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4 years ago

What is one common way that a charge can accumulate on a object?

Physics

1 answer:

Damm [24]4 years ago

5 0

Answer:

Friction

Explanation:

Friction is rubbing two objects together. When this happens electrons from one object goes to the other. When one object loses electrons, it will have more protons, so it will be positively charged. When one object gains electrons it becomes negatively charged.

This is what happens when you rub a rubber balloon on your hair, or even on the wall. Because one will be negatively charged and the other positive, the objects then attract each other and they stick together, like what happens when you rub the balloon on your hair, you hair will stick to the balloon, or vice-versa.

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When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner

Paul [167]

Answer:

1353.38 Watt

Explanation:

T₁ = Initial temperature of the house = 35°C

T₂ = Final temperature of the house = 20°C

Δt = Time taken to cool the house = 38 min = 38×60 = 2280 s

m = mass of air in the house = 800 kg

Cv = Specific heat at constant volume = 0.72 kJ/kgK

Cp = Specific heat at constant pressure = 1.0 kJ/kgK

Heat removed

q = mCvΔT

⇒q = 800×720×(35-20)

⇒q = 8640000 J

Average rate of hear removal

$Q=\frac{q}{\Delta t}\\\Rightarrow Q=\frac{8640000}{2280}\\\Rightarrow Q=3789.47\ W$

$COP=\frac{Q}{W}\\\Rightarrow W=\frac{Q}{COP}\\\Rightarrow W=\frac{3789.47}{2.8}\\\Rightarrow W=1353.38\ W$

∴ Power drawn by the air conditioner is 1353.38 Watt

6 0

3 years ago

Read 2 more answers

The temperature light will come on in your vehicle when

nikitadnepr [17]

If your engine is not at the right temprature, usually when it is too hot.

5 0

3 years ago

The legislative branch has the power to (3 points)

maw [93]

Hi there!

The Legislative Branch of government includes the House of Representatives and the Senate. This branch makes laws. Its powers include passing laws, originating spending bills (House), impeaching government officials (Senate), and approving treaties (Senate).

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!

3 0

3 years ago

Read 2 more answers

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T

zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f) The two trains avoid collision if the acceleration of the freight train is at least $0.35 m/s^2$

Explanation:

We can describe the position of the passenger train at time t with the equation

$x_p(t)=u_p t + \frac{1}{2}at^2$

where

$u_p = 25.0 m/s$ is the initial velocity of the passenger train

$a=-0.100 m/s^2$ is the deceleration of the train

On the other hand, the position of the freight train is given by

$x_f(t)=x_0 + v_f t$

where

$x_0=200 m$ is the initial position of the freight train

$v_f = 15.0 m/s$ is the constant velocity of the train

The collision occurs if the two trains meet, so

$x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0$

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

$x_f(t)=x_0 +v_ft$

And substituting t = 22.5 s, we find:

$x_f(22.5)=200+(15)(22.5)=537.5 m$

We can verify that the passenger train is at the same position at the time of the collision:

$x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m$

So, the two trains collide at x = 537.5 m.

In the graph in attachment, the position-time graph of each train is represented. We have:

The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

In order to avoid the collision, the freight train must have a initial position of

$x_0'$

such that the two trains never meet.

We said that the two trains meet if:

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t$

Re-arranging,

$\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0$

Substituting the values for the acceleration and the velocity,

$0.05t^2-10t+x_0'=0$

The solution of this equation is given by the formula

$t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}$

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

$10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m$

Therefore, the freight train must have a head start of 500 m.

In this case, we want to find the acceleration $a'$ of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t$

Re-arranging

$\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0$

Substituting,

$-0.5at^2-10t+200=0$

The solution to this equation is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}$

Again, the two trains never meet if the discriminant is negative, so

$10^2-4\cdot (-0.5a') \cdot (200)$

So, the deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

$x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2$

where $a_f$ is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2$

Re-arranging,

$\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0$

And the solution is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}$

Again, the two trains avoid collision if the discriminant is negative, so

$10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2$

Learn more about accelerated motion:

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brainly.com/question/2506873

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#LearnwithBrainly

8 0

3 years ago

Describe the interior structure of Earth and Earth’s crust as divided into tectonic plates riding on top of the slow moving curr

Dafna1 [17]

Hello There!

The mantle is the layer located directly under the sima. It is the largest layer of the Earth, 1800 miles thick. The mantle is composed of very hot, dense rock. This layer of rock even flows like asphalt under a heavy weight. This flow is due to great temperature differences from the bottom to the top of the mantle. The movement of the mantle is the reason that the plates of the Earth move! The temperature of the mantle varies from 1600 degrees Fahrenheit at the top to about 4000 degrees Fahrenheit near the bottom!

Hope This helped!

HyperZ ^_^

7 0

3 years ago