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3 years ago

Questions E6 a& b and E7 a&b?

Physics

1 answer:

erica [24]3 years ago

8 0

Explanation:

6a) Work = force × distance

W = Fd

W = (60 N) (10 m)

W = 600 J

6b) Change in energy = work

ΔKE = 600 J

7a) Kinetic energy is half the mass times the square of the velocity.

KE = ½ mv²

KE = ½ (0.4 kg) (25 m/s)²

KE = 125 J

7b) Work = change in energy. When the ball is stopped, it has zero kinetic energy.

W = ΔKE

W = 0 J − 125 J

W = -125 J

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No. We don't know of the existence of another universe besides
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4 years ago

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Which of the following describes a displacement vs. time graph that looks like this?

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constant non zero acceleration

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3 years ago

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

r = distance of corner of the square from it center

$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

$V = 0.815 V$

now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

$r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}$

$r_2 = 11.2 nm$

now potential is given as

$V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}$

$V = -0.576 - 0.257 = -0.833 V$

now final potential energy is given as

$U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J$

Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

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3 years ago

Q4. Consider the skier on a slope shown in the figure below. Her mass including equipment is 55.0 kg.

Shkiper50 [21]

Answer:

Part a)

When there is no friction then acceleration is

$a = 4.14 m/s^2$

Part b)

if there is friction force along the inclined then acceleration is

$a = 3.33 m/s^2$

Explanation:

Part a)

As we know that the skier is on inclined plane

So here if there is no friction then net force along the inclined plane is given as

$F = mg sin\theta$

now acceleration of the skier is given as

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

$a = g sin\theta$

$a = 9.81(sin25)$

$a = 4.14 m/s^2$

Part b)

if there is friction force along the inclined then net force along the inclined plane is given as

$F = mg sin\theta - F_f$

now acceleration of the skier is given as

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

$a = g sin\theta - \frac{F_f}{m}$

$a = 9.81(sin25) - \frac{45}{55}$

$a = 3.33 m/s^2$

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4 years ago

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victus00 [196]

A line that is falling towards the x axis represents an object that is negatively accelerating, or slowing down. When the line hits the x axis, the object has stopped moving. If the graph continues below the x axis, the object has changed direction and is moving backwards at increasing velocity.

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3 years ago