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valkas [14]
3 years ago
12

The direction you are traveling and the speed at which you travel

Physics
1 answer:
Bess [88]3 years ago
3 0
The answer is Velocity
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75 kilometers per hour is an example of:<br> A. speed.<br> B. velocity<br> C. acceleration
kherson [118]
The answer is A. speed
hope this helps! :D

5 0
3 years ago
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1. A heat lamp is designed to keep food and other things warm. Would it also make a good tanning lamp? Why or why not?
NemiM [27]
1. becuase it is so hot that it could tan or be used to keep warm.
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What is the magnitude of the kinetic frictional force
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The magnitude of the kinetic friction force, ƒk, on an object is. Where μk is called the kinetic friction coefficient and |FN| is the magnitude of the normal force of the surface on the sliding object. The kinetic friction coefficient is entirely determined by the materials of the sliding surfaces. hope it helps

4 0
3 years ago
Two particles, each of charge Q, are fixed at opposite corners of a square that lies in the plane of the page. A positive test c
amid [387]

Answer:

The magnitude of the net force is √2F.

Explanation:

Since the two particles have the same charge Q, they exert the same force on the test charge; both attractive or repulsive. So, the angle between the two forces is 90° in any case. Now, as we know the magnitude of these forces and that they form a 90° angle, we can use the Pythagorean Theorem to calculate the magnitude of the resultant net force:

F_N=\sqrt{F^{2}+F^{2}}\\\\F_N=\sqrt{2F^{2}}\\\\F_N=\sqrt{2}F

Then, it means that the net force acting on the test charge has a magnitude of √2F.

7 0
3 years ago
A single-turn circular loop of wire of radius 45 mm lies in a plane perpendicular to a spatially uniform magnetic field. During
Lina20 [59]

Answer:

Magnitude of induced emf is 0.00635 V

Explanation:

Radius of circular loop r = 45 mm = 0.045 m

Area of circular loop A=\pi r^2

A=3.14\times 0.045^2=0.00635m^2

Magnetic field is increases from 250 mT to 350 mT

Therefore change in magnetic field dB=250-350=100mT

Emf induced is given by

e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}

e=-0.00635\times \frac{100\times 10^{-3}}{0.10}=-0.00635V

Magnitude of induced emf is equal to 0.00635 V

7 0
3 years ago
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