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2 years ago

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A stone of weight 10N falls from the top of a 250m high cliff. a) Calculate how much work is done by the force of gravity in pul

ling the stone to the foot of the cliff. b) How much energy is transferred to the stone?

1 answer:

tiny-mole [99]2 years ago

4 0

Answer:

work done = ( force × displacement)

(a)The force acting on the block is it's self weight and displacement is equal to height of the tower.

work done by gravity = (250 × 10) = 2500 joule

(b) The work done by gravity 2500 joule is transferred to the object in the form of it's kinetic energy.

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(a) What is the escape speed on a spherical asteroid whose radius is 500. km and whose gravitational acceleration at the surface

navik [9.2K]

Answer:

a) v= 1732.05m/s

b) d=250000m

c) v= 1414.214m/s

Explanation:

Notation

M= mass of the asteroid

m= mass of the particle moving upward

R= radius

v= escape speed

G= Universal constant

h= distance above the the surface

Part a

For this part we can use the principle of conservation of energy. for the begin the initial potential energy for the asteroid would be $U_i =-\frac{GMm}{R}$ .

The initial kinetic energy would be $\frac{1}{2}mv^2$ . The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

$-\frac{GMm}{R}+\frac{1}{2}mv^2=0$ (1)

Dividing both sides by m and replacing $\frac{GM}{R}$ by $a_g R$

And the equation (1) becomes:

$-a_g R+\frac{1}{2} v^2=0$ (2)

If we solve for v we got this:

$v=\sqrt{2 a_g R}=\sqrt{2x3\frac{m}{s^2}x500000m}=1732.05m/s$

Part b

When we consider a particule at this surface at the starting point we have that:

$U_i=-\frac{GMm}{R}$

$K_i=\frac{1}{2}mv^2$

Considering that the particle is at a distance h above the surface and then stops we have that:

$U_f=-\frac{GMm}{R+h}$

$K_f=0$

And the balance of energy would be:

$-\frac{GMm}{R}+\frac{1}{2}mv^2 =-\frac{GMm}{R+h}$

Dividing again both sides by m and replacing $\frac{GM}{R}$ by $a_g R^2$ we got:

$-a_g R+\frac{1}{2}v^2 =-\frac{a_g R^2}{R+h}$

If we solve for h we can follow the following steps:

$R+h=-\frac{a_g R^2}{-a_g R+\frac{1}{2}v^2}$

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

$h=\frac{2a_g R^2}{2a_g R-v^2}-R$

Replacing:

$h=\frac{2x3\frac{m}{s^2}(500000m)^2}{2(3\frac{m}{s^2})(500000m)-(1000m/s)^2}- 500000m=250000m$

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

$U_i=-\frac{GMm}{R+h}$

$K_i=0$

And after the particle reach the asteroid we have this:

$U_f=-\frac{GMm}{R}$

$K_f=\frac{1}{2}mv^2$

So the balance of energy would be:

$-\frac{GMm}{R+h}=-\frac{GMm}{R}+\frac{1}{2}mv^2$

Replacing again $a_g R^2$ instead of GM and dividing both sides by m we have:

$-\frac{a_g R^2}{R+h}=-a_g R+\frac{1}{2}v^2$

And solving for v:

$a_g R-\frac{a_g R^2}{R+h}=\frac{1}{2}v^2$

Multiplying both sides by two and taking square root:

$v=\sqrt{2a_g R-\frac{2a_g R^2}{R+h}}$

Replacing

$v=\sqrt{2(3\frac{m}{s^2})(500000m)-\frac{2(3\frac{m}{s^2}(500000m)^2}{500000+1000000m}}=1414.214m/s$

3 0

3 years ago

The ground heats the air through what

Tema [17]

Explanation:

The atmosphere is heated in several ways - heat from the core of the Earth, by radiation from the Sun, conduction from contact with warm land and water, convection to even out the temperature and by absorption of infrared radiation from the warm land and water. The core of the Earth is very hot.

4 0

3 years ago

What type of pressure is associated with more dense, sinking air?

Verizon [17]

B I hope it helps you

3 0

3 years ago

Read 2 more answers

While taking a shower, you notice that the shower head is made up of 44 small round openings, each with a radius of 2.00 mm. You

Julli [10]

To solve the problem it is necessary to apply the concepts related to the calculation of discharge flow, Bernoulli equations and energy conservation in incompressible fluids.

PART A) For the calculation of the velocity we define the area and the flow, thus

$A = \pi r^2$

$A = pi (2*10^{-3})^2$

$A = 12.56*10^{-6}m^2$

At the same time the rate of flow would be

$Q = \frac{1L}{2s}$

$Q = 0.5L/s = 0.5*10^{-3}m^3/s$

By definition the discharge is expressed as

$Q = NAv$

Where,

A= Area

v = velocity

N = Number of exits

Q = NAv

Re-arrange to find v,

$v = \frac{Q}{NA}$

$v = \frac{0.5*10^{-3}}{44*12.56*10^{-6}}$

$v = 0.9047m/s$

PART B) From the continuity equations formulated by Bernoulli we can calculate the speed of water in the pipe

$P_1 + \frac{1}{2}\rho v_1^2+\rho gh_1 = P_2 +\frac{1}{2}\rho v^2_2 +\rho g h_2$

Replacing with our values we have that

$1.5*10^5 + \frac{1}{2}(1000) v_1^2+(1000)(9.8)(0) = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)$

$v_1^2 = 10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)$

$v_1 = \sqrt{10^5 +\frac{1}{2}(1000)(0.9047)^2 +(1000)(9.8)(5.7)-1.5*10^5 - \frac{1}{2}(1000)}$

$v_1 = 3.54097m/s$

PART C) Assuming that water is an incomprehensible fluid we have to,

$Q_{pipe} = Q_{shower}$

$v_{pipe}A_{pipe}=v_{shower}A_{shower}$

$3.54097*A_{pipe}=0.9047*12.56*10^{-6}$

$A_{pipe} = \frac{0.9047*12.56*10^{-6}}{3.54097}$

$A_{pipe = 3.209*10^{-6}m^2$

3 0

3 years ago

Can anyone help me with this question

7nadin3 [17]

Answer:

6650km

Explanation:

Distance = (6.4 × 10³ + 250) km

= 6650 km

8 0

3 years ago