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artcher [175]
3 years ago
11

As an acapulco cliff diver drops to the water from a height of 42.0 m, his gravitational potential energy decreases by 27,500 j.

what is the diver's weight in newtons?
Physics
1 answer:
seraphim [82]3 years ago
3 0

We know that potential energy is simply energy due to position which has the formula:

PE = mass * gravity * height

We know that mass * gravity is weight, therefore:

PE = weight * height

27500 J = weight * 42 m

<span>weight = 654.76 Newtons</span>

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A 2.0 kg book is lying on a 0.78 m -high table. You pick it up and place it on a bookshelf 2.1 m above the floor.Part ADuring th
zloy xaker [14]

Given data:

* The mass of the book is 2 kg.

* The initial height of the book is 0.78 m.

* The final height of the book is 2.1 m.

Solution:

(A). The work done by the gravity on the book is,

W=mg(h_f-h_i)

where m is the mass, g is the acceleration due to gravity, h_i is the initial height and h_f is the final height,

The work is done in moving the object in upward direction where as the gravitational force is acting in the down ward direction. Thus, the value g (acceleration due to gravity) is taken as negative in this case.

Substituting the known values,

\begin{gathered} W=2\times(-9.8)\times(2.1-0.78) \\ W=-19.6\times1.32 \\ W=-25.9\text{ J} \end{gathered}

Thus, the work done by the gravitational force is -25.9 J.

(B). The work done by the hand on the moving the book is,

\begin{gathered} W_1=-W \\ =25.9\text{ J} \end{gathered}

Thus, the work done by the hand on the book is 25.9 J.

7 0
1 year ago
Please help me it's URGENT (#10 btw and also how do I solve for time​
Afina-wow [57]

Answer:

t = 12s

Explanation:

Given:

v-initial = 0 m/s

x = 360 m

a = 5.0 m/s^2

Solve:

x = (v-initial)t + 1/2(a*t^2)

360 = 0t + 1/2 (5.0t^2)

360 = 2.5 t^2

144 = t^2

t = sqrt(144) = 12

Therefore, it takes 12 seconds.

5 0
3 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
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