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3 years ago

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When the cart moves on an incline at constant speed, it is in equilibrium; i.e., the net force on it is zero. Does it require mo

re, less, or the same tension in the string to pull the cart up the track at constant speed or down the track at constant speed?

1 answer:

Amanda [17]3 years ago

5 0

Answer:

It requires more tension to pull up the track

Explanation:

Net force must be zero to maintain constant velocity.

Weight force will always be pointed down the slope. Call it W

Friction force (Call it Ff) will be down slope when movement is up slope.

Friction force will be up slope when movement is down slope.

W and Ff are always positive numbers

Call the pulling force T

If Up slope is considered the positive direction

Moving up slope

Tu - Ff - W = 0

Tu = Ff + W

Moving down slope

Td + W - Ff = 0

Td = Ff - W

Ff + W > Ff - W therefore Tu > Td

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Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart.

serious [3.7K]

Answer:

$1.6\times 10^{-7}$ N

$2.4\times 10^{-7}$ N

Explanation:

$i_{1}$ = 1 A

$i_{2}$ = 4 A

$r$ = distance between the two wire = 5 m

$F$ = Force per unit length acting between the two wires

Force per unit length acting between the two wires is given as

$F = \frac{\mu _{o}}{4\pi }\frac{2i_{1}i_{2}}{r}$

$F = (10^{-7})\frac{2(1)(4)}{5}$

$F = 1.6\times 10^{-7}$ N

$r'}$ = distance of each wire from the midpoint = 2.5 m

Magnetic field midway between the two wires is given as

$B = \frac{\mu _{o}}{4\pi } \left \left ( \frac{2i_{2}}{r'} \right - \frac{2i_{1}}{r'} \right \right ))$

$B = (10^{-7}) \left \left ( \frac{2(4)}{2.5} \right - \frac{2(1)}{2.5} \right \right ))$

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3 years ago

A woman is applying 300N/m2 of pressure on to door with her hand. Her hand has area of 0.02m2. Work out the force being applied

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Answer:

6N

Explanation:

Given parameters:

Pressure applied by the woman = 300N/m²

Area = 0.02m²

Unknown:

Force applied = ?

Solution:

Pressure is the force per unit area on a body

Pressure = $\frac{force}{area}$

Force = Pressure x area

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3 years ago

A heavy solid disk rotating freely and slowed only by friction applied at its outer edge takes 120 seconds to come to a stop.

alisha [4.7K]

Answer:

The time is 16 min.

Explanation:

Given that,

Time = 120 sec

We need to calculate the moment of inertia

Using formula of moment of inertia

$I=\dfrac{1}{2}MR^2$

If the disk had twice the radius and twice the mass

The new moment of inertia

$I'=\dfrac{1}{2}\times2M\times(2R)^2$

$I'=8I$

We know,

The torque is

$\tau=F\times R$

We need to calculate the initial rotation acceleration

Using formula of acceleration

$\alpha=\dfrac{\tau}{I}$

Put the value in to the formula

$\alpha=\dfrac{F\times R}{\dfrac{1}{2}MR^2}$

$\alpha=\dfrac{2F}{MR}$

We need to calculate the new rotation acceleration

Using formula of acceleration

$\alpha'=\dfrac{\tau}{I'}$

Put the value in to the formula

$\alpha=\dfrac{F\times R}{8\times\dfrac{1}{2}MR^2}$

$\alpha=\dfrac{2F}{8MR}$

$\alpha=\dfrac{\alpha}{8}$

Rotation speed is same.

We need to calculate the time

Using formula angular velocity

$\Omega=\omega'$

$\alpha\time t=\alpha'\times t'$

Put the value into the formula

$\alpha\times120=\dfrac{\alpha}{8}\times t'$

$t'=960\ sec$

$t'=16\ min$

Hence, The time is 16 min.

5 0

3 years ago