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Bogdan [553]
3 years ago
5

A 50kg box is pulled up a distance of 5.0m up a 30 degree incline by a 300. N force. If there is friction between the box and th

e incline, how much work is performed (net) on the box?
Physics
1 answer:
Finger [1]3 years ago
5 0

Answer:

                   

Explanation:

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Explain the benefits of understanding sound waves properties in real life.
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Answer:

The vibrations from sound waves cause our ears to send signals to our brains to create sound. The speed of sound waves will determine the sound's pitch, or how high or low something sounds. Sound waves are important because they allow us to hear important messages and emergency signals to protect ourselves.

Explanation:

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A soccer ball is kicked with an initial velocity 13 m/s at an angle of 35 degrees from the horizontal. What is the maximum heigh
Nana76 [90]

Answer:

The maximum height reached by the ball is 2.84 m

Explanation:

Given;

initial velocity of the soccer, u = 13 m/s

angle of projection, θ = 35°

The maximum height reached by the ball = ?

Apply the following kinematic equation, to determine the maximum height reached by the ball.

Maximum height (H) is given as;

H = \frac{U^2sin^2\theta}{2g} \\\\H = \frac{(13)^2(sin \ 35^0)^2}{2*9.8}\\\\H = 2.84 \ m

Therefore, the maximum height reached by the ball is 2.84 m

3 0
3 years ago
A sprinter accelerates at 7.5 m/s from rest in 2.0 s, what distance did she go? (15 m)
a_sh-v [17]

Answer:

<em>The sprinter traveled a distance of 7.5 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the rate of change of the velocity of an object is constant.

The equation that rules the change of velocities is:

v_f=v_o+at\qquad\qquad [1]

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

The distance traveled by the object is given by:

\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]

Using the equation [1] we can solve for a:

\displaystyle a=\frac{v_f-v_o}{t}

The sprinter travels from rest (vo=0) to vf=7.5 m/s in t=2 s. Computing the acceleration:

\displaystyle a=\frac{7.5-0}{2}

a=3.75\ m/s^2

Now calculate the distance:

\displaystyle x=0*2+\frac{3.75*2^2}{2}

\displaystyle x=7.5\ m

The sprinter traveled a distance of 7.5 m

8 0
3 years ago
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