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4 years ago

Calculate the electric potential energy in a capacitor that stores 4.0 10-10 C of charge at 250.0 V

Physics

2 answers:

Arada [10]4 years ago

7 0

Q = C.v
v = Q/C
v = 4 × 10^(-10)/250
= 4 × 10^(-10)/2.5 × 10^2
= 1.6 × 10^(-12) volt

Bogdan [553]4 years ago

5 0

Answer:

1.6 x 10^(-12)F

Explanation:

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. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the b

Arte-miy333 [17]

Answer:

elative magnitude of the two forces is the same and they are applied in a constant direction.

Explanation:

Newton's second law states that the sum of the forces is equal to the mass times the acceleration

∑ F = m a

in this case there are two forces on the x axis

F_applied - fr = 0

since they indicate that the velocity is constant, consequently

F_applied = fr

the relative magnitude of the two forces is the same and they are applied in a constant direction.

3 0

3 years ago

Find the velocity, acceleration, and speed of a particle with the given position function. r(t) = t i + t2 j + 3 k v(t) = a(t) =

stepladder [879]

Answer:

Velocity

v= i + 2 j

Acceleration

a= 2 j

Explanation:

Given that

r(t) = t i + t² j + 3 k

We know that

Velocity v given as

v= dr/dt

dr/dt= i + 2 t j +0

v= i + 2 t j

At t= 1

v= i + 2 x 1 j

v= i + 2 j

Acceleration ,a

a= dv/dt

v= i + 2 t j

dv/dt= 2 j

a= 2 j

6 0

3 years ago

Work and Energy

Sergeeva-Olga [200]

a) $W_1 = 2332 J$ , $W_2= 2332 J$

The work done by the student in each trial is equal to the gravitational potential energy gained by the student:

$W=mg\Delta h$

where

m = 68 kg is the mass of the student

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h$ is the gain in height of the student

For the first student, $\Delta h = 3.5 m$ , so the work done is

$W_1 = (68)(9.8)(3.5)=2332 J$

The second student runs up to the same height (3.5 m), so the work done by the second student is the same:

$W_2 = (68)(9.8)(3.5)=2332 J$

2) $P_1 = 204.6 W$ , $P_2 = 274.4 W$

The power exerted by each student is given by

$P=\frac{W}{t}$

where

W is the work done

t is the time taken

For the first student, $W_1 = 2332 J$ and $t=11.4 s$ , so the power exerted is

$P_1 = \frac{W_1}{t_1}=\frac{2332 J}{11.4 s}=204.6 W$

For the second student, $W_2 = 2332 J$ and $t=8.5 s$ , so the power exerted is

$P_2 = \frac{W_2}{t_2}=\frac{2332 J}{8.5 s}=274.4 W$

5 0

3 years ago

In a physics laboratory experiment, a coil with 250 turns enclosing an area of 10.5 cm2 is rotated during the time interval 3.10

olchik [2.2K]

I guess the problem is asking for the induced emf in the coil.

Faraday-Neumann-Lenz states that the induced emf in a coil is given by:
$\epsilon = -N \frac{\Delta \Phi}{\Delta t}$
where
N is the number of turns in the coil
$\Delta \Phi$ is the variation of magnetic flux through the coil
$\Delta t$ is the time interval

The coil is initially perpendicular to the Earth's magnetic field, so the initial flux through it is given by the product between the magnetic field strength and the area of the coil:
$\Phi_i = BA=(5.30 \cdot 10^{-5}T)(10.5 \cdot 10^{-4} m^2)=5.57 \cdot 10^{-8} Wb$
At the end of the time interval, the coil is parallel to the field, so the final flux is zero:
$\Phi_f = 0$

Therefore, we can calculate now the induced emf by using the first formula:
$\epsilon = -N \frac{\Delta \Phi}{\Delta t}=- (250) \frac{5.57 \cdot 10^{-8} Wb - 0}{3.10 \cdot 10^{-2} s} = -4.5 \cdot 10^{-4} V$

7 0

4 years ago

Which of these materials is permeable?

kogti [31]

Correct answer is letter B. sandstone

8 0

3 years ago