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2 years ago

14

Jesse wants to know how well a particular brand of car wax protects his car from dirt. Which of the following is a variable that

must be kept constant?
a. The use of the car wax
b. How dirty the car gets
c. The kind of car

1 answer:

sweet [91]2 years ago

4 0

B but I'm not fo sure .

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A ball is dropped from a window and takes two seconds to reach the ground .it starts from rest and reaches a final speed of 20m/

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The ball can't reach the speed of 20 m/s in two seconds, unless you THROW it down from the window with a little bit of initial speed. If you just drop it, then the highest speed it can have after two seconds is 19.6 m/s .

If an object starts from rest and its speed after 2 seconds is 20 m/s, then its acceleration is 20/2 = 10 m/s^2 .

(Gravity on Earth is only 9.8 m/s^2.)

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Describe a situation when you might travel at a high velocity, but with low acceleration

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Reading a book in your warm, comfy seat ... in Row-27 of a
passenger airliner cruising at 450 miles per hour.

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2 years ago

A force of constant magnitude pushes a box up a vertical surface, as shown in the figure.

Ray Of Light [21]

The work done on the box by the applied force is zero.

The work done by the force of gravity is 75.95 J

The work done on the box by the normal force is 75.95 J.

<h3>The given parameters:</h3>

Mass of the box, m = 3.1 kg
Distance moved by the box, d = 2.5 m
Coefficient of friction, = 0.35
Inclination of the force, θ = 30⁰

<h3>What is work - done?</h3>

Work is said to be done when the applied force moves an object to a certain distance

The work done on the box by the applied force is calculated as;

$W = Fd cos(\theta)\\\\W = (ma)d \times cos(\theta)$

where;

a is the acceleration of the box

The acceleration of the box is zero since the box moved at a constant speed.

$W = (0) d \times cos(30)\\\\W = 0 \ J$

The work done by the force of gravity is calculated as follows;

$W = mg \times d\\\\W = 3.1 \times 9.8 \times 2.5 \\\\W = 75.95 \ J$

The work done on the box by the normal force is calculated as follows;

$W = (F_n) \times d\\\\W = (mg + F sin\theta) \times d\\\\W = (mg + 0) \times d\\\\W = mgd\\\\W = 3.1 \times 9.8 \times 2.5\\\\W = 75.95 \ J$

Learn more about work done here: brainly.com/question/8119756

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1 year ago

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

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2 years ago

Do true blackbodies exist in nature?<br><br> Yes or No??

MaRussiya [10]

Answer:

no

Explanation:

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3 years ago

Read 2 more answers