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Over [174]
3 years ago
14

A hot lump of 35.8 g of copper at an initial temperature of 70.7 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to

reach thermal equilibrium. What is the final temperature of the copper and water, given that the specific heat of copper is 0.385 J/(g·°C)? Assume no heat is lost to surroundings.

Physics
1 answer:
AURORKA [14]3 years ago
4 0

I upload the answer via Image because Brainly warned me that there were inappropriate links or words.

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In a third class lever, the distance from the effort to the fulcrum is ____________ the distance from the load/resistance to the
padilas [110]
The complete sentence is:
In a third class lever, the distance from the effort to the fulcrum is SMALLER the distance from the load/resistance to the fulcrum.
In fact, in a third class lever, the fulcrum is on one side of the effort and the load/resistance is on the other side, so the effort is located somewhere between the two of them. This means that the distance effort-fulcrum is smaller than the distance load-fulcrum.
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3 years ago
please tell th answer fast suppose A ball of mass M is thrown vertically upward with initial speed be its speed is continuously
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Answer:.

the ball would go down and speed of it would not strike so that wouldnt be an example of the conversation momentum

Explanation:

The answer I think

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3 years ago
Braddy connected the loose wire to the battery and created an electromagnet. He picked up 45 thumb tacks with his electromagnet,
Katarina [22]

Answer:

C) Use two batteries instead of one.

Explanation:

-Strength of an electromagnet depends on the electrical current which flows through the wires.

-Two batteries have a higher current than one and thus higher strength in the electromagnet.

3 0
3 years ago
Smooth surfaces produce a __________________ reflection.
Galina-37 [17]

Answer:

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6 0
3 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

3 0
4 years ago
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