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erma4kov [3.2K]
3 years ago
14

Lifting a bale of cotton is no easy task. Farmer Jones wanted to store the cotton in the loft of his barn but he could not budge

the bale. He consulted the high school's science teacher for help. She suggested a simple machine.
Farmer Jones' options included an inclined plane, a fixed pulley, a block and tackle, and a lever. If he had to lift the bale of cotton to a height of fifteen feet, which would be the MOST practical solution?

A) lever
B) screw
C) inclined plane
D) block and tackle (pulley)
Physics
2 answers:
qwelly [4]3 years ago
8 0

Answer:

d

Explanation:

julia-pushkina [17]3 years ago
6 0
An inclined plane would not be practical because he would still have to push it and the amount of effort and energy would increase. A lever would not be practical because he would have to use his weight to make the bale rise 15 feet. He would also need to make sure that he has a 15 foot ladder in the first place. A screw would not be practical because he would have to drill a hole through the medium and use rotation.

A pulley would be the most practical way of moving the cotton bale. A pulley supports movement and will help the farmer move the bale.

\sf\ Answer:\ D)\ BLOCK\ AND\ TACKLE\ (PULLEY)
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In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
Daniel [21]

Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

5 0
3 years ago
When a solid is placed in a container and heat is applied, a phase change occurs. Watch the video and identify which of the foll
goldenfox [79]

Answer:

a) false b) true c) true d) false and e) true

Explanation:

a) false.   All the energy applied is used for the phase change, so the temperature remains constant.

b)   true.   The kinetic energy is associated with the speed of the particles and they have more mobility in the liquid, therefore, more kinetic energy.

c)   true.   Since energy is used for state change

d)  false.   In general, mobility and temperature are proportional

e)  true.  Heat is the source of energy for the change of state

4 0
3 years ago
Read 2 more answers
A plane has an airspeed of 142 m/s. A 30.0 m/s wind is blowing southward at the same time as the plane is flying. What must be t
const2013 [10]

Answer:

\theta=12.19^{\circ}

Explanation:

Given that

The speed of the airplane ,v= 142 m/s

The speed of the air ,u = 30 m/s

Lets take angle make by airplane from east direction towards north direction is θ .

Now by using diagram ,we can say that

sin\theta =\dfrac{u}{v}

Now by putting the values in the above equation we get

sin\theta =\dfrac{30}{142}

sin\theta=0.21

\theta=12.19^{\circ}

Therefore the angle will be 12.19° .

 

4 0
3 years ago
Technician A says that wheel cylinder dust boots keep dust and moisture out of the cylinder bore. Technician B says that it is n
timurjin [86]

Answer:

i think its technician A

7 0
3 years ago
Two identical twins hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about the
777dan777 [17]

Answer:

Part a)

L = 2683.2 kg m^2/s

Part b)

v' = 8.60 m/s

Part c)

W = 4326.7 J

Explanation:

Part a)

As we know that there is no external torque on the system of two twins

so here we will use

L = mv r + mvr

L = 2(78 \times 4.30 \times 4)

L = 2683.2 kg m^2/s

Part b)

Since angular momentum is conserved here as there is no external torque

so we will have

2(m v r) = 2( m v' \frac{r}{2})

v' = 2v

v' = 8.60 m/s

Part c)

Work done by both of them = change in kinetic energy

so we have

W = 2(\frac{1}{2}mv'^2 - \frac{1}{2}mv^2)

W = m(v'^2 - v^2)

W = 78(8.60^2 - 4.3^2)

W = 4326.7 J

5 0
3 years ago
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