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4 years ago

13

A system releases 255 cal of heat to the surroundings while delivering 428 cal of work. what is the change in internal energy of

the system (in cal)?

1 answer:

vichka [17]4 years ago

7 0

Answer:-683 cal

Explanation:

Given

Heat released by system Q=-255 cal

as heat released is taken as negative and vice-versa

Work done by system W=428 cal

From First law of thermodynamics

$Q=\Delta +W , where \Delta U$ =change in internal Energy

$\Delta U=-255-428=-683$

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The rings of Saturn occupy the region inside Saturn's Roche limit.

Oliga [24]

I suppose this is a true or false question and that sentence is true.

7 0

3 years ago

(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

Two spectators at a soccer game in Montjuic Stadium see, and a moment later hear, the ball being kicked on the playing field. Th

NikAS [45]

Answer:

a) 68.943 m

b) 41.846 m

c) 80.648 m

Explanation:

Given:

Delay in time for spectator A, t₁ = 0.201 s

Delay in time for spectator B, t₂ = 0.122 s

the delay in sound heard is the due to the distance being traveled by the sound from the kicker to the spectator

thus,

a) Distance of the kicker from A,

d₁ = speed of sound × time taken

d₁ = 343 m/s × 0.201 s = 68.943 m

b) Distance of the kicker from B,

d₂ = speed of sound × time taken

d₂ = 343 m/s × 0.122 = 41.846 m

c) Since the angle between the two spectators for the player is 90°

thus, a right angles triangle is formed.

where, the distance between the spectators is the hypotenuses (s) of the so formed triangle

Therefore,

s² = d₁² + d₂²

on substituting the values, we get

s² = 68.943² + 41.846²

or

s² = 6504.22

or

s = √6504.22

or

s = 80.648 m

hence, the distance between the spectators is 80.648 m

3 0

3 years ago

What effect does increasing the current have on a solenoid?

marysya [2.9K]

Ampere’s law states that the magnetic field of a solenoid is directly proportional to the current supplied. This is expressed mathematically as:

B = μ n I

where,

B = magnetic field

μ = magnetic permeability

n = number of turns per length

I = current

<span>Therefore, increasing the current will increase the magnetic field of a solenoid.</span>

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4 years ago

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Diano4ka-milaya [45]

Answer:

True

Explanation:

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