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3 years ago

14

A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at

a rate of 2ft/sec. Find the velocity of the top of the ladder at time t=1.

1 answer:

stellarik [79]3 years ago

5 0

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

$\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft$

Also, at any instant t

$\Rightarrow l^2=x^2+y^2$

differentiate w.r.t.

$\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s$

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Answer:24.47

Explanation:

Given

$L_1=30 m$

$d_1=3 mm$

$R_1=25 \Omega$

$L_2=40 m$

$d_2=3.5 mm$

we know Resistance $R=\frac{\rho L}{A}$

Where R=resistance

$\rho =resistivity$

L=Length

A=area of cross-section

$A=\frac{\pi d^2}{4}$

Thus $R\propto \frac{L}{d^2}$

therefore

$R_1\propto \frac{L_1}{d_1^2}$ --------1

$R_2\propto \frac{L_2}{d_2^2}$ --------2

divide 1 and 2 we get

$\frac{R_1}{R_2}=\frac{L_1}{L_2}\times \frac{d_2^2}{d_1^2}$

$\frac{R_1}{R_2}=\frac{30}{40}\times \frac{3.5^2}{3^2}$

$R_2=25\times 0.734\times 1.33$

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3 years ago

A solenoid that is 78.8 cm long has a cross-sectional area of 15.9 cm2. There are 914 turns of wire carrying a current of 8.25 A

Harlamova29_29 [7]

Answer:

(a) Energy density will be equal to $57.31J/m^3$

(b) Total energy will be equal to 0.0718 J

Explanation:

It is given that length of solenoid l = 78.8 cm = 0.788 m

Cross sectional area $A=15.9cm^2=15.9\times 10^{-4}m^2$

Number of turns of the wire N = 914

Current in the solenoid i = 8.25 A

Inductance of the wire is equal to $L=\frac{\mu _0N^2A}{l}=\frac{4\pi \times 10^{-7}\times 914^2\times 15.9\times 10^{-4}}{0.788}=2.117\times 10^{-3}H$

(b) Total energy stored in magnetic field $U=\frac{1}{2}Li^2=\frac{1}{2}\times 2.11\times 10^{-3}\times 8.25^2=0.0718J$

(a) Energy density will be equal to

$U_b=\frac{0.0718}{15.9\times 10^{-4}\times 0.788}=57.31J/m^3$

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2 years ago

A child is riding a merry-go-round that has an instantaneous angular speed of 12 rpm. If a constant friction torque of 12.5 Nm i

sammy [17]

Answer:

$-0.25 rad/s^2$

Explanation:

The equivalent of Newton's second law for rotational motions is:

$\tau = I \alpha$

where

$\tau$ is the net torque applied to the object

I is the moment of inertia

$\alpha$ is the angular acceleration

In this problem we have:

$\tau = -12.5 Nm$ (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

$I=50.0 kg m^2$ is the moment of inertia

Solving for $\alpha$ , we find the angular acceleration:

$\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2$

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Georgie was pulling her brother (of mass 16 kg) in a 9.4 kg sled with a constant force of 39 N for one block (128 m). How much w

Genrish500 [490]

Answer:

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While driving down the road, a grasshopper strikes the windshield of a bus and makes a mess in front of the face of the driver.

nlexa [21]

Answer:

The force on the grass hopper is equal and opposite to the force on the wind shield.

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It is a known fact every action has an equal and opposite reaction. Here the action force is the grasshopper hitting the windshield.

The reaction force is the force exerted by the windshield on the grasshopper.

If the force of windshield on the grasshopper is more, grasshopper would have been thrown away. If the grasshopper's force on the windshield is greater, the windshield would have broken.

But here these constitute equal and opposite action-reaction pair of forces and so the grasshopper is able to repeatedly strike the windshield.

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