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kramer
3 years ago
12

What are the three components of the equation for the second law of motion?

Physics
1 answer:
noname [10]3 years ago
6 0

force= mass × acceleration

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a 25 kg object is pushed with a horizontal force of 5 N esat across a table. If the force of friction is 2 N, what is the accele
Julli [10]

Answer:

Apply Newton's second law in the moving direction.

Explanation:

\begin{aligned}F &= ma\\5N-2N &= 25kg\times a\\a & = \frac{3N}{25kg}\\&= 0.12ms^{-2}\end{aligned}

Friction force applies in the opposite direction of motion; as a restriction.

8 0
4 years ago
A 400g sample of water absorbs 500j of energy. how did the water temperature change if the specific heat of water is 4.18j/g©. S
Mashutka [201]

In general, the quantity of heat energy, Q, required to raise a mass m kg of a substance with a specific heat capacity of <span>c </span>J/(kg °C), from temperature t1 °C to t2 °C is given by:

<span>Q </span>= <span>mc(t</span><span>2 </span><span>– t</span>1<span>) joules</span>

<span>So:</span>

(t2-t1) =Q / mc

<span>As we know:
Q = 500 J </span>

<span>m = 0.4 kg</span>

<span>c = 4180 J/Kg </span>°c

<span>We can take t1 to be 0</span>°c

t2 - 0 = 500 / ( 0.4 * 4180 )

t2 - 0 = 0.30°c

6 0
3 years ago
Help with this I can’t come up with anything
kakasveta [241]
Let’s say you have a spring. You press on the spring with your finger. The spring goes down. This is the action force. Then, the spring goes back up after you take your finger off of it. This is known as the reaction force.
3 0
3 years ago
Of the following systems, which contains the most heat?
densk [106]

Answer:d

Explanation:

Given systems are state of matter and do not contain any heat instead Heat is required to change Phase or raise the temperature of the particular system.

For example 600 kg of ice at 0^{\circ}C

Heat Required to convert it to water at 0^{\circ}C is

Q=m\times L

Where L=latent heat of Fusion =334 J/gm

Q=600\times 334\times 1000

Q=200.4 MJ

8 0
3 years ago
A student walks (2.9±0.1)m, stops and then walks another (3.9 ±0.2)m in the same directionWith the given uncertainties, what is
Anika [276]

Answer:

The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.

Explanation:

For 2 quantities A and B represented as

A\pm \Delta A and B\pm \Delta B

The sum is represented as

Sum=(A+B)\pm (\Delta A+\Delta B)

For the the values given to us the sum is calculated as

Sum=(2.9+3.9)\pm (0.1+0.2)

Sum=6.8\pm 0.3

Now the since the uncertainity inthe sum is \pm 0.3

The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity

Thus closest distance equals 6.8-0.3=6.5meters

3 0
4 years ago
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