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Vilka [71]
2 years ago
8

A student drops a ball from the top of a tall building; it takes 2.9 s for the ball to reach the ground.

Physics
1 answer:
AlexFokin [52]2 years ago
4 0

<h2><u>We have</u>,</h2>

  • Initial velocity (u) = 0 m/s
  • Time taken (t) = 2.9s
  • Acceleration due to gravity (g) = + 10 m/s² [Down]

<h2><u>To calculate</u>,</h2>

  • Final velocity (v)
  • Height (h)

<h2><u>Solution</u><u>,</u></h2>

→ v = u + gt

→ v = 0 + 10(2.9)

→ v = 29 m/s \qquad … ( Ans )

And,

→ h = ut + ½gt²

→ h = 0(2.9) + ½ × 10 × (2.9)²

→ h = 5 × 8.41

→ h = 42.05 m \qquad … ( Ans )

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zubka84 [21]
The solution for this problem would be:(10 - 500x) / (5 - x) 
so start by doing: 
x(5*50*2) - xV + 5V = 0.02(5*50*2) 
500x - xV + 5V = 10 
V(5 - x) = 10 - 500x 
V = (10 - 500x) / (5 - x) 
(V stands for the volume, but leaves us with the expression for x)
3 0
3 years ago
If my weight on Earth is 140lbs, what is my mass?
SashulF [63]

Answer:

63.57 kg

Explanation:

weight = 140 lbs

Let the mass  is m.

1 lbs = 4.45 N

The weight of an object is defined as the force with which our earth attracts the body towards its centre.

Weight is the product of mass of the body and the acceleration due to gravity of that planet.

W = m x g

On earth surface g = 9.8 m/s^2

Now convert lbs in newton

So, 140 lbs = 140 x 4.45 = 623 N

So, m x 9.8 = 623

m = 63.57 kg

Thus, the mass is 63.57 kg.

5 0
3 years ago
Failure to accomplish Erikson’s psychosocial task of late adulthood leads to despair.
Y_Kistochka [10]
False is the answer
6 0
3 years ago
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A tug boat pulls a ship with a constant net horizontal force of 5.00•10*3 N and causes the ship to move through a harbor. How mu
Murljashka [212]

The work done on the ship is 1.5\cdot 10^7 J

Explanation:

The work done by a force on an object is given by:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and of the displacement

In this problem, we have:

F=5.00\cdot 10^3 N (force acting on the ship)

d = 3.00 km = 3000 m (displacement of the ship)

\theta=0^{\circ} (because the force is horizontal, and the displacement is horizontal as well)

Therefore, the work done on the ship is

W=(5.00\cdot 10^3)(3000)(cos 0^{\circ})=1.5\cdot 10^7 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

8 0
3 years ago
The joule (J) is a unit of energy. Recall that energy may be converted between many different forms such as mechanical energy, t
REY [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is  W = -177.275J

Explanation:

From the question we are told that

      The initial Volume is  Vi = 0.160 L

      The final volume is  V_f = 0.510L

      The external pressure is  P = 5.00 \ atm

Generally the change in volume is

           \Delta V = V_f - V_i

Substituting values we have

           \Delta V = 0.510 -0.160

                 = 0.350L

Generally workdone is mathematically represented as

           W = -P \Delta V

W is negative because the working is done on the environment by the system which is indicated by volume increase

     Substituting values

                W = - 5* 0.350

                    = -1.75 \ L \ \cdot atm

Now  1 \  L \cdot atm = 101.3J

  Therefore  W = -1.75* 101.3

                          = -177.275J

   

7 0
3 years ago
Read 2 more answers
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