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ankoles [38]
3 years ago
5

How are moving pulleys different from fixed-position pulleys? (Points : 3) Moving pulleys have more force than fixed-position pu

lleys. Moving pulleys are easier to use than fixed-position pulleys. Moving pulleys are attached to the object and move with the object. Fixed-position pulleys are connected to a fixed point and do not move. Moving pulleys are attached to the object and are stationary. Fixed-position pulleys are connected to a fixed point and move freely.
Physics
2 answers:
Lubov Fominskaja [6]3 years ago
8 0
"Fixed-position pulleys are connected to a fixed point and do not move" is the one among the following choices given in the question that explains how the moving<span> pulleys are different from fixed-position pulleys. The correct option among all the options that are given in the question is the fourth option.</span>
meriva3 years ago
4 0

Answer : Option D) Fixed-position pulleys are connected to a fixed point and do not move.

Explanation : As the name suggests the moving pulleys move freely whereas the fixed pulleys as per their name is attached to a fixed position and at a fixed point and they can not move freely.

Movable pulleys are ones which has one part of the rope attached to a fixed object, like a bar or a beam. Whereas when both the parts of a rope are attached to a fixed object, this is called as a fixed pulley.

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4. Going back to the dog whistle in question 1, what is the minimum riding speed needed to be able to hear the whistle? Remember
jeyben [28]

Answer:

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency is 15.7  m/s

Explanation:

The Doppler shift equation is given as follows;

f' = \dfrac{v - v_o}{v + v_s} \times f

Where:

f' = Required observed frequency = 20.0 kHz

f = Real frequency = 21.0 kHz

v = Sound wave velocity = 330 m/s

v_o = Observer velocity = X m/s

v_s = Source velocity = 0 m/s (Assuming the source is stationary)

Which gives;

20 = \dfrac{330- v_o}{330+0} \times 21

330 - v_o = (20/21)*330

v_o = 330 - (20/21)*330 = 15.7 m/s

The minimum riding speed relative to the whistle (stationary) to be able to hear the sound at 21.0 kHz frequency = 15.7  m/s.

8 0
3 years ago
A)If your torsion balance deflects to an angle of 10° when your two spheres are 40cm apart, what angle will it deflect to when t
svp [43]

Answer:

Explanation:

Given

for \theta=10^{\circ}

Sphere are d=40\ cm

when sphere d_2=10\ cm apart suppose deflection is \theta _2

We know

F=k_t\cdot \theta

Where F=force between charged particle

\theta =Deflection

F=\frac{kQ_1Q_2}{r^2}=k_t\cdot \theta

\theta =\frac{k}{k_t}\times \frac{Q_1Q_2}{r^2}----1

thus \theta \propto \frac{1}{r^2}

for \theta _2

\frac{\theta _1}{\theta _2}=(\frac{r_2}{r_1})^2

\theta _2=16\times \theta _1

\theta _2=160^{\circ}

(b)for 10^{\circ} deflection Potential v_1=8\ kV

Electric Potential is V=\frac{kQ}{r}

Q=\frac{V\cdot r}{k}

where V=voltage

k=constant

r=distance between charges

Put value of Q in equation 1

\theta =\frac{k}{k_t}\times \frac{V^2r^2}{k^2}

\theta =\frac{V^2r^2}{k\cdot k_t}

thus \theta \propto V^2

therefore

\frac{\theta _1}{\theta _2}=(\frac{V_1}{V_2})^2

\frac{10}{\theta _2}=(\frac{8}{4})^2

\theta _2=\frac{10}{4}=2.5^{\circ}

5 0
3 years ago
A thermodynamic system undergoes a process in which its internal energy decreases by 500 joules. At the same time, 220 joules of
Mkey [24]

Answer:

- 720 Joules\\

Explanation:

As per the first law of thermodynamics,

d U = Q + W\\

U is  change in the internal energy of the system and W is work done on the system.

Q = d U - W \\= - 500 J - 220 J\\= -720 Joules\\

7 0
3 years ago
Read 2 more answers
An object of mass 30KG is falling in air and experiences a force due to air resistance of 50 newtons determine the net force act
ryzh [129]

Answer:

Assume that g = 10\; \rm N \cdot kg^{-1}. The net force on this object will be 250\; \rm N (downwards.) The acceleration of this object will be approximately 8.3\; \rm m \cdot s^{-2} (also downwards.)

Explanation:

<h3>Net force</h3>

The object is falling towards the ground because of gravity. The size of the gravitational force on this object depends on its mass and the strength of the gravitational field at its location.

Near the surface of the earth, the gravitational field strength is approximately 10\; \rm N \cdot kg^{-1}. In other words, approximately 10\; \rm N of gravitational force acts on each kilogram of mass near the surface of the earth.

The mass of this object is given as m = 30\; \rm kg. Therefore, the size of the gravitational force on it will be:

W = m \cdot g \approx 30 \; \rm kg \cdot 10\; N \cdot kg^{-1} = 300\; \rm N.

Near the surface of the earth, gravitational forces point towards the ground. On the other hand, the direction of air resistance on this object will be opposite to its direction of motion. Since this objects is moving towards the ground, the air resistance on it will be directed in the opposite direction. That's exactly the opposite of the direction of the gravitational force on this object. The net force on this object will be:

300\; \rm N - 50\; \rm N =250\; \rm N.

<h3>Acceleration</h3>

Let a denote the acceleration on this object. Apply Newton's Second Law of motion:

\begin{aligned} a &= \frac{F(\text{net force})}{m} \approx \frac{250\; \rm N}{30\; \rm kg} \approx 8.3\; \rm m \cdot s^{-2}\end{aligned}.

Note that the acceleration of this object and the net force on it should be in the same direction.

8 0
3 years ago
Can anyone wanna help me
FrozenT [24]

I'd say the answer's D.

7 0
3 years ago
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