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professor190 [17]
3 years ago
10

The mass of a certain neutron star is 2.5x10^30kg and the radius 7000m. what is the force of gravity on a 1kg object of the surf

ace of the neutron star
Physics
1 answer:
Makovka662 [10]3 years ago
8 0

Answer:

3.42N

Explanation:

*not too sure bc i left my physics notes at school so it might not be 100% accurate :p*

Use the equation: F = (GMm)/(r^2)

F = force of gravity

G = gravitational constant (6.7x10^-11)

M = mass1 (2.5x10^30kg)

m = mass2 (1kg)

r = radius (7000m)

Plug it in: F = ((6.7x10^-11)(2.5x10^30)(1)) / (7000^2)

F = (1.675x10^20) / (4.9x10^7)

F = 3.4183673x10^12

F = 3.42N

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Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 349 m/s,
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Answer:

Time period between the successive beats will be 0.1703 sec

Explanation:

We have given speed of the sound v = 349 m/sec

Wavelength of piano A\lambda _A=0.766m

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So frequency of piano A f_1=\frac{v}{\lambda _1}=\frac{349}{0.766}=455.61Hz

Frequency of piano B f_2=\frac{v}{\lambda _1}=\frac{349}{0.776}=449.74Hz

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So time period T=\frac{1}{f}=\frac{1}{5.87}=0.1703sec

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3 years ago
A scientist reports a measurement of the temperature of the surface of a newly discovered planet as negative 20 Kelvin. What con
jarptica [38.1K]

Answer:

The temperature of this newly discovered planet violates the third law of thermodynamic, there is a mistake in this value.

Explanation:

The third law of the Thermodynamic says:

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So we can say there is no thermodynamic system that has temperature values less than 0 K.

The conclusion of the report will be.

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I hope it helps you!

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2 years ago
Ron is on a Ferris wheel of radius 30 ft that turns counterclockwise at a rate of one revolution every 12 seconds. The lowest po
alexgriva [62]

Answer:

x = 30cos\frac{\pi}{6}t

y = 30sin\frac{\pi}{6}t + 45

Explanation:

1 full revolution is 2\pi. let \theta be the angle of Ron's position.

At t = 0. \theta = 0

one full revolution occurs in 12 sec, so his angle at t time is

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x = rcos\theta

y = rsin\theta

for r = 30 sec

x = 30cos\frac{\pi}{6}t

y = 30sin\frac{\pi}{6}t

however, that is centered at (0,0) and the positioned at time t = 0 is (30,0). it is need to shift so that the start position is (30,45). it can be done by adding to y

x = 30cos\frac{\pi}{6}t

y = 30sin\frac{\pi}{6}t + 45

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3 years ago
Equation for pressure at a depth H inside a fluid PLSS URGENTT
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Answer:

We begin by solving the equation P = hρg for depth h: h=Pρg h = P ρ g . Then we take P to be 1.00 atm and ρ to be the density of the water that creates the pressure.

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