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jarptica [38.1K]

2 years ago

7

When Brett and Will ride the carousel, Brett always selects a horse on the outside row, whereas Will prefers the row closest t

o the center. These rows are 19 ft 1 in.19 ft 1 in. and 11 ft 11 in.11 ft 11 in. from the center, respectively. The angular speed of the carousel is 2.72.7 revolutions per minute. What is the difference, in miles per hour, in the linear speeds of Brett and Will?

1 answer:

sp2606 [1]2 years ago

7 0

Answer:

the difference, in miles per hour, in the linear speeds of Brett and Will;

∆v = 1.38 mph

Explanation:

Given;

Angular speed w = 2.7 revolutions per minute

Converting to revolutions per hour

w = 2.7 × 60 revolutions per hour

w = 162 rev/hour

Linear speed v = angular speed × 2πr

the difference, in miles per hour, in the linear speeds of Brett and Will;

∆v = w × 2π(r1 - r2)

r1 = Brett radius in miles

r2 = Will radius in miles

r1 = 19ft 1in = (19×12 + 1) = 229 in

r1 = 229 × 1.57828283 × 10^-5 miles

r2 = 11 ft 11 in = (11×12 + 11) = 143 in

r2 = 143 × 1.57828283 × 10^-5 miles

Substituting the values;

∆v = 162 × 2π × (229-143)×1.57828283 × 10^-5 mph

∆v = 1.38 mph

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Answer:

Ratio of length will be $\frac{L_2}{L_1}=1.108$

Explanation:

We have given time period of the pendulum when length is $L_1$ is $T_1=0.95sec$

And when length is $L_2$ time period $T_2=1sec$

We know that time period is given by

$T=2\pi \sqrt{\frac{L}{g}}$

So $0.95=2\pi \sqrt{\frac{L_1}{g}}$ ----eqn 1

And $1=2\pi \sqrt{\frac{L_2}{g}}$ -------eqn 2

Dividing eqn 2 by eqn 1

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8 0

2 years ago

(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500 - μC charge and flies due west at a sp

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(a) $2.64\cdot 10^{-8} N$ north

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$F=qvB sin \theta$

where

$q=0.500 \mu C = 0.500\cdot 10^{-6} C$ is the charge on the plane

v = 660 m/s is the velocity

$B=8.00\cdot 10^{-5} T$ is the magnitude of the magnetic field

$\theta=90^{\circ}$ is the angle between the direction of motion of the jet and of the magnetic field

Substituting,

$F=(0.5\cdot 10^{-6})(660)(8.0\cdot 10^{-5})=2.64\cdot 10^{-8} N$

The direction can be found by using Fleming's left hand rule. We have:

- index finger: magnetic field direction (straight up)

- middle finger: velocity of the plane (due west)

- force: thumb --> north

(b) Not negligible

As we can see from part (a), the magnitude of the force is not really big, so the effects are negligible.

For instance, we can compare this force with the weight of a plane. If we take a Boeing 737, its mass is about 80,000 kg, so its weight is

$W=mg=(80000)(9.8)=784,000 N$

As we can see, this is several orders of magnitude bigger than the magnetic force calculated at point (a), so the effects of the magnetic force are negligible.

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2 years ago

Two charges, qA and qB, are separated by a distance, d, and exert a force, F, on each other. Analyze Coulomb's law and answer th

maria [59]

Answer: a. F doubled

b. F reduced by one-quarter i.e

1/4*(F)

c. 1/9*(F)

d. F increased by a factor of 4 i.e 4*F

e. F reduces 3/4*(F)

Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is

F = k*(qA*qB)/d²

a. If qA is doubled therefore the force is doubled since they are directly proportional.

b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4

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e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,

3/4*F

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2 years ago

Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the b

Salsk061 [2.6K]

Answer:

Explanation:

GIVEN DATA:

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angle made by keisha toside building 30 degree

height of her friend monique is 1.5 m

from the figure

$\Delta ACB$

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= 14.376 + 1.5

$= 15.876 \simeq 16 m$

therefore option c is correct

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2 years ago

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