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valina [46]
3 years ago
11

Positive charge is distributed uniformly throughout a non-conducting sphere. The highest electric potential occurs: A. Far from

the sphere B. Just outside the surface C. At the surface D. Halfway between the center and surface E. At the center
Physics
1 answer:
Ilya [14]3 years ago
6 0

Answer:E

Explanation:

At the center

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Position yourself with a clear line-of-sight so you can search the traffic environment about __________ seconds ahead.
maxonik [38]

Answer:

30 seconds ahead.

Explanation:

Its about 90% of the decisions which are made by someone while driving based on what they see. The vision must be good to make the decision accurately and timely before reacting to the environment.

Line of sight is define as the view which you can see without any obstruction. The best condition of proper line of sight is that someone is needed to see in all direction front, behind, left and right.

To establish the clean line of sight the person should search the traffic environment about 30 seconds ahead.

7 0
4 years ago
Read 2 more answers
The air in a room has a mass of 50kg and a specific heat of 1,000 J/(kg.C). What is the change in the thermal energy of the air
dem82 [27]
<span>We know the change of thermal energy is proportional to the specific heat of the matter, its mass and the change of temperatures, according the following formula: Q = C*m*(Tf-Ti) where: Q is the heat added or removed (in joules, J) C is the specific heat of the matter, in our case air 1,000 J/(Kg*C) m is the mass, in our case 50 kg of air Tf is final temperature Ti is initial temperature then: Q = 1000 * 50 * (30-20) = 500,000 Joules That means that 500,000 joules need to be added to increase temperature of the room 10 ÂşC (from 20 to 30 ÂşC)</span>
5 0
4 years ago
Two identical charged pith balls are brought together to touch each other. They are then
sergejj [24]

Answer:

-17.5 nC

Explanation:

charge A = -30 nC

charge B = -5 nC

After adding them it would be the average of the two charges because of the getting same voltage difference. so

c = (-30+(-5)) / 2 nC

c= -17.5 nC

answer is -17.5 nC

4 0
3 years ago
A ski jumper travels down a slope and leaves
Serhud [2]

Question seems to be missing. Found it on google:

a) How long is the ski jumper airborne?

b) Where does the ski jumper land on the incline?

a) 4.15 s

We start by noticing that:

- The horizontal motion of the skier is a uniform motion, with constant velocity

v_x = 28 m/s

and the distance covered along the horizontal direction in a time t is

d_x = v_x t

- The vertical motion of the skier is a uniformly accelerated motion, with initial velocity u_y = 0 and constant acceleration g=9.8 m/s^2 (where we take the downward direction as positive direction). Therefore, the vertical distance covered in a time t is

d_y = \frac{1}{2}gt^2

The time t at which the skier lands is the time at which the skier reaches the incline, whose slope is

\theta = 36^{\circ} below the horizontal

This happens when:

tan \theta = \frac{d_y}{d_x}

Substituting and solving for t, we find:

tan \theta = \frac{\frac{1}{2}gt^2}{v_x t}= \frac{gt}{2v_x}\\t = \frac{2v_x}{g}tan \theta = \frac{2(28)}{9.8} tan 36^{\circ} =4.15 s

b) 143.6 m

Here we want to find the distance covered along the slope of the incline, so we need to find the horizontal and vertical components of the displacement first:

d_x = v_x t = (28)(4.15)=116.2 m

d_y = \frac{1}{2}gt^2 = \frac{1}{2}(9.8)(4.15)^2=84.4 m

The distance covered along the slope is just the magnitude of the resultant displacement, so we can use Pythagorean's theorem:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(116.2)^+(84.4)^2}=143.6 m

7 0
3 years ago
A tire is filled with air at 10 ∘C to a gauge pressure of 250 kPa. Part A If the tire reaches a temperature of 45 ∘C, what fract
Alex_Xolod [135]

Answer:

The air fraction to be removed is 0.11

Given:

Initial temperature, T = 10^{\circ} = 283 K

Pressure, P = 250 kPa

Finally its temperature increases, T' = 45^{\circ} = 318 K

Solution:

Using the ideal gas equation:

PV = mRT

where

P = Pressure

V = Volume

m = no. of moles of gas

R = Rydberg's Constant

T = Temperature

Now,

Considering the eqn at constant volume and pressure, we get:

mT = m'T'

Thus

\frac{m}{m'} = \frac{T'}{T}                      (1)

Now, the fraction of the air to be removed for the maintenance of pressure at 250 kPa:

y = \frac{m - m'}{m} = 1 - \frac{m'}{m}

From eqn (1):

y = 1 - \frac{T}{T'}

y = 1 - \frac{283}{318} = 0.11

6 0
4 years ago
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