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1 year ago

A set of electric trains are powered by a 9 V battery. What is the resistance of the trains if they draw 4.0 A of current?

Physics

1 answer:

Dimas [21]1 year ago

8 0

Answer:

2.25 Ω

Explanation:

Standard equation

V = IR re-arrange to

V /I = R then sub in the values given

9 / 4 = 2.25 Ω

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The figure shows an uncharged conducting cubic shell and a positive point charge Q outside the shell. There are two lettered poi

Nataliya [291]

If the point charge is placed at point $E_W$ is equal to $E_X$ .

<h3>Electric field inside a conducting cubic shell</h3>

The electric field inside a conducting cubic shell is zero. When a point charge is placed outside the shell, the electric field outside the shell is greater than the electric field inside the shell which is zero.

We can deduce the following;

Electric field inside the shell is zero, thus $E_W = E_X =0$
Electric field outside the shell is greater than electric field inside the shell

Thus, the only correct statement is, If the point charge is placed at point $E_W$ is equal to $E_X$ .

Learn more about electric field inside a shell here: brainly.com/question/539905

4 0

2 years ago

Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi

nirvana33 [79]

Answer:

a) $F = 882.63\,N$ , b) $\dot W= 4413.15\,W$ , c) $\eta = 15.216\,\%$ .

Explanation:

a) Let assume that car travel on a horizontal surface. The equations of equilibrium of the car are:

$\Sigma F_{x} = F - \mu_{r}\cdot N = 0$

$\Sigma F_{y} = N - m\cdot g = 0$

After some algebraic handling, the following expression for the propulsion force is constructed:

$F = \mu_{r}\cdot m \cdot g$

$F = (0.06)\cdot (1500\,kg)\cdot (9.807\,\frac{m}{s^{2}} )$

$F = 882.63\,N$

b) The power require to move the car at a speed of 5 meters per second is:

$\dot W = F\cdot v$

$\dot W = (882.63\,N)\cdot (5\,\frac{m}{s} )$

$\dot W= 4413.15\,W$

c) The efficiency of the car is:

$\eta = \frac{(882.63\,N)\cdot (15\,mi)\cdot (\frac{1609\,m}{1\,mi} )}{(1.4\times 10^{8}\,J)} \times 100\,\%$

$\eta = 15.216\,\%$

3 0

3 years ago

V125BC [204]

Answer:

All the planets orbit the Sun in nearly the same plane.

7 0

3 years ago

A 200g air-track glider is attached to a spring. The glider is pushed in 10cm and released. A student with a stopwatch finds tha

cricket20 [7]

The spring constant will be k= 5.5N/m for a 200g air track glider attached to a spring.

<h3>What is spring constant?</h3>

The spring constant, k, is a measure of the stiffness of the spring. It is different for different springs and materials.

Calculation for What is the spring constant

First step is to calculate the time period

T = 12 second/10

T = 1.2 second

Now let calculate the spring constant using this formula

$k=\dfrac{4\pi ^2m}{T^2}$

Where,

m=0.2kg

T=1.2second

k represent spring constant=?

Let plug in the formula

$k=\dfrac{4 \pi \times 0.2}{(1.2)^2}$

$k=\dfrac{39.48\times 0.2}{1.44}$

$k=\dfrac{7.90}{1.44}$

k=5.48 N/m

k=5.5 N/m ( Approximately)

Therefore the spring constant will be 5.5 N/m

To know more about spring constant follow

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4 0

2 years ago

Find the weight of an object of mass 5 kg

Elis [28]

Answer:

weight on earth is mg

which is 5*9.8

49 Newton

weight on moon is 1/6 th of weight on earth

1/6*49

8.166 Newton..

3 0

3 years ago

Read 2 more answers