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3 years ago

What is the change in internal energy if 60 J of heat are released from a

Physics

1 answer:

lana [24]3 years ago

7 0

Answer:

60 J of heat are released from the system: Q=-60 J

20 J of work is done on the system: W=-20 J

∆U=Q-W

∆U=-60-(-20)

∆U=-40 J

Explanation:

If the heat is released from the system, Q should be negative.

If the heat is gained by the system, Q should be positive.

If the work is done on the system, W should be negative.

If the work is done by the system, W should be positive.

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Which stars have the lowest absolute brightness?

Finger [1]

Answer:

DWARFS

Explanation:

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2 years ago

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Dwight uses a spring (k = 70 N/m) on a horizontal surface to make a launcher for model cars. The spring is attached to a holder

kramer

Part A)

As we know that spring force is given by

F = kx

here x = stretch in the spring from natural length

So here when spring reaches to its natural length

Force due to spring = 0

so acceleration = 0

Part b)

When spring is compressed from its natural length it will have elastic potential energy in it

so it is given by

$U = \frac{1}{2}kx^2$

now we know that there is no friction in it so maximum kinetic energy of the launcher must be equal to the elastic potential energy of the spring

$KE = \frac{1}{2}kx^2$

here we have

k = 70 N/m

x = 0.4 m

$KE = \frac{1}{2}(70)(0.4)^2$

$KE = 5.6 J$

Part c)

Now to find the speed we know that

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

$5.6 = \frac{1}{2}0.3v^2$

$v = 6.11 m/s$

so its speed is 6.11 m/s

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2 years ago

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Whats an astral projection

Dominik [7]

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3 years ago

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$