All categories

2 years ago

F. You get inside and you have to push the cart across a thick carpet. Describe

Physics

1 answer:

aev [14]2 years ago

5 0

Answer:

Forces Due to Friction and Newton's Third Law

The forces on the cart include the forward force the horse exerts on the cart and the backward force due to friction at the ground, acting on the wheels. At rest, or at constant velocity, these two are equal in size, because the acceleration of the cart is zero.

You might be interested in

Can an eagle reach exosphere through flying?

nignag [31]

Simply no.

And the problem wouldn't be only the lack of gravity, it would be also the lack of air.

Flying is an action not to far from swiming: something is moving trough some fluid. Of course that the fluid will incredibly matter, but the basics are the same.

When a eagle flies, it is pushing its wings against the air, and the air is holding that pressure against the wing, and, also for a bunch of anatomic reasons, the eagle is able to fly.

Without air and gravity, it will just keep floating around until one of the gravitational pulls wins the battle and attrack it to a painfull fate.

I hope I helped!

3 0

1 year ago

Read 2 more answers

(a) When the displacement in SHM is one-sixth the amplitude xm, what fraction of the total energy is kinetic energy? (b) What fr

andreev551 [17]

Answer:

Explanation:

$KE = \frac{1}{2}\times K\times (A^2-X^2)$

at $X = \frac{1}{6}A\\\\
KE = \frac{1}{2}\times K\times ( A^2 - A^{\frac{2}{36})\\\\
fraction = \frac{35}{36} = 0.9722
$

b) PE fraction = 1/36 = 0.02777

c) 1/2*KX^2 = 1/2*K*(A^2-X^2)

=> X = A/sqrt(2)

3 0

2 years ago

Two point charges of equal magnitude are 7.3 cm apart. At the midpoint of the line connecting them, their combined electric fiel

Delicious77 [7]

Answer:

q₁ = q₂ = Q = 14.8 pC

Explanation:

Given that

q₁ = q₂ = Q = ?

Distance between charges = r =7.3 cm = 0.073 m

Combined electric field = E₁ + E₂ = E = 50 N/C

Using formula

$E=2\frac{kQ}{r^2}$

Rearranging for Q

$Q= \frac{Er^2}{2k}\\\\Q=\frac{(50)(.005329)}{2\times 9\times 10^9}$

$Q=14.8\times 10^{-12}\, C\\\\Q=14.8\, pC$

4 0

2 years ago

When a wire is attached to the negative terminal of a battery, what happens?

Leokris [45]

Answer: The current will flow from the negative to the positive terminal as fast as possible. This will wear out the battery quickly.

Explanation:

3 0

2 years ago

Read 2 more answers

The 64.5-kg climber in is supported in the “chimney” by the friction forces exerted on his shoes and back. The static coefficien

kobusy [5.1K]

Answer:

The minimum force the climber must exert is about 439N.

Explanation:

We use the relationship between friction and normal force to answer this question:

$F_{friction} = \mu_{static} \cdot F_{normal}\implies F_{normal}=\frac{F_{friction}}{\mu_{static}}$

We are given the static coefficients of friction but need to determine the friction force. To do that we consider the totality of forces acting on this hapless gentleman stuck in a chimney. There is the gravity acting downward (+), then there are two friction forces acting upward (-), namely through his shoes and his back. The horizontal force exerted by the climber on both walls of the chimney is the same and is met with equally opposing normal force. Since the climber is not falling the net force in the vertical direction is zero:

$F_{net} = 0 = F_g - F_{shoes}-F_{back}= mg - \mu_{shoes}F_{norm}-\mu_{back}F_{norm}\\F_{norm}=\frac{mg}{\mu_{shoes}+\mu_{back}}=\frac{64.5kg\cdot 9.8\frac{m}{s^2}}{0.8+0.64}\approx 438.96N\\$

The normal force in this equilibrium is about 439N and because we are told that the static friction forces are both at their maximum, this value is at the same time the <em>minimum</em> force needed for the climber to avoid starting slipping down the chimney.

7 0

2 years ago