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Lerok [7]
3 years ago
8

Two students are simultaneously pushing a glider on a frictionless air track, each with a constant horizontal force. In the exam

ples that follow, assume that student 1 always does positive work on the block (Ws, > 0 and that student 2 always does negative work (Ws2 0 Suppose that the glider moves to the left with constant speed. 1. In which direction is each student pushing the block? Use the definition of work to answer
Physics
1 answer:
Gala2k [10]3 years ago
4 0

Answer:

<em>The student 1 pushes to the left, the student 2 pushed to the right</em>

Explanation:

<u>Mechanical Work</u>

The work done by a force F on an object moving a distance X and making an angle \theta with the force is given by

W=F.X.cos\theta

If the force and the distance point in the same direction, the angle is 0 and the work is positive.

If the force and the distance point in opposite directions, the angle is 180° and the work is negative.

We know the glider moves to the left and the student 1 always does positive work, thus he must be pushing the block to the left to have the same direction.

The same can be applied to the second student who does a negative work when the block goes to the left, so he must be pushing to the right direction.

Summary: The student 1 pushes to the left, the student 2 pushed to the right

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Two gliders on an air track collide in a perfectly elastic collision. Glider A has a mass of 1.1 kg and is initially travelling
Eva8 [605]

m1= mass 1 = 1.1 kg

Vi1 = initial velocity 1 = 2.7 m/s

m2= 2.4 kg

V2i = -1.9 m/s

We assume east as positive and west as negative.

Apply the formulas:

Vf1 = ?

vf1=(\frac{m1-m2}{m1+m2})Vi1+(\frac{2m2}{m1+m2})Vi2

Replacing:

Vf1=\frac{(1.1-2.4)}{(1.1+2.4)}2.7+\frac{(2\times2.4)}{(1.1+2.4)}-1.9Vf1=(\frac{-1.3}{3.5})2.7+(\frac{4.8}{3.5})-1.9Vf1=-1-2.6=-3.6\text{ m/s}

Answer: 3.6 m/s west

6 0
1 year ago
A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be consi
vagabundo [1.1K]

Answer:1.7 rev/s

Explanation:

Given

Frequency of wheel N_1=2\ rev/s

angular speed \omega_1=2\pi N_1=4\pi\  rad/s

mass of wheel m_1=4.5\ kg

diameter of wheel d_1=0.30\ m=30\ cm

radius of wheel r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm

mass of clay m_2=2.8\ kg

the radius of the chunk of clay r_2=8\ cm

Moment of inertia of Wheel

I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2

Combined moment of inertia of wheel and clay chunk

I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2

Conserving angular momentum

\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi

Common frequency of wheel and chunk of clay is

\Rightarrow N_2=\dfrac{4\pi \times 0.849}{2\pi}=1.698\approx 1.7\ rev/s

5 0
3 years ago
A typical adult human has a mass of about 70 kg.
Misha Larkins [42]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2 
<span>= 2.40 x 10-3 N

b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N 
FMFE 2.40 x 10-3N685 N= 0.0004%</span>
3 0
3 years ago
Insulators have very high .
Vlad1618 [11]

Answer:

Resistance to electrical currents

Explanation:

Conductors have low resistance to electrical currents, and are used to "conduct" the flow of electricity.

Insulators have very high resistance and are used to protect us from the flow of electricity.

5 0
3 years ago
Which will produce a magnetic field?
Yanka [14]

Answer:

Answer:

a magnet

3 0
4 years ago
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