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3 years ago

What does the magnitude of centripetal acceleration depend on ?

Physics

1 answer:

MariettaO [177]3 years ago

3 0

Explanation:

Centripetal acceleration ac is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity v and has the magnitude ac=v2r;ac=rω2 a c = v 2 r ; a c = r ω 2 .

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The velocity of a 1.3 kg block sliding down a frictionless inclined plane is found to be 1.26 m/s. 1.10 s later, it has a veloci

nasty-shy [4]

Answer:

$\theta = 25.3^\circ$

Explanation:

The acceleration of the block can be found by the kinematics equations:

$v = v_0 + at\\5.88 = 1.26 + a(1.1)\\a = 4.2~m/s^2$

Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.

$F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ$

7 0

3 years ago

A 0.25 kg skeet (clay target) is fired at an angle of 30 degrees to the horizon with a speed of 25 m/s. When it reaches the maxi

kozerog [31]

Answer:

6.51 m and 37.13 m

Explanation:

from the question we were given

mass of skeet = 0.25 kg

speed of skeet = 25 m/s

angle = 30 degrees to the horizon

mass of pellet = 15 g

speed of pellet = 2000 m/s

without being hit by the pellet, the x and y components of the skeet velocity are

Vx = 25 cos 30 = 21.65

Vy = 25 sin 30 = 12.5

now

V = U + (a x t)

where V = final velocity, U = initial velocity , a = acceleration, t = time and s = distance

-25 sin 30 = 25 sin 30 + (-9.8 x t)

-12.5 = 12.5 - 9.8 t

t = 2.55 s

also

V^2 = U^2 + 2as ( s = vertical distance and V = 0 )

0 = (25 sin 30)^2 + 2 x (-9.8) x Y

19.6 Y = 156.25

Y =7.97 m

the distance traveled without the pellet hitting the skeet can be gotten using distance = speed x time

distance = 21.65 x 2.55 = 55.2 m

applying the conservation of linear momentum

on the x axis : (Ms x Us) + (Mp x Up) = (Ms x Vx) + (Mp x Vx) ...equation 1

on the y axis : (Ms x Us) + (Mp x Up) = (Ms x Vy) + (Mp x Vy) ...equation 2

(0.25 x 25 cos 30) + 0 = (0.25 +0.015) Vx

Vx = 20.42m/s

0 + (0.015 x 200) = (0.25 + 0.015) Vy

Vy = 11.32 m/s

now V^2 = U^2 + 2 as

0 = 11.3^2 + (2 x (-9.8) x s)

s = 6.51 m

to find the extra distance moved after collision we apply

s = ut + 1/2 at^2

-7.98 = 11.32t + 1/2 (-9.8) t^2

4.9 t^2 - 11.32t + 7.98

t = 3.17 s

recall that distance = speed x time

distance = 20.42 x 3.17 = 64.73 m

the distance of the skeet before being hit would be half of the distance it travels without being hit, this is because the skeet was hit at its maximum height = 55.2 /2

= 27.6 m

therefore the extra distance traveled would be the change in distance = 64.73 -27.6 = 37.13 m

5 0

3 years ago

A 25kg chair initially at rest on a horizontal floor requires 165 N force to set it in motion. Once the chair is in motion, a 12

bazaltina [42]

The coefficient of static friction between the chair and the floor is 0.67

Explanation:

Given:

Weight of the chair = 25kg

Force = 165 N (F_applied)

Force = 127 N (F_max)

To find: Coefficient of static friction

The “coefficient of static friction” between a chair and the floor is defined as the ration of maximum force to the normal force acting on the chair

μ_s= $F_{max}/F_{n}$

The F_n is equal to the weight multiplied by its gravity

∴ $F_{n}$ =mg

Thus the coefficient of static friction changes as

μ_s= $F_{max}/mg$

μ_{s} = $=165N/((25kg)\times(9.80 m/s^2 ) )$

= 0.67

3 0

3 years ago

If each of the charges is increased by two times and the distance between them is also increased by two times, the electromagnet

PIT_PIT [208]

Answer: The force does not change.

Explanation:

The force between two charges q₁ and q₂ is:

F = k*(q₁*q₂)/r^2

where:

k is a constant.

r is the distance between the charges.

Now, if we increase the charge of each particle two times, then the new charges will be: 2*q₁ and 2*q₂.

If we also increase the distance between the charges two times, the new distance will be 2*r

Then the new force between them is:

F = k*(2*q₁*2*q₂)/(2*r)^2 = k*(4*q₁*q₂)/(4*r^2) = (4/4)*k*(q₁*q₂)/r^2 = k*(q₁*q₂)/r^2

This is exactly the same as we had at the beginning, then we can conclude that if we increase each of the charges two times and the distance between the charges two times, the force between the charges does not change.

8 0

3 years ago

What is physics and important of physics

Arte-miy333 [17]

Answer:

Physics is the answer to all the forces in nature and it is important to know all of that in a day to day life to succeed in life

Explanation:

4 0

3 years ago

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