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nadezda [96]
2 years ago
10

A 0.40-kg cart with charge 4.0 x 10-5 C starts at rest on a horizontal frictionless surface 0.50 m from a fixed object with char

ge 2.0 x 10-4 C. When the cart is released, it moves away from the fixed object. (a) How fast is the cart moving when very far (infinity) from the fixed charge
Physics
1 answer:
maw [93]2 years ago
4 0

Answer:

26.82m/s

Explanation:

Given

Mass = m= 0.4kg

Initial Velocity = u = 0

Charge = 4.0E-5C

Distance= d = 0.5m

Object Charge = 2E-4C

First, we'll calculate the initial energy (E)

E = Potential Energy

PE = kQq / d

Where k = coulomb constant = 8.99E9Nm²/C²

Energy is then calculated by;

PE = 8.99E9 * 4E-5 * 2E-4 / 0.5

PE = 143.84J

Energy = Potential Energy = Kinetic Energy

K.E = ½mv² = 143.84J

½mv² = ½ * 0.40 * v² = 143.85

0.2v² = 143.85

v² = 143.85/0.2

v² = 719.25

v = √719.25

v = 26.81883666380777

v = 26.82m/s

Hence, the object is 26.82m/s fast when the cart moving is very far (infinity) from the fixed charge

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2 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

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Suppose you want to make a scale model of a hydrogen atom. You choose, for the nucleus, a small ball bearing with a radius of 1.
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Answer:

A)  x _electron = 0.66 10² m , B)   x _Eart = 1.13 10² m , C)  d_sphere = 1.37 10⁻² mm

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                x_Earth = 1.5 10¹¹ / 6.96 10⁸  1.5

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