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2 years ago

Which situation describes the highest rate of power?

Physics

1 answer:

Darya [45]2 years ago

8 0

Answer : The correct option is, (D) A machine does 400 joules of work in 5 seconds.

Explanation :

Power : It is defined a the rate of doing work per unit time.

Formula used :

$P=\frac{w}{t}$

where,

P = power

w = work done

t = time

Now we have to determine the rate of power for the following options.

(A) A machine does 200 joules of work in 10 seconds.

$P=\frac{200}{10}=20W$

(B) A machine does 400 joules of work in 10 seconds.

$P=\frac{400}{10}=40W$

(C) A machine does 200 joules of work in 5 seconds.

$P=\frac{200}{5}=40W$

(D) A machine does 400 joules of work in 5 seconds.

$P=\frac{400}{5}=80W$

From this we conclude that, a machine does 400 joules of work in 5 seconds has the highest rate of power.

Hence, the correct option is, (D)

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An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$

And re-arranging for F,

$F=m\frac{w^2}{R}-N sin \theta$ (3)

But from eq.(2) in the previous part we know that

$N sin \theta = m \frac{v^2}{R}$

So, susbtituting into eq.(3),

$F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)$

4 0

1 year ago

Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for

Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0

2 years ago

g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobb

Law Incorporation [45]

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g

Buoyant force on lead = .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115 / 33.5

= .36

= 36 %

4 0

1 year ago

A nonlinear spring is used to launch a toy car. The car is pushed against the spring, compressing the spring 2.5 cm. The force t

dybincka [34]

Answer:

1.56 J

Explanation:

given,

Spring compression, x = 2.5 cm

Force exerts by the spring,

F = - k x

k = 5000 N/m

Potential energy stored = ?

energy stored in the spring

$PE = \dfrac{1}{2}kx^2$

$PE = \dfrac{1}{2}\times 5000\times 0.025^2$

PE = 1.56 J

Hence, the potential energy stored in the car is equal to 1.56 J.

5 0

2 years ago

We know that every object exerts an attraction on every other object and the heavier the object ___ the attraction.

hichkok12 [17]

I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question?

6 0

2 years ago