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2 years ago

Alculate the potential difference if 20J of energy are transferred by 8C of charge.

Physics

1 answer:

sveta [45]2 years ago

6 0

Answer:

V = 2.5 J/C

Explanation:

Given:

Energy = E = 20 J

Charge = Q = 8 C

Required:

Potential Difference = V = ?

Formula:

V = $\frac{E}{Q}$

Solution:

V = 20/8

V = 2.5 J/C

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Two astronauts of identical mass are connected by a taut cable of negligible mass, as shown in the figure above, and are initial

PolarNik [594]

Answer:

The right answer is "The center of mass doesn't move".

Explanation:

It generates a voltage throughout the cable while the astronaut falls on either the wire. At other ends of the spectrum or cable, the tension will be similar. As such, with both astronauts, there would be the same energy, although throughout the opposite way.
Thus, the net force seems to be essentially negative on the machine. And therefore the mass center stays stationary.

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2 years ago

The atmosphere pressure can support mercury in a tube, which the upper end is closed, up to 0.76 meter. If the mercury is replac

Leni [432]

Answer:

Maximum height the atmosphere pressure can support the

water=10.336 m

Explanation:

We know that ,

$Pressure = h\cdot\rho\cdot g$

Case 1 - Mercury in the tube

$Density\ of\ mercury =\rho_1\\and\ height\ attained\ for\ mercury\ column = h_1$

Case 2 - Water in the tube

$Density\ of\ water =\rho_2\\and\ height\ attained\ for\ water\ column = h_2$

Since atmospheric pressure is same

. $P=h_1\cdot\rho_1\cdot g = h_2\cdot\rho_2\cdot g$

or, $h_2=\frac{h_1\rho_1}{\rho_2}$

$Given\ h_1= 0.76\ m,\rho_1=13.6\cdot\rho_2$

∴ $h_2=0.76\cdot13.6=10.336\ m$

Hence height of the water column =10.336 m

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2 years ago

A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .

rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

$-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0$

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

$v=u+at$

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

$v = 15.5 +(-9.8)(2.64)=-10.4 m/s$

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

$-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0$

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

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2 years ago

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A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical accel

Mazyrski [523]

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What are two parts that make up a vector

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Answer:

Vectors have both magnitude and direction.

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2 years ago