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son4ous [18]
3 years ago
8

A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation. (1) the force of the ho

rse 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 Suppose that the horse and cart have started from rest; and as time goes on, their speed increases in the same direction. Which one of the following conclusions is correct concerning the magnitudes of the forces mentioned above?
Physics
1 answer:
expeople1 [14]3 years ago
3 0

Complete Question: A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation. (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 Suppose that the horse and cart have started from rest; and as time goes on, their speed increases in the same direction. Which one of the following conclusions is correct concerning the magnitudes of the forces mentioned above?

A) Force 1 exceeds force 2.

B) Force 2 is less than force 3.

C) Force 2 exceeds force 4.

D) Force 3 exceeds force 4.

E) Forces 1 and 2 cannot have equal magnitudes.

Answer: B) Force 2 is less than force 3.

Explanation: From the given four forces to be considered, Force 2 is less than Force 3 is correct option.

After evaluating the forces one by one. Force 3 and Force 4 are the same, both have equal magnitude.

And Force 3 is greater than Force 2, i.e., the force of the horse pushing on the road is much higher than the cart pulling the horse. Hence, more forces is required to pull the cart while on the road towards forward direction.

More so, the force of the road pushing on the horse is greater than the force of the cart pulling on the horse.

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A car of mass 1600 kg can just be lifted what is the least force that the electronicmagnet must use to lift the car ?
Mkey [24]

The car's mass is 1600 kg.

Its weight is (mass) x (gravity).  

On Earth, that's (1600 kg) x (9.8 m/s²)  =  15,680 Newtons.

At the moment, that's the only force acting on the car, directed downward and provided by gravity.

If you want to lift the car, then the net force has to be directed upward, and must either exactly cancel or exceed the force of gravity.

So the minimum force required to lift the car is <em>15,680 Newtons</em>, directed vertically upward.

5 0
3 years ago
That 1.5 kg brick falls to the ground. What is the kinetic energy when the brick is moving at 26 m/s? Formula: Variables: Work w
Tresset [83]

Answer:

<h2>507 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

k =  \frac{1}{2} m {v}^{2}  \\

m is the mass

v is the velocity

From the question we have

k =  \frac{1}{2}  \times 1.5 \times  {26}^{2}  \\  =  \frac{1}{2}  \times 1.5 \times 676 \\  = 1.5 \times 338  \\ = 507

We have the final answer as

<h3>507 J</h3>

Hope this helps you

5 0
3 years ago
PLEASE HELP! I'LL GIVE BRAINLEST​
melisa1 [442]

Answer:

Weight = 8.162 Newton.

Explanation:

Given the following data;

Mass = 2.2 kg

Acceleration due to gravity = 3.71 N/kg

To find the weight of the textbook;

Weight = mass * acceleration due to gravity

Weight = 2.2 * 3.71

Weight = 8.162 N

Therefore, the weight of the science textbook in mars is 8.162 Newton.

3 0
3 years ago
Light is incident on an air-water interface at an angle of 30 degree to the normal. What angle does the refracted ray make with
Vikentia [17]

Answer:

22 degree

Explanation:

Angle of incidence, i = 30 degree

the refractive index of water with respect to air is 4/3.

As the ray of light travels from rarer medium to denser medium, that mean air to water, the refraction takes place.

According to Snell's law,

Refractive index of water with respect to air = Sin i / Sin r

Where, r be the angle of refraction

4 / 3 = Sin 30 / Sin r

0.75 = 2 Sin r

Sin r = 0.375

r = 22 degree

Thus, the angle of refraction is 22 degree.

6 0
3 years ago
Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat
Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

The volume of of a sphere is

v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}

The surface of area of the balloon is(S) = 4\pi r^2

S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}

Putting the value of \frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}

\Rightarrow \frac{dS}{dt} =\frac{20}{ r}

Given that r = 5 cm

[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0
4 years ago
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