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Charra [1.4K]
3 years ago
5

What was the change in internal energy (chemical energy plus thermal energy) of the person pulling the block?

Physics
1 answer:
enyata [817]3 years ago
7 0

The historical method includes what steps?


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Which circuit is a series circuit
posledela

Answer:

The answer is A

Explanation:

7 0
3 years ago
A 2kg watermelon is dropped from a 4m tall roof a) use the appropriate kinematic equations to determine the instantaneous veloci
Ivenika [448]

Answer:

8.85m/s

Explanation:

The potential energy the watermelon held before dropping is Ep=mgh=2*9.8*4=78.4J.

When it strikes the ground, all of its Ep will transfer into Ek, so 1/2*m*v^2=78.4.

We already knew that m=2, so insert that in, we will get the V^2=78.4 m/s, V=8.85 m/s

6 0
3 years ago
Which force stops the car from moving?
Alex73 [517]

Answer:

The force of friction.

Explanation:

Gravity keeps the car on the ground.

Motion Allows the car to move.

The force of speed doesnt make sense.

Friction would cause the car to stop moving.

8 0
2 years ago
Read 2 more answers
iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms
saw5 [17]

Answer:

8.60 g/cm³

Explanation:

In the lattice structure of iron, there are two atoms per unit cell. So:

\frac{2}{a^{3} }  = \frac{N_{A} }{V_{molar} } where V_{molar}  = \frac{A}{\rho } an and A is the atomic mass of iron.

Therefore:

\frac{2}{a^{3} } = \frac{N_{A} * p }{A}

This implies that:

A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} }  }

  = \frac{4}{\sqrt{3} }r

Assuming that there is no phase change gives:

\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3}   }

  = 8.60 g/m³

3 0
3 years ago
A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is
Umnica [9.8K]

Answer:

<h2>Magnitude of the second charge is -4.17*10^{-7}C</h2>

Explanation:

According to columbs law;

F = kq1q2/r^{2}

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

0.03=9*10^{9}*  -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C

6 0
3 years ago
Read 2 more answers
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