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3 years ago

If an object travels at a constant speed in a circular path, the acceleration of the object is:

Physics

1 answer:

Vilka [71]3 years ago

6 0

Answer:

1- The acceleration of the object is larger in magnitude the smaller the radius of the circle.

Explanation:

The acceleration of an object in a circular path is

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

As can be seen from the equation, if the radius of the circle is decreases, the magnitude of the acceleration increases.

As for the direction of the acceleration, it is always towards the center, and it is always perpendicular to the direction of the velocity.

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3 years ago

A 500 N force accelerates an object at 20 m s-2. What is its mass?

avanturin [10]

Answer: The mass of the object is 25kg.

The given question deals with Newton's second law of motion and its applications.

Explanation: Given force, F=500N

acceleration, a=20 m/ $s^{2}$

From Newton's 2nd law of motion , we have

F=ma where m=mass of the object

⇒500=m×20

⇒m=500/20=25

∴ Mass of the object is 25 kg .

Reference Link: brainly.com/question/1141170

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8 0

1 year ago

How much kinetic energy does a 50 kg rock has if its moving at a velocity of 2m/s

maria [59]

Answer:

KE = 100 J

Explanation: Should be correct

5 0

2 years ago

There's an electric field in some region of space that doesn't change with position. An electron starts moving with a speed of 2

tangare [24]

Answer:

Explanation:

Given

speed of Electron $u=2\times 10^7\ m/s$

final speed of Electron $v=4\times 10^7\ m/s$

distance traveled $d=1.2\ cm$

using equation of motion

$v^2-u^2=2as$

where v=Final velocity

u=initial velocity

a=acceleration

s=displacement

$(4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}$

$a=5\times 10^{16}\ m/s^2$

acceleration is given by $a=\frac{qE}{m}$

where q=charge of electron

m=mass of electron

E=electric Field strength

$5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}$

$E=248.3\ kN/C$

5 0

3 years ago

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

gayaneshka [121]

Force applied on the car due to engine is given as

$F_1 = 300 N$ towards right

Also there is a force on the car towards left due to air drag

$F_2 = 150 N$ towards left

now the net force on the car will be given as

$\vec F_{net} = \vec F_1 + \vec F_2$

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

$F_{net} = F_1 - F_2$

$F_{net} = 300 - 150$

$F_[net} = 150 N$

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0

3 years ago