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hichkok12 [17]
3 years ago
13

A flying saucer lifts the Physical Science building 10,000 ft into the air before discovering it is useless and discards the rem

ains. If the building weighs 1,000,000 pounds and the ascent takes 20 seconds, what is one saucer power?
Physics
1 answer:
scZoUnD [109]3 years ago
7 0

Answer:

500000000 lbft/s

Explanation:

F = Force or weight = 1000000 lbf

s = Displacement = 10000 ft

t = Time taken = 20 seconds

Work done is given by

W=Fs\\\Rightarrow W=1000000\times 10000\\\Rightarrow W=10000000000\ lb-ft

Power is given by

P=\dfrac{W}{t}\\\Rightarrow P=\dfrac{10000000000}{20}\\\Rightarrow P=500000000\ lbft/s

One Saucer power is 500000000 lbft/s

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How will a wave change as it moves from a layer of higher velocity into a lower layer of lower velocity?
denis-greek [22]

Explanation:

Wave is defined as a disturbance or oscillation that travels through space-time, accompanied by a transfer of energy. Wave motion transfers energy from one point to another, often with no permanent displacement of the particles of the medium.

The velocity of wave is equal to the product of its wavelength and frequency (number of vibrations per second). Longitudinal waves like sound waves travel through a medium.

Therefore, a wave move from a layer of high velocity to that of a lower velocity the wavelength changes (that is, decreases) as it moves.

6 0
3 years ago
Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too gr
IgorLugansk [536]

(a) 273.9 V

The power rating of the resistor is given by

P=\frac{V^2}{R}

where

P is the power rating

V is the potential difference across the resistor

R is the resistance

If the maximum power rating is P=5.0 W, and the resistance of the resistor is R=15 k\Omega = 15000 \Omega, then we can find the maximum potential difference across the resistor by re-arranging the previous equation for V:

V=\sqrt{PR}=\sqrt{(5.0 W)(15000 \Omega)}=273.9 V

(b) 1.6 W

In this case, we have:

R=9.0 k\Omega = 9000 \Omega is the resistance of the resistor

V=120 V is the potential difference across the resistor

So we can find the power rating by using the same formula of part (a):

P=\frac{V^2}{R}=\frac{(120 V)^2}{9000 \Omega}=1.6 W

(c) Maximum voltage: 14.1 V; Rate of heat: 2.00 W and 3.00 W

Here we have two resistors of

R_1 = 100 \Omega\\R_2 = 150 \Omega

and each resistor has a power rating of

P = 2.00 W

So the greatest potential difference allowed in the first resistor is

V=\sqrt{PR_1}=\sqrt{(2.00 W)(100 \Omega)}=14.1 V

While the greatest potential difference allowed in the second resistor is

V=\sqrt{PR_2}=\sqrt{(2.00 W)(150 \Omega)}=17.3 V

So the greatest potential difference allowed not to overheat either of the resistor is 14.1 V.

In this condition, the power dissipated on the first resistor is 2.00 W, while the power dissipated on the second resistor is

P_2 = \frac{V^2}{R_2}=\frac{(14.1 V)^2}{150 \Omega}=1.33 W

And this corresponds to the rate of heat generated in the first resistor (2.00 W) and in the second resistor (1.33 W).

4 0
3 years ago
II Force on a tennis ball. The record speed for a tennis ball that is served is 73.14 m/s. During a serve, the ball typically st
AveGali [126]

Answer:

F=248.5W N

Explanation:

Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

a=\frac{\Delta x}{\Delta t}=\frac{v-v_0}{t-t_0}

Since we start counting at 0s and the ball departs from rest, this is just a=\frac{v}{t}

So we can write:

F=ma=\frac{mv}{t}=\frac{gmv}{gt}

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

F=\frac{Wv}{gt}=\frac{v}{gt}W

Substituting our values:

F=\frac{(73.14m/s)}{(9.81m/s^2)(30\times10^{-3}s)}W=248.5W N

Where the average force exerted has been written it terms of the tennis ball's weight W.

8 0
3 years ago
A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,
egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

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4 years ago
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<span>A spatial pattern arranges main points according to physical direction or location.</span>
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