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3 years ago

It requires 2,500 joules to raise a certain amount of water (c = 4.186 J/g C) from 20 C to 60 C

Physics

1 answer:

Ksivusya [100]3 years ago

7 0

The mass of the water is 14.9 g

Explanation:

When a certain amount of a susbtance is heated, the temperature of the substance increases according to the equation

$Q=mC_s \Delta T$

where

Q is the amount of energy supplied to the substance

m is the mass of the substance

$C_s$ is its specific heat capacity

$\Delta T$ is the change in temperature

In this problem, we have:

Q = 2500 J of energy supplied to the water

$C_s = 4.186 J/gC$ is the specific heat capacity of water

$\Delta T=60 C - 20 C = 40^{\circ}C$ is the change in temperature of the water

Therefore we can solve for m to find the mass of the water:

$m=\frac{Q}{C_s \Delta T}=\frac{2500}{(4.186)(40)}=14.9 g$

Learn more about specific heat capacity:

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A pendulum consists of a 2.0-kg block hanging on a 1.5-m length string. A 10-g bullet moving with a horizontal velocity of 900 m

pav-90 [236]

Answer:

The maximum height above its initial position is:

$h_{max}=1.53\: m$

Explanation:

Using momentum conservation:

$m_{b}v_{ib}=m_{B}v_{fB}+m_{b}v_{fb}$ (1)

Where:

m(b) is the mass of the bullet
m(B) is the mass of the block
v(ib) is the initial velocity of the bullet
v(fb) is the final velocity of the bullet
v(fB) is the final velocity of the block

Let's find v(fb) using equation (1)

$m_{b}(v_{ib}-v_{fb})=m_{B}v_{fB}$

$v_{fB}=\frac{m_{b}(v_{ib}-v_{fb})}{m_{B}}$

$v_{fB}=\frac{0.1(900-300)}{2}$

$v_{fB}=30\: m/s$

We need to find the maximum height, it means that all kinetic energy converts into gravitational potential energy.

$\frac{1}{2}m_{B}v_{fB}=m_{B}gh_{max}$

$h_{max}=\frac{1}{2g}v_{fB}$

$h_{max}=\frac{1}{2(9.81)}30$

$h_{max}=1.53\: m$

I hope it helps you!

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3 years ago

A 40 kg-skier starts at the top of a 12-meter high slope. At the bottom, she is traveling 10 m/s. How much energy does she lose

julia-pushkina [17]

2704 j is my answer i did the math. good luck

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3 years ago

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Charge X has twice as much charge as particle Y. The two charges are placed near each other. Compared to the force on particle X

Art [367]

The force on charge Y is the same as the force on charge X

Explanation:

We can answer this problem by applying Newton's third law of motion, which states that:

"When an object A exerts a force on object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"

In this problem, we can identify object A as charge X and object B as charge Y. The magnitude of the electrostatic force between them is given by

$F=k\frac{q_x q_y}{r^2}$ (1)

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_x, q_y$ are the two charges

r is the separation between the two charges

According to Newton's third law, therefore, the magnitude of the force exerted by charge X on charge Y is the same as the force exerted by charge Y on charge X (and it is given by eq.(1)), however their directions are opposite.

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3 years ago

A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic field str

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Answer:

$B_2 = 1.71 mT$

Explanation:

As we know that the magnetic field near the center of solenoid is given as

$B = \frac{\mu_0 N i}{L}$

now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

So the magnetic field at the center of the solenoid is 2 mT

now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

$\frac{B_1}{B_2} = \frac{L_2}{L_1}$

now plug in all values in it

$\frac{2.0 mT}{B_2} = \frac{21}{18}$

$B_2 = 1.71 mT$

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Answer:

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Explanation:

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