What could you do to increase the electric potential energy between two

Physics

2 answers:

slamgirl [31]3 years ago

7 0

Answer: The correct answer would be, Reduce the distance by a factor of 16.

Explanation: A P E X.

SSSSS [86.1K]3 years ago

3 0

Answer:

Increase the charge of one particle by a factor of 16

Explanation:

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an open cart is moving along a straight frictionless horizontal track. When rain starts falling vertically into the cart, what h

fgiga [73]

Answer:

The speed of the cart decreases

Explanation:

The track is frictionless, so we can consider the cart as isolated system (no external forces acting on it): therefore, the momentum of the cart is conserved

$p=mv=const.$

where m is the mass and v the speed of the cart.

When it starts raining, some water falls into the cart: this means that the mass of the cart, m, increases. But we said that the momentum, p, must remain constant: in order for that to happen, this means that the speed of the cart, v, must decrease (because the relationship $mv=const.$ is an inverse relationship, so if m increases, v must decrease).

6 0

3 years ago

In anticipation of a long 10 upgrade, a bus driver accelerates at a constant rate of 5 ft/s2 while still on a level section of a

fgiga [73]

Answer:

$S = 20903.4$ ft

Explanation:

First we will determine the acceleration of the bus while it is moving upward

The equilibrium equation would be

$F - W sin\theta = ma'\\ma - W sin\theta = ma'\\m = \frac{W}{g} \\\frac{W}{g}*a - W sin\theta = \frac{W}{g} * a'\\\frac{a}{g} - sin\theta = \frac{a'}{g}\\\frac{x}{y} \frac{5}{32.2} - sin 10 = \frac{a'}{32.2} \\\a' = -0.59$

Let the displacement be $S_0$

As per newton's third law of motion

$v^2 -u^2 = 2as\\u = 80 \frac{mi}{h} = 117.33\frac{ft}{s}\\v = 50 \frac{mi}{h} = 73.33\frac{ft}{s}\\73.33 ^ 2 - 117.33^2 = 2 * (-0.59) * (S-S_0)\\\\S_0 = 0\\S = 20903.4$