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____ [38]
3 years ago
15

A horse pulls a cart along a flat road. They start from rest and as time goes on their speed increases. Consider the following f

our forces:
The force of the horse pulling on the cart;

The force of the cart pulling on the horse;

The force of the horse pushing on the road

The force of the road pushing on the horse.

Which of the following statements is true concerning these forces?

Force 2 exceeds force 4.

Force 2 is less than force 3.

Force 3 exceeds force 4.

Forces 1 and 2 cannot have equal magnitudes.

Force 1 exceeds force 2.
Physics
1 answer:
Anna11 [10]3 years ago
5 0

Answer:

Forces 1 and 2 cannot have equal magnitudes.

Force 1 exceeds force 2.

Explanation:

Numbering the given statements:

  1. The force of the horse pulling on the cart.
  2. The force of the cart pulling on the horse.
  3. The force of the horse pushing on the road .
  4. The force of the road pushing on the horse.
  • When a horse pulls a cart along a flat road starting from the state of rest, the horse applies a force on the ground in downward-back direction diagonally and as its reaction it moves forward. During this action there is also a reaction force by the road on the horse which exceeds the force of the horse on the ground and as a result the horse moves forward.
  • For pulling the cart the horse applies the force on  the cart  in the forward direction and as a reaction the cart applies a force on the horse in the backward direction which is less than the force of the horse on the cart resulting in the forward motion of the cart along with the horse.

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A 400.0 ohm resistor has a potential difference of 20.0 volts. What is the magnitude of the power dissipated by the resistor
Sergio [31]

P=IV, where P is power, I is resistance, and V is voltage.  Plug in and solve:

P=400(20)

P=8000W

Hope this helps!!

3 0
3 years ago
If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system ar
AnnyKZ [126]

Answer:

a. The angular frequency is doubled.

e. The period is reduced to one-half of what it was.

Explanation:

Angular frequency is given as;

ω = 2πf

\frac{\omega _1}{f_1} = \frac{\omega _2}{f_2}

when the frequency is doubled

\frac{\omega _1}{f_1} = \frac{\omega _2}{(2f_1)} \\\\\omega _1 = \frac{\omega _2}{2}\\\\\omega _2 = 2\omega _1

Thus, the angular frequency will be doubled.

Amplitude in simple harmonic motion is the maximum displacement.

Frequency is related to period in simple harmonic motion as given in the equation below;

f = \frac{1}{T} \\\\f_1T_1= f_2T_2\\\\T_2 = \frac{f_1T_1}{f_2}

when the frequency is doubled;

T_2 = \frac{f_1T_1}{2f_1} \\\\T_2 = \frac{T_1}{2}

Thus, the period will be reduced to one-half of what it was.

5 0
3 years ago
What happens to density when you change temperature of an object?
solong [7]
Most of the substances have higher density in solid state. When we heated solids it change its state to liquid. After a few minutes the liquid will boil and after that it will change to gas. Gaseous state of matter has the lowest density. From this we can conclude that density decreases with increase in temperature.

Note the point: Water has higher density in 4℃, at its liquid state
4 0
3 years ago
1. If you push a 40.0kg mass object with a force of 145.ON, what will be the object's
horrorfan [7]

Newton said . . . F = m a

Divide each side by  m . . .  a = F / m

Acceleration = (force) / (mass)

Acceleration = (145.O N) / (40.0 kg)

<em>Acceleration = 3.625 m/s²</em>

4 0
3 years ago
What do I have to do to figure out " What % of an object’s mass is above the water line if the object’s density is 0.82g/ml " wi
blsea [12.9K]

Answer:

18%

Explanation:

There are two equal and opposite forces on a floating object: weight and buoyancy.

W = B

The weight of an object is its mass times gravity: W = mg

Buoyancy is the weight of the displaced fluid: W = mf g

Plugging in:

mg = mf g

m = mf

Mass is density times volume:

ρV = ρf Vf

Solving for the ratio of Vf / V:

Vf / V = ρ / ρf

Given that ρ = 0.82 g/mL and ρf = 1.00 g/mL:

Vf / V = 0.82

That means 82% of the object's volume (and therefore, 82% of its mass, assuming uniform density) is submerged.  Which means that 18% is above the water line.

4 0
3 years ago
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