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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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A science class puts a balloon containing 1.25 l of air at 101 kpa into a bell jar. using an air pump, the class removes some of
storchak [24]
The first thing you should know in this case is the following definition:
 PV = nRT
 Then, as the temperature is constant, then:
 PV = k
 Then, we have two states:
 P1V1 = k
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 We can then equalize both equations:
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 Substituting the values:
 (1.25) * (101) = (2.25) * (P2)
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 answer:
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8 0
3 years ago
3. A pendulum with a 1.0-kg weight is set in motion from a position 0.04 m above the lowest point on the path of the weight.
gavmur [86]

Answer: K.E = 0.4 J

Explanation:

Given that:

M = 1.0 kg

h = 0.04 m

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K.E = P.E

K.E = mgh

K.E = 1 × 9.81 × 0.04

K.E = 0.3924 Joule

The kinetic energy of the pendulum at the lowest point is 0.39 Joule

6 0
3 years ago
g At some point the road makes a right turn with a radius of 117 m. If the posted speed limit along this part of the highway is
user100 [1]

Answer:

Ф = 28.9°

Explanation:

given:

radius (r) = 117m

velocity (v) = 25.1 m/s

required: angle Ф

Ф = inv tan (v² / (r * g))      we know that g = 9.8

Ф = inv tan (25.1² / (117 * 9.8))

Ф = 28.9°

6 0
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What is the similarities of a furnace and the sun?
Svetlanka [38]

Answer:

They are both extremely hot, they both produce a form of light, they both have/use fire(typically)

Explanation:

6 0
2 years ago
A 50g ball is released from rest 1.0 above the bottom of thetrack
ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2

v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}

v=\sqrt{19.6}

v=4.42\ m/s

We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

y=0.996\ m

Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

x=1.99\ m\ \approx 2 m

Hence, The maximum height of the ball is 2 m.

4 0
2 years ago
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