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3 years ago

What is it called when a neutral object is given a temporary charge?

Physics

1 answer:

fgiga [73]3 years ago

4 0

It's called induced charge separation I believe.

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A river flows due east with a speed of 3.00 m/s relative to earth. The river is 80.0 m wide. A woman starts at the southern bank

babymother [125]

Answer:

48 m

Explanation:

As she travels at the rate of 5m/s due north, the amount of time it would take for her to cross the 80m wide river would be

t = 80 / 5 = 16 seconds

This is also the time it takes for the river to push her to the east side at the rate of 3m/s. So after 16 seconds, she would reach the opposite point at a horizontal distance from her starting of

s = 16*3 = 48 m

4 0

4 years ago

E) Thermal energy is released during

vichka [17]

Answer:

e) True, f) False

Explanation:

e) Let consider a close system, that is, a system with no mass interactions with surroundings. Then, we get the following expression by the First Law of Thermodynamics:

$Q_{net,in} - W_{net, out} = \Delta U$ (1)

Where:

$Q_{net, in}$ - Net input heat, measured in joules.

$W_{net, out}$ - Net output work, measured in joules.

$\Delta U$ - Change in thermal energy, measured in joules.

Please notice that work comprises all kind of work (i.e. mechanical, electric, magnetic), whereas heat comprises all heat interactions including chemical and radioactive phenomena.

If thermal energy is released, then $\Delta U < 0$ , which is caused by three scenarios:

(i) $Q_{net,in} < 0$ , $W_{net, out} < 0$ , $|Q_{net,in}|>|W_{net,out}|$

(ii) $Q_{net, in} > 0$ , $W_{net,out} > 0$ , $|Q_{net,in}|$

(iii) $Q_{net,in}< 0$ , $W_{net, out}>0$

In the case $Q_{net,in} > 0$ , $W_{net, out}$ , the thermal energy of the system is increased. Therefore, thermal energy is released during some energy conversions. Answer: True

f) A liquid solidifies when temperature goes below point of fusion, meaning a realease of heat with no work interactions. That is:

$Q_{net, in} = \Delta U$ , $Q_{net, in} < 0$ (2)

If $Q_{net, in} < 0$ , then $\Delta U < 0$ . Then, if a liquid absorbs heat energy, then thermal energy is increase and the liquid does not solidifies. Answer: False.

7 0

3 years ago

Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

mihalych1998 [28]

Answer:

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

7 0

3 years ago

In order for a ship to stay afloat, its buoyant force must be

Kruka [31]

The buoyant force must be greater than water.

4 0

4 years ago

Why should the substage condenser not be included in computing the magnification?

Leni [432]

The reason as to why the substage condenser does not need to be included in computing the magnification and the only component needed is the ocular lens and the objective lenses is because the condenser is only responsible for gathering light and it does not contribute with the magnification of the object under the microscope.

5 0

3 years ago

Read 2 more answers