (m/k)

T / 2π = √ (m/k)

T√k = 2π √m

(T√k) / 2π = √m

( (T√k) / 2π)^2 = m

m = ( (7.5√82.5) / 2π)^2

m = 117.66 Kg

3 0

3 years ago

Apply a force of 50N to the left describe the motion of the box

GenaCL600 [577]

When a force is applied to the box , this will cause an acceleration to the box.
(force =mass×acceleration)

So the box has a constant acceleration and a changing velocity.

4 0

3 years ago

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f

Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ( $r$ ), measured in kilometers, can be represented by a right triangle:

$r = \sqrt{x^{2}+y^{2}}$ (1)

Where:

$x$ - Horizontal distance between the rocket and the observer, measured in kilometers.

$y$ - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ( $\theta$ ), measured in sexagesimal degrees, is defined by the following trigonometric relation:

$\tan \theta = \frac{y}{x}$ (2)

If we know that $x = 5\,km$ , then the expression is:

$\tan \theta = \frac{y}{5}$

And the rate of change of this angle is determined by derivatives:

$\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y$

$\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}$

$\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}$

$\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}$

Where:

$\dot \theta$ - Rate of change of the angle of elevation, measured in sexagesimal degrees.

$\dot y$ - Vertical speed of the rocket, measured in kilometers per hour.

If we know that $y = 4\,km$ and $\dot y = 400\,\frac{km}{h}$ , then the rate of change of the angle of elevation is:

$\dot \theta = 48.780\,\frac{\circ}{s}$

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0

3 years ago

I am moving and had to pack all of my things. I pushed the boxes with a force of 15N to the left, and there was a frictional for

devlian [24]

Answer:

8N

Explanation:

Applied force - Frictional force = net force

6 0

3 years ago