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3 years ago

14

an apple with the mass of 0.95 kilograms hangs from a tree branch at 3.0 meters above the ground. what is the potential energy?

1 answer:

Liula [17]3 years ago

8 0

The potential energy is calculated by the formula
E = mgh = 0.95 * 10 * 3 J = 28.5 J

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A machine produces 4,000 J of work in 5 seconds. how much power does the machine produce.

ad-work [718]

Explanation:

Power is defined as the work done per unit time or

$P = \dfrac{\Delta W}{\Delta t}$

$\;\;\;\;= \dfrac{4000\:\text{J}}{5\:\text{s}} = 800\:\text{Watts}$

5 0

2 years ago

telo118 [61]

Ik what it is hit me up for it

5 0

3 years ago

If you hold your arm outstretched with palm upward, the force to keep your arm from falling comes from your deltoid muscle. assu

Maurinko [17]

Your question seems to be incorrect. Please check below:

What force must the deltoid muscle provide to keep the arm in this position? By what factor does this force exceed the weight of the arm?<span>If you hold your arm outstretched with palm upward, as in (Figure 1) , the force to keep your arm from falling comes from your deltoid muscle. Assume that the arm has mass 4 kg and the distances and angles shown in (Figure 1) .

F=?
F/w= ?

The answer is </span><span>339 N</span><span>

</span>

4 0

3 years ago

What is the minimum force required to increase the energy of a car by 69 J over a distance of 42 m? Assume the force is constant

sineoko [7]

ANSWER: 1.6N (forth option)

EXPLANATION:
E= Force x displacement
69=F(42)
69/42 = F
F=1.64 N

Answer : 1.6N

3 0

3 years ago

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

3 years ago