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3 years ago

A car starts from rest with an acceleration of 5 ft/s. What is its velocity after it has gone 600 ft?

Physics

1 answer:

Soloha48 [4]3 years ago

3 0

Answer:

First, let’s correct the question. Acceleration is the rate of change in velocity. Its unit therefore is ft/sec/sec. If S is the distance traveled for a given duration, S = Vot + (1/2)at^2 where Vo is the initial velocity, a is the acceleration and t is the time. For Vo = 0, a = 6m/sec/sec and t = 3 sec. The distance traveled is S = 0 + (1/2) x 6 x 3^2 = 27 meters

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As the temperature of water drops, it continues to give off thermal energy until it freezes. This is an example of.

jeka57 [31]

Answer:

The answer should be C

Explanation:

An exothermic process releases heat, causing the temperature of the immediate surroundings to rise. An endothermic process absorbs heat and cools the surroundings.

Please tell me if it is wrong

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7 0

3 years ago

Read 2 more answers

3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a

serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as $\mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}$ .

Where, ${V}_{i}$ is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

${V}_{i}$ =3 m/s, V = 0 m/s, and ∆s = 2 m

Substitute the values in the above formula,

$0=3^{2}-2 \times 2 \times a$

0 = 9 - 4a

4a = 9

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration.

Calculating Coefficient of friction:

$\mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}$

Compare the above equation

$\mu \times m \times g=m \times a$

Cancel "m" common term in both L.H.S and R.H.S

$\text { Equation becomes, } \mu \times g=a$

$\text { Coefficient of friction } \mu=\frac{a}{g}$

$\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}$

$\mu=\frac{2.25}{9.8}$

$\mu=0.229$

Hence coefficient of friction is 0.229.

calculating force:

$\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })$

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

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3 years ago

We know that the law of conservation of energy states that energy can not be created or destroyed. It only changes form. Conside

Pavel [41]

The conservation of energy always holds true even when not clearly observable in machines that are less than 100% efficient. More often than not a machine will suffer energy losses (e.g. consider for a cooling fan: friction between the rotating blades, drag resistance in the air the fan is pushing around, resistance in the wire, and heat radiating/conducting away from the circuitry).

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4 years ago

Read 2 more answers

At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the ti

34kurt

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r = 0.89 x 0. 5 = 0.445 m

Angular velocity

$\omega =\frac{72}{0.445}=161.8rad/s$

Frequency

$f=\frac{2\pi}{\omega}=\frac{2\times \pi}{161.8}=0.0388rev/s=2.33rev/min$

Revolutions per minute = 2.33

b) Centripetal acceleration

$a=\frac{v^2}{r}$

Here linear velocity = 72 m/s

Radius, r = 0.445 m

Substituting

$a=\frac{72^2}{0.445}=11649.44m/s^2$

Centripetal acceleration = 11649.44m/s²

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3 years ago

light waves are first transmitted through the ________ at the front of the eye and enter an opening called the ________ before s

viva [34]

The transmission of light waves is usually done through cornea of the eyes, then move through another opening which is regarded as pupil before it will get to the retina.

Light waves can be regarded as moving energy which contains microscopic particles known as photons.

The vision of the eye can be completed through the light wave passing through the components of the eyes and this process goes thus;

Light will move through the (cornea) which is situated at the front area of the eyes into lens.

Then both the cornea and the lens give room for the focusing of the light rays to the retina which is situated at the back of the eye .

Then through the help of the cells in the retina, the light will be absorbed and then be converted to electrochemical impulses and then transfer it to the brain as well as optic nerve.

Therefore, light wave are form of tiny microscopic particles.

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2 years ago