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3 years ago

Ill mark as brainliest.

Physics

1 answer:

rewona [7]3 years ago

8 0

Answer: 75V

Explanation:

Given that,

total resistance (Rtotal) = 150Ω

Current (I) = 0.5A

Change in electric potential (V) = ?

Recall that potential difference is the product of amount of current and the amount of resistance in the circuit. And its unit is volts.

So, apply the formula V = I x Rtotal

V = 0.5A x 150Ω

V = 75V

Thus, the change in electric potential across the circuit is 75 Volts

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A) $2.4\cdot 10^{-6} J$

The energy stored in a capacitor is given by:

$E=\frac{1}{2}\frac{Q^2}{C}$

where

Q is the charge stored

C is the capacitance

The capacitance of a parallel-plate capacitor is

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=6.0 cm \cdot 6.0 cm=36.0 cm^2=36\cdot 10^{-4} m^2$ is the area of each plate

$d=1.5 mm=0.0015 m$ is the distance between the plates

Substituting,

$C=\frac{(8.85\cdot 10^{-12} F/m)(36\cdot 10^{-4} m^2)}{0.0015 m}=2.1\cdot 10^{-11} F$

The charge stored on the capacitor is

$Q=10 nC=10\cdot 10^{-9}C$

So, the energy stored is

$E=\frac{1}{2}\frac{(10\cdot 10^{-9}C)^2}{2.1\cdot 10^{-11} F}=2.4\cdot 10^{-6}J$

B) $2.6\cdot 10^{-6}J$

This time, the separation between the plates is

d = 1.7 mm = 0.0017 m

So, the new capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(36\cdot 10^{-4} m^2)}{0.0017 m}=1.9\cdot 10^{-11} F$

And so, the new energy stored is

$E=\frac{1}{2}\frac{(10\cdot 10^{-9}C)^2}{1.9\cdot 10^{-11} F}=2.6\cdot 10^{-6}J$

Energy must be conserved, so the difference between the initial energy of the capacitor and its final energy is just equal to the work done to increase the separation between the two plates from 1.5 mm to 1.7 mm (in fact, the two plates of the capacitor attract each other since they have opposite charge, so work must be done in order to increase their separation)

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3 years ago

True or false: Ultimate tensile strength increases as the thickness of a solid material sample increases.

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Answer:

False.

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How much work is done against gravity when lifting a 2-kg sack of groceries a distance of 2.5 meters?

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W = F x d/x = (m x Ag) x h, therefore, mass (2kg x 9.8) x 2.5m = 49J

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The two regions of the electromagnetic spectrum where the Earth's atmosphere is transparent (radiation can get in) are visible l

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The two regions of the electromagnetic spectrum where the Earth's atmosphere is transparent are visible light and some radio waves.

<h3>What is electromagnetic spectrum?</h3>

The electromagnetic spectrum consists of the range of all types of EM radiation. Radiation is packet of energy that travels and spreads out. It has radio waves, infrared waves, visible light waves, ultraviolet rays and microwaves.

They possess high energy even to penetrate deep within the material. The electromagnetic spectrum has longer wavelength and less frequency.

Thus, the two regions of the electromagnetic spectrum where the Earth's atmosphere is transparent are visible light and some radio waves.

Learn more about electromagnetic spectrum.

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3 years ago

Can someone please help?? I don’t understand this material!!!

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Answer:

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At the beginning,

$p_i=0$

after throwing the ball, the total momentum is the sum of the momentum of the person and of the ball:

$p_f=p_p + p_b$

Since momentum is conserved,

$p_i = p_f\\0=p_p+p_b$

$p_p = -p_b$

Therefore, the person has equal momentum (in magnitude) but opposite direction to the ball, so the person rolls backward.

However, if the person hold to the ball, then they will have same momentum (moving in the same direction). In order to conserve the total momentum (which was zero at the beginning), the only possible solution is that

$p_p=p_b=0$

which means that both the person and the ball will remain at rest. This is because there are no external forces acting on the system, so the system cannot move.

The change in momentum of an object is given by

$\Delta p=m(v-u)$

where

m is the mass of the object

v is its final velocity

u is the initial velocity

For the clay ball in this problem, we have:

m = 50 g = 0.050 kg

v = 0 m/s (it sticks on the wall)

u = 1 m/s

So its change in momentum is

$\Delta p_c=(0.050)(0-1)=-0.050 kg m/s$

For the superball, we have:

m = 50 g = 0.050 kg

v = -0.8 m/s (it bounces back)

u = 1 m/s

So its change in momentum is

$\Delta p_s = (0.050)(-0.8-1)=-0.09 kg m/s$

So, the superball has a greater change in momentum (in magnitude).

3a)

According to Newton's third law of motion:

"When an object A exerts a force (action force) on an object B, then object B exerts an equal and opposite force (reaction force) on object A".

Here, we have a Hummer and a Beetle colliding head-on: we can identify them as object A and object B. Therefore, according to Newton's third law:

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And according to the Law, the two forces are equal in magnitude: so, the two vehicles experience the same force of impact.

3b)

The change in momentum of each vehicle during the collision can be written as

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where

$\Delta p$ is the change in momentum

$F$ is the force experienced by the vehicle

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in part 3a), we said that the two vehicles experience the same force in the collision.

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3c)

According to Newton's second law of motion, the acceleration of an object is given by:

$a=\frac{F}{m}$

where

F is the force experienced by the object

m is its mass

a is its acceleration

In part 3a), we stated that the force experienced by the Beetle and the Hummer is the same. However, the mass of the Beetle is smaller than the mass of the Hummer: from the equation we see that the acceleration is inversely proportional to the mass, therefore the Beetle will experience a greater acceleration.

4a)

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$\Delta p$ is the change in momentum

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F is the force experienced by the climber

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4c)

This technique is used to exploit the "push" given by the second car of the train to the first car when the brakes are applied.

At first, the engine is started, and the first car starts accelerating, pulling the second car (and the following cars). Then, the brakes are applied on the first car: however, the second car keeps moving by inertia, so then it gives a push forward on the first car. Then, this action is repeated several times, so that this push exerted by the second car is exploited several times.

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