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3 years ago

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Superman is standing 393m horizontally away from Lois Lane. A villian drops a rock 4 m directly above Lois. I f Superman is to i

ntervene and ctath the brock

1 answer:

Nat2105 [25]3 years ago

5 0

You have to draw a triangle

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A circuit containing an inductor and a capacitor in series is designed to have a resonant frequency of 4511 Hz. If the inductor

OLEGan [10]

Answer:

(e) $6.835\times 10^{-7}F$

Explanation:

At resonance we know that $X_l=X_C$

That is $\omega L=\frac{1}{\omega C}$

$\omega ^2=\frac{1}{LC}$

$\omega =\frac{1}{\sqrt{LC}}$

$f=\frac{1}{2\pi \sqrt{LC}}$

We have given resonance frequency f =4511 Hz and inductance L=1.82 mH

So $4511=\frac{1}{2\pi \sqrt{LC}}$

$LC=\frac{1}{4\pi ^2\times 4511^2}$

$LC=1.244\times 10^{-9}$

$C=\frac{1.244\times 10^{-9}}{1.82\times 10^{-3}}=0.6835\times 10^{-6}=6.835\times 10^{-7}F$

So option e is the correct answer

3 0

3 years ago

A rock is sitting at the edge of a flat merry-go-round at a distance of 1.6 meters from the center. The coefficient of static fr

PSYCHO15rus [73]

Answer:

ω = 2.1 rad/sec

Explanation:

As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
Now, we need to ask ourselves: what supplies this force?
In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
We know that this force can be expressed as follows:

$f_{frs} = \mu_{s} * F_{n} (1)$

where μs = coefficient of static friction between the rock and the merry-

go-round surface = 0.7, and Fn = normal force.

In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
Fn = m*g (2)
This static friction force is just the same as the centripetal force.
The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:

$F_{c} = m* \omega^{2}*r (3)$

Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:

$\omega = \sqrt{\frac{\mu_{s} * g}{r} } = \sqrt{\frac{0.7*9.8m/s2}{1.6m}} = 2.1 rad/sec$

This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.

4 0

3 years ago

Which of the following fields of science has more scientific theories and fewer scientific laws?

Sindrei [870]

Answer:

Explanation:

Newton's first law of motion.

Newton's second law of motion.

Newton's law of universal gravitation.

Law of conservation of mass.

Law of conservation of energy.

Law of conservation of momentum

5 0

3 years ago

Read 2 more answers

What has a larger wavelength the color iolet or the color red?explain?

horsena [70]

Red has the longer wavelength. Wavelength and frequency are inversely related, so the longer the wavelength, the lower the frequency. This lower frequency is them perceived by our eye as the color red.

4 0

3 years ago

Read 2 more answers

In the Hunger Games movie, Katniss Everdeen fires a 0.0200-kg arrow from ground level to pierce an apple up on a stage. The spri

egoroff_w [7]

Answer:

a) $v=99.8584\ m.s^{-1}$

b) $v'=99.366\ m.s^{-1}$

Explanation:

Given:

mass of the arrow, $m=0.02\ kg$

stiffness constant of the bow, $k=330\ N,m^{-1}$

distance of pulling back the arrow on the bow from its mean position, $\Delta x=0.55\ m$

height of the apple targeted, $h=5\ m$

<u>Force on the arrow due to the stiffness of the bow:</u>

$F=k.\Delta x$

$F=330\times 0.55$

$F=181.5\ N$

Now the acceleration of the arrow upwards:

$a=\frac{F}{m}$

$a=\frac{181.5}{0.02}$

$a=9075\ m.s^{-2}$

a) For the course of motion when the arrow leaves the bow after the stretch is relaxed we consider that the arrow left the bow after its string goes to the mean position. During this phase the arrow also faces gravity in the downward direction.

Using the equation of motion:

$v^2=u^2+2(a-g).\Delta x$

where:

$v=$ velocity with which the arrow leaves the bow

$u=$ initial velocity of the arrow after it left

$v^2=0^2+2\times (9075-9.81)\times 0.55$

$v=99.8584\ m.s^{-1}$

b) Now when the arrow travels up then it is under a constant gravitational force acting opposite to the motion.

<u>Using eq. of motion:</u>

$v'^2=v^2-2\times g.h$

where:

$v'=$ final velocity when the arrow hits the target

$v=$ initial velocity after the arrow has been launched

$v'^2=99.8584^2-2\times 9.81\times 5$

$v'=99.366\ m.s^{-1}$

3 0

3 years ago