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3 years ago

What is a Pulley why is it used

Physics

1 answer:

aleksley [76]3 years ago

4 0

Answer:

A simple pulley is a wheel with a rope that allows you to pull one end and have it lift whatever is on the other end. A modern, common example of this is a crane, often used in construction.

Explanation:

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The force of gravity is _m/s2

larisa86 [58]

The force of gravity is positive 9.8m/s^2.

6 0

3 years ago

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

the resistivity of gold is 2.44×10−8Ω⋅m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries a cur

cricket20 [7]

Answer:

0.03605 V/m is the electric field in the gold wire.

Explanation:

Resistivity of the gold = $\rho = 2.44\times 10^{-8} \Omega.m$

Length of the gold wire = L = 14 cm = 0.14 m ( 1 cm = 0.01 m)

Diameter of the wire = d = 0.9 mm

Radius of the wire = r = 0.5 d = 0.5 × 0.9 mm = 0.45 mm = $0.45\times 0.001 m$

( 1mm = 0.001 m)

Area of the cross-section = $A=\pi r^2=\pi r^(0.45\times 0.001 m)^2$

Resistance of the wire = R

Current in the gold wire = 940 mA = 0.940 A ( 1 mA = 0.001 A)

$R=\rho\times \frac{L}{A}$

$V(voltage)=I(current)\times R(Resistance)$ ( Ohm's law)

$\frac{V}{I}=\rho\times \frac{L}{A}$

We know, Electric field is given by :

$E=\frac{dV}{dr}$

$E=\frac{V}{L}$

$E=\frac{V}{L}=\rho\times \frac{I}{A}$

$E=2.44\times 10^{-8} \Omega.m\times \frac{0.940 A}{\pi r^(0.45\times 0.001 m)^2}=0.03605 V/m$

0.03605 V/m is the electric field in the gold wire.

3 0

4 years ago

A solid cylinder is mounted above the ground with its axis of rotation oriented horizontally. A rope is wound around the cylinde

Romashka [77]

Answer:

(a)10.5 rad/s2

(b) 20.9 rev

Explanation:

As the block of mass 53 kg is falling and pulling on the rope. The tension force on the rope must be equal to the gravity acting on the block according to Newton's 3rd law

T = mg = 53*9.81 = 519.93 N

Since this tension force would rotate the cylinder freely without any friction. The torque created by this tension force is

To = TR = 519.93 * 0.36 = 187.17 Nm

This solid cylinder would have a moment of inertia around it's rotating axis of:

$I = \frac{mR^2}{2} = \frac{275 * 0.36^2}{2} = 17.82kgm^2$

(a)We can use Newton's 2nd law to calculate the angular acceleration exerted by such torque on the solid cylinder

$\alpha = \frac{To}{I} = \frac{187.17}{17.82} = 10.5 rad/s^2$

(b) With such constant angular acceleration, the angle it would make after 5s is

$\theta = \frac{\alphat^2}{2} = \frac{10.5*5^2}{2} = 131.3 rad$

Since each revolution equals to $2\pi rad$ of angle, we can calculate the number of revolution it makes

$\frac{\theta}{2\pi} = \frac{131.3}{6.28} \approx 20.9 rev$

(c) Assume the thickness of the rope is negligible (and its wounded radius is unchanging), we can calculate the rope length unwinded after rotating 131.3rad

$\theta R = 131.3*0.36 = 47.27 m$

3 0

3 years ago

What is the approximate average speed of the players run to first base

wolverine [178]

Answer:

90 ft/s is what i put. Let me know if its wrong

5 0

3 years ago

What is a Pulley why is it used​

What is a Pulley why is it used