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RideAnS [48]
3 years ago
6

The gravitational constant G was first measured accurately by Henry Cavendish in 1798. He used an exquisitely sensitive balance

to measure the force between two lead spheres whose centers were 0.29 m apart. One of the spheres had a mass of 148 kg, while the mass of the other sphere was 0.63 kg.What was the ratio of the gravitational force between these spheres to the weight of the lighter sphere?
Physics
1 answer:
djverab [1.8K]3 years ago
7 0

Answer:

1.198*10^{-8}

Explanation:

The law of gravitation has to be used in this question along with Newton's second law

Ratio = \frac{Gravitational force}{weight of the smaller sphere}= \frac{\frac{6.674*10^{-11*148*0.63} }{0.29^{2} } }{0.63 kg * 9.8} = 1.198*10^{-8}

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Astrophysics is a term used to describe knowledge about the universe. 
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3 years ago
A car drives on a circular road of radius R. The distance driven by the car is given by = + [where a and b are constants, and t
pishuonlain [190]

Answer:

The question is incomplete, the complete question is "A car drives on a circular road of radius R. The distance driven by the car is given by d(t)= at^3+bt [where a and b are constants, and t in seconds will give d in meters]. In terms of a, b, and R, and when t = 3 seconds, find an expression for the magnitudes of (i) the tangential acceleration aTAN, and (ii) the radial acceleration aRAD3"

answers:

a.18a m/s^{2}

b. a_{rad}=\frac{(27a +b)^{2}}{R}

Explanation:

First let state the mathematical expression for the tangential acceleration and the radial acceleration.

a. tangential acceleration is express as

a_{tan}=\frac{d|v|}{dt} \\

since the distance is expressed as

d=at^{3}+bt

the derivative is the velocity, hence

V(t)=\frac{dd(t)}{dt}\\V(t)=3at^{2}+b\\

hence when we take the drivative of the velocity we arrive at

a_{tan}=\frac{dv(t)}{dt}\\ a_{tan}=6at\\t=3 \\we have \\a_{tan}=18a m/s^2

b. the expression for the radial acceleration is expressed as

a_{rad}=\frac{v^{2}}{r}\\a_{rad}=\frac{(3at^{2} +b)^{2}}{R}\\t=3\\a_{rad}=\frac{(27a +b)^{2}}{R}

4 0
3 years ago
A train is moving in the positive direction down a track. First the train speeds up, and then it slows down. What is its acceler
tatuchka [14]
C.

The train first sped up, giving it a positive acceleration in the beginning. This eliminates D since that choice states that it begins with a negative acceleration. This also eliminates B since that choice states that the train only had a negation acceleration.

Next, the train slows down, giving it a negative acceleration. We’re looking for the answer choice that starts with a positive acceleration and ends with a negative one. This makes C the correct answer. Hope this helps!
8 0
3 years ago
Read 2 more answers
The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr
WARRIOR [948]
In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
y_n= \frac{n \lambda D}{a} (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
\lambda is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem, 
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while the distance between the first and the fifth minima is
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inna [77]

Answer:

Your real dad according to Science and your DNA codes.

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