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3 years ago

10

A freight car moves along a frictionless level railroad track at constant speed. The freight car is open on the top. A large loa

d of sand is suddenly dumped into the freight car. What happens to the speed of the freight car?

1 answer:

Fittoniya [83]3 years ago

3 0

The added weight of the sand puts more downward pressure on the wheels contacting the rails, which would cause the trains speed to decrease.

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1. The resistance of an electric device is 40,000 microhms. Convert that measurement to ohms.

lidiya [134]

Answer:

The answer would be 0.04ohms.

Explanation:

Hopefully this helps

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3 years ago

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What is the ACCELERATION of a 50 kg object pushed with a force of 500 newtons *

Anon25 [30]

Answer:

The answer is

<h2>10 m/s²</h2>

Explanation:

To find the acceleration of an object given the force and mass we use the formula

<h3> $acceleration = \frac{force}{mass}$ </h3>

From the question

mass of object = 50 kg

force = 500 N

So the acceleration is

<h3> $acceleration = \frac{500}{50} \\ = \frac{50}{5}$ </h3>

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

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3 years ago

If a fuse melts, does it create an open circuit, a closed circuit, or a short circuit?

Andrew [12]

Answer:

short circuit

7 0

3 years ago

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A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

6 0

4 years ago

A small plastic ball with a mass of 7.00 10-3 kg and with a charge of +0.155 µC is suspended from an insulating thread and hangs

adell [148]

Answer:

the magnitude of the charge Q on each plate is $3.053 *10^{-8} \ C$

Explanation:

Given that :

mass (m) = $7.00 *10 ^{-3} \ kg$

charge (q) = +0.155 µC = $+0.155 *10^{-6}\ C$

angle $\theta = 30^0 \ C$

Area A on each plate = 0.0135 m²

From the diagram below;

$tan \ \theta = \frac{Eq}{mg}$ ----- equation (1)

Also by using Gauss Law ;

$Q = \epsilon_0 \phi$

$Q = \epsilon_0EA$ ----- equation (2)

Combination equation 1 and 2 together ; we have

$Q = \frac{\epsilon_0\ * \ m\ *\ g \ \ * \ tan \theta \ * \ A}{q}$

$Q = \frac{(8.85*10^{-12}C^2/N.m^2 )\ * \ (7.00*10^{-3} kg)\ *\ (9.8 m/s^2) \ \ * \ tan \(30 \ * \ (0.0135 m^2)}{0.155*10^{-6}\ C}$

$Q = 3.053 *10^{-8} \ C$

8 0

3 years ago