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3 years ago

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A bowling (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as uniform

spheres. What is the magnitude of the maximum gravitational force that each can exert on the other?

1 answer:

cestrela7 [59]3 years ago

7 0

Hope this helps you!

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How much power is required to do 40 J of work on an object in 5 seconds?

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3 years ago

Which circuit has the largest equivalent resistance?.

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The series circuit has the higher equivalent resistance, but to find the larger equivalent resistance you have to use Ohm's law (V2=I2R2), the power dissipated by the resistor can also be found using P2=I22R2=V22R2. To find the equivalent resistance of the circuit, notice that the parallel connection of R2 and R3 is in series with R1, so the equivalent resistance is Req=R1+(1R2+1R3)−1=1.00Ω+(16.00Ω+113.00Ω)−1=5.10Ω.

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2 years ago

AysviL [449]

Answer:

The wind will blow from the higher pressure over the water to lower pressure over the land causing the sea breeze. The sea breeze strength will vary depending on the temperature difference between the land and the ocean. At night, the roles reverse. The air over the ocean is now warmer than the air over the land.

Sea breezes occur during hot, summer days because of the unequal heating rates of land and water. During the day, the land surface heats up faster than the water surface. Therefore, the air above the land is warmer than the air above the ocean. Now, recall that warmer air is lighter than cooler air.

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4 years ago

Which object would most like absorb sound?

vredina [299]

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3 years ago

Read 2 more answers

While visiting the Albert Michelson exhibit at Clark University, you notice that a chandelier (which looks remarkably like a sim

Fudgin [204]

Answer:

a) 0,1613 Hz

b) 1,01342 rad/sec

c) 9.5422 m

d) $9.4314 m/sec^2$

Explanation:

In the Albert Michelson exhibit at Clark University, we know the period of oscillation of the chandelier is T = 6.2 seconds

The chandelier will be modeled as a simple pendulum

a) Since the frequency is the reciprocal of the period, we have

$f=\frac{1}{T}=\frac{1}{6.2sec}=0,1613 Hz$

b) The angular frequency is computed as

$w=2\pi f=2\pi (0,1613) = 1,01342\ rad/sec$

c) The period of a simple pendulum is given by

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

Where L is its length and g is the local acceleration of gravity (assumed 9.8 $m/sec^2$ for this part)

We want to know the length of the pendulum, so we isolate L

$L=\frac{T^2g}{4\pi^2}$

$L=\frac{(6.2)^2(9.8))}{4\pi^2}=9.5422 m$

d) While hanging out in J.J. Thompson’s House O’ Blues, the new period is 6.2+0.11=6.32 sec. Since the chandelier is the very same (same length), we can assume the gravity is slightly different. We use again the formula for T, but now we'll isolate g as follows

$g=\frac{4\pi^2L}{T^2}=\frac{4\pi^2(9.5422m))}{(6.32sec)^2}$

Which results

$g=9.4314\ m/sec^2$

8 0

4 years ago