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MaRussiya [10]
3 years ago
7

v}^{2} = mgh" align="absmiddle" class="latex-formula">
g= 9.8 m/s^2
h= 875
v= ?
Physics
1 answer:
Katyanochek1 [597]3 years ago
8 0

You said 0.5 · m · v² = m · g · h

Divide each side by 'm' :  0.5 · v² = g · h

Multiply each side by 2 :  v² = 2 · g · h

Square root each side :   v = √(2 · g · h)

You said that  g = 9.8 m/s²  and  h = 875 units

So  v = √(9.8 m/s² · 875 units)

v = √(8,575 m·unit/s²)

v = 92.6 / s² · √(m · unit)

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One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side
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Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

L1^2=L2^2=L^2=2^2+(H/2)^2  Replacing this value in the previous equation:

\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34  Solving for H:

H=\sqrt{52}

We can now, calculate the angle between L1 and the 2m segment:

\alpha = atan(\frac{H/2}{2})=60.98°

If we make a sum of forces in the midpoint of the rope we get:

-2*T*cos(\alpha ) + F = 0  where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

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7 0
3 years ago
14. a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s. a. how much later does the ball
Burka [1]

The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s

<h3>Motion Under Gravity</h3>

The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.

Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.

a. how much later does the ball hit the ground?

The time can be calculated by considering the vertical component of the motion with the use of formula below.

h = ut + 1/2gt²

Where

  • Height h = 75 m
  • Initial velocity u = 0 ( vertical velocity )
  • Acceleration due to gravity g = 9.8 m/s²
  • Time t = ?

Substitute all the parameters into the formula

75 = 0 + 1/2 × 9.8 × t²

75 = 4.9t²

t² = 75/4.9

t² = 15.30

t = √15.3

t = 3.9 s

b. how far from the building will it land?

The range can be found by using the formula

R = ut

Where u = 4.6 m/s ( horizontal velocity )

R = 4.6 × 3.9

R = 18 m

c. what is the velocity of the ball just before it hits the ground?

The final velocity will be

v = u + gt

v = 4.6 + 9.8 × 3.9

v = 4.6 + 38.22

v = 42.82 m/s

Therefore, the answers are 3.9 s, 18 m and 42.82 m/s

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#SPJ1

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