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horrorfan [7]
2 years ago
13

A horse is trotting along pulling a sleigh through the snow. To move the 225 kg sleigh straight ahead at a constant speed, the h

orse must pull with a force of 123N.
a. What is the net force acting on the sleigh?

b. What is the coefficient of kinetic friction between the sleigh and the snow?
Physics
1 answer:
RoseWind [281]2 years ago
5 0

Answer:

a) The net force acting on the sleigh is zero because it is moving at a constant speed.

b) F_{fr}= \mu mg = F_{pull}, then \mu = \frac{F_{pull}}{mg} = 0.0557.

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You pull a 70-kg crate at an angle of 30° above the horizontal. If you pull with a force of 600N and the coefficient of kinetic
Alenkinab [10]

Answer:

Explanation:

Force of friction acting on the body = μ mg cosθ

= .4  x 70 x 9.8 x cos30

= 237.63 N

component of weight = mgsinθ

= 70 x 9.8 x sin30

= 343 N  

Net upward force = 600 - mgsinθ - μ mg cosθ

= 600 - 343 - 237.63

= 105.37 N

acceleration in upward direction = 105.37 / 70

= 1.5 m /s²

s = ut + 1/2 a t²

= 0 + .5 x 1.5 x 3²

= 6.75 m .

5 0
3 years ago
Help!! <br> A. 0.5 s <br> B. 1.0 s<br> C. 1.5 s<br> D. 2.0 s
algol13

Answer:

c.    1.5 s

Explanation:

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3 years ago
Jim thinks that he can throw a baseball six meters per second. What does the six meters refer to?
SVETLANKA909090 [29]

Answer:

Six meters refers to the distance the baseball travels in one second.

Explanation:

6 m/s means that every second, the baseball travels 6m. So, 6 meters is the distance traveled in one second.

3 0
2 years ago
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A book is sitting on a desk. What best describes the normal force acting on the book?
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3 years ago
A student makes a short electromagnet by winding 300 turns of wire around a wooden cylinder of diameter d 5.0 cm. The coil is co
kupik [55]

Answer:

A) μ = A.m²

B) z = 0.46m

Explanation:

A) Magnetic dipole moment of a coil is given by; μ = NIA

Where;

N is number of turns of coil

I is current in wire

A is area

We are given

N = 300 turns; I = 4A ; d =5cm = 0.05m

Area = πd²/4 = π(0.05)²/4 = 0.001963

So,

μ = 300 x 4 x 0.001963 = 2.36 A.m².

B) The magnetic field at a distance z along the coils perpendicular central axis is parallel to the axis and is given by;

B = (μ_o•μ)/(2π•z³)

Let's make z the subject ;

z = [(μ_o•μ)/(2π•B)] ^(⅓)

Where u_o is vacuum permiability with a value of 4π x 10^(-7) H

Also, B = 5 mT = 5 x 10^(-6) T

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Solving this gives; z = 0.46m =

3 0
3 years ago
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