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qaws [65]
3 years ago
11

A person drives for 1.00 h at 10.0 km/h and 2.00 h at 20.0 km/h. What is their average speed, in SI units?

Physics
1 answer:
Veronika [31]3 years ago
6 0
I believe it is b if i’m correct
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The dome of a Van de Graaff generator receives a charge of 0.00011 C. The radius of the dome is 5.2 m. Find the strength of the
wel

Answer:

Answer:

Explanation:

Given that

K=8.98755×10^9Nm²/C²

Q=0.00011C

Radius of the sphere = 5.2m

g=9.8m/s²

1. The electric field inside a conductor is zero

εΦ=qenc

εEA=qenc

net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero

This surface encloses no charge, and thus qenc=0. Gauss’ law.

Since it is inside the conductor

E=0N/C

2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as

F=kq1/r²

F=kQ/r²

F=8.98755E9×0.00011/5.2²

F=36561.78N/C

The electric field at the surface of the conductor is 36561N/C

Since the charge is positive the it is outward field

3. Given that a test charge is at 12.6m away,

Then Electric field is given as,

E=kQ/r²

E=8.98755E9 ×0.00011/12.6²

E=6227.34N/C

5 0
3 years ago
A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.05 m/s. If the roof is pitched at
vekshin1

1) Time in the air: 0.78 s

The motion of the ball is a projectile motion, which consists of two independent motions:

- A horizontal motion with constant horizontal velocity

- A vertical motion with constant downward acceleration of

g=-9.8 m/s^2 (acceleration of gravity)

The initial vertical velocity is

u_y = u sin \theta = (3.05)(sin(-40^{\circ}))=-1.96 m/s

where the negative sign means the direction is downward.

The vertical position of the ball is given by

y(t) = h + u_y t + \frac{1}{2}gt^2

where

h = 4.50 m is the initial heigth of the ball when it starts falling down

The ball reaches the ground when y = 0, so we have:

0 = 4.50 -1.96t-4.9t^2

This is a second-order equation; solving for t, we get

t = -1.18 s

t = 0.78 s

We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.

2) Horizontal distance: 1.83 m

For this second part of the problem, we just have to consider the horizontal motion of the ball.

As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by

v_x = u cos \theta = (3.05)(cos (-40.0^{\circ}))=2.34 m/s

where u = 3.05 m/s is the initial speed and \theta the angle of projection.

For a uniform motion, we can use the following relationship between distance covered and velocity:

d=v_x t

and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground:

d=(2.34)(0.78)=1.83 m

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