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4 years ago

Which statement is true about an object moving in a circular motion due to centripetal force, F, when the mass is doubled?

Physics

1 answer:

marshall27 [118]4 years ago

4 0

Ans) A) Centripetal force will be doubled.

See centripetal force F = mv^2/r

That means centripetal force is directly proportional to the mass of the particle

So, if we double the mass, centripetal force will be increased by twofolds.

So, option A) is correct.

Now, looking at the other options,

B) says centripetal force is unaltered which is incorrect as centripetal force has been altered and increased twofold.

Option C) and D) reduces centripetal force which are also not possible here.

So, only Option A) is correct

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Atmospheric pressure is due to the weight _______

Kryger [21]

Of the gravitational pull and other things like mass. Planet earth it,s self as you said sir.

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3 years ago

Consider f(x) = -4x2 + 24x + 3. Determine whether the function has a maximum or minimum value. Then find the

murzikaleks [220]

Answer:

The function has a maximum in $x=3$

The maximum is:

$f(3) = 39$

Explanation:

Find the first derivative of the function for the inflection point, then equal to zero and solve for x

$f(x)' = -4*2x + 24=0$

$-4*2x + 24=0$

$8x=24$

$x=3$

Now find the second derivative of the function and evaluate at x = 3.

If $f (3) ''< 0$ the function has a maximum

If $f (3) '' >0$ the function has a minimum

$f(x)''= 8$

Note that:

$f(3)''= -8$

the function has a maximum in $x=3$

The maximum is:

$f(3)=-4(3)^2+24(3) + 3\\\\f(3) = 39$

4 0

3 years ago

Three point charges, two positive and one negative, each having a magnitude of 20 C are placed at the vertices of an equilateral

Daniel [21]

The resultant force on the positive charge is mathematically given as

X=40N

<h3>What is the magnitude of the electrostatic force on the negative charge?</h3>

Question Parameters:

Three-point charges, two positive and one negative, each having a magnitude of 20

Generally, the -ve charge is mathematically given as

$Q+=\sqrt{x^2+x^2+2x.xcos120}\\\\Q+=\sqrt{2x^2+2x*(1/2)}$

Q+=X

Therefore

$x=\frac{Kq1q2}{r2}\\\\x=\frac{9*10^9*20*10^{-6}*20*10^{-6}}{(30*10^-2)^2}$

X=40N

For more information on Force

brainly.com/question/26115859

5 0

2 years ago

A car accelerates from rest at 3.6 m/s 2 . How much time does it need to attain a speed of 5 m/s?

Olenka [21]

car starts from rest

$v_i = 0$

final speed attained by the car is

$v_f = 5 m/s$

acceleration of the car will be

$a = 3.6 m/s^2$

now the time to reach this final speed will be

$t = \frac{v_f - v_i}{a}$

$t = \frac{5 - 0}{3.6}$

$t = 1.39 s$

so it required 1.39 s to reach this final speed

6 0

3 years ago

If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular

erastova [34]

Answer;

D. The car would begin to move in the direction it was headed in a straight line.

Explanation;

-Centripetal force is any net force causing uniform circular motion. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration.

-The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

-Therefore,If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced then the car would begin to move in the direction it was headed in a straight line.

6 0

3 years ago

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