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lidiya [134]
3 years ago
12

During circular motion, the force that is perpendicular to the velocity and toward the center of the circle is the ... ?

Physics
1 answer:
Ivanshal [37]3 years ago
5 0
That's the "centripetal" force.  It produces the centripetal acceleration
that pulls the object away from a straight path into a bent path.
You might be interested in
(A) The figure shows the setup which is used to observe an image formed wen a lighted candle is kept in front of a bi convex len
GREYUIT [131]

Answer:

<h3>1. In the case of candle flame , the object is placed beyond c , that means the image is formed or focused between c and f .in second case the object or sun is at infinity , so the image will be </h3><h3>formed at focus.this means the distance between image and lens has decreased in the second </h3><h3>case. either we have to move the screen towards the lens or the lens towards the screen.</h3>

Explanation:

I have posted the drawing.

in the second image

it is project c and characteristics

hope it helps u :)

6 0
2 years ago
A 6 kg block initially at rest is pulled to East along a horizontal surface with coefficient of kinetic friction μk=0.15 by a co
ozzi

Answer:

1.8 m/s

Explanation:

Draw a free body diagram of the block.  There are four forces:

Normal force Fn up.

Weight force mg down.

Applied force F to the east.

Friction force Fn μ to the west.

Sum the forces in the y direction:

∑F = ma

Fn − mg = 0

Fn = mg

Sum the forces in the x direction:

F − Fn μ = ma

F − mg μ = ma

a = (F − mg μ) / m

a = (12 N − 6 kg × 9.8 m/s² × 0.15) / 6 kg

a = 0.53 m/s²

Given:

Δx = 3 m

v₀ = 0 m/s

a = 0.53 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (0.53 m/s²) (3 m)

v = 1.8 m/s

8 0
3 years ago
1. Calculate the mechanical advantage of a machine that has an input force of 13 N and
Helga [31]

Answer:

52 N output

Explanation: 65-13=52

6 0
3 years ago
Read 2 more answers
1. A truck with a mass of 8, 000 kg is traveling at 26.8 m/s when it hits the brakes. A.)What is the momentum of the truck befor
NikAS [45]

Answer:

1. A.) The moment of the truck before it hits the brakes is 214,400 kg·m/s

B.) The force it takes to stop the truck is approximately 17,290.4 N

Explanation:

1. A.) The given parameters are;

The mass of the truck, m = 8,000 kg

The velocity of the truck when it hits the brakes, u = 26.8 m/s

Momentum = Mass × Velocity

The moment of the truck = The mass of the truck × The velocity of the truck

Therefore;

The moment of the truck before it hits the brakes = 8,000 kg × 26.8 m/s = 214,400 kg·m/s

B.) The amount of momentum lost when the truck comes to a stop = The initial momentum of the truck

The time it takes the truck to come to a complete stop, t = 12.4 s

The deceleration, "a" of the truck is given by the following kinematic equation of motion

v = u - a·t

Where;

v = The final velocity of the truck = 0 m/s

u = The initial velocity = 26.8 m/s

a = the deceleration of the truck

t = The time of deceleration of the truck = 12.4 s

Substituting the known values gives;

0 = 26.8 - a × 12.4

Therefore;

26.8 = a × 12.4

a = 26.8/12.4 ≈ 2.1613

The deceleration (negative acceleration) of the truck, a ≈ 2.1613 m/s²

Force = Mass × Acceleration

The force required to stop the truck = The mass pf the truck × The deceleration (negative acceleration) given to the truck

∴ The force it takes to stop the truck = 8,000 kg × 2.1613 m/s² ≈ 17,290.4 N.

8 0
3 years ago
A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial
Nataly_w [17]

Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

qvB=m\frac{v^2}{r}

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

r=\frac{mv}{qB}

For the ion of oxygen-16, we have:

m_A=2.66\cdot 10^{-26}kg

q_A = 1.6\cdot 10^{-19}C (it is singly charged)

v_A=2.90\cdot 10^6 m/s

B_A=1.30 T

So the radius of its orbit is

r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m

For the ion of oxygen-18, we have:

m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg

q_B = 1.6\cdot 10^{-19}C (it is singly charged)

v_B=2.90\cdot 10^6 m/s

B_B=1.30 T

So the radius of its orbit is

r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

d=2(r_B-r_A)=2(0.417-0.371)=0.092 m

3 0
3 years ago
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