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Alexxx [7]
1 year ago
7

Cavendish used the attraction between large and small lead balls to measure an experimental value for the gravitational constant

, G. Is it true or false?
Physics
1 answer:
irina1246 [14]1 year ago
8 0

The gravitational constant was experimentally measured by W Cavendish using the attraction between big and small lead balls. is true

The correct answer is true

<h3>How do you define gravitational constant?</h3>

the strength of gravity. a factor in use in Newton's gravity law to relate the strength of the gravitational pull between two bodies with their masses and distance from one another. 6.67259 X 10-11 newtons per square kilogram is roughly the gravitational constant. G is its identifier.

<h3> where is the strongest gravity is?</h3>

The gravitational pull of the earth is greatest near sea level, normally, and weakens as you get further from the center, such as to the summit of Mt. Everest. Because the obloid earth was slightly wider, but only by a minor ratio, the gravity just at poles is stronger than that at the equator.

To know more about gravitational constant visit:

brainly.com/question/858421
#SPJ9

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dusya [7]
Displacement equals (Velocity times Time) plus half times the (acceleration times time squared). =. (48 * 4) + 1/2 * (12 *12^2) = 288meters
3 0
3 years ago
The density of gold is 19.3 g cm³. What is the mass of a bar of gold in Kg that measures 6 cm x 4 cmx to 2 cm ?
klemol [59]

Answer: 0.9264 kg

Explanation: [I'll use "cc" for cubic centimeter, instead of cm^3.

The volume is 6cm*4cm*2cm = 48 cm^3 (cc).

Density of Au is 19.3 g/cc

Mass of gold = (48 cc)*(9.3 g/cc) = 926.4 grams Au

1 kg = 1,000 g

(926.4 grams Au)*(1 kg/1,000 g) = 0.9264 kg, 0.93 kg to 2 sig figs

At gold's current price of $57,500/kg, this bar is worth $53,268. Keep it hidden from your lab partner (and instructor).

3 0
3 years ago
A ball is thrown with a velocity of 35 meters per second at an angle of 30° above the horizontal. which quantity has a magnitude
enot [183]
The quantity that has a magnitude of zero when the ball is at the highest point in its trajectory is the vertical velocity.

In fact, the motion of the ball consists of two separate motions:
- the horizontal motion, on the x-axis, which is a uniform motion with constant velocity v_x=v_0 cos 30^{\circ}, where v_0=35 m/s
- the vertical motion, on the y-axis, which is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2 directed downwards, and with initial velocity v_y=v_= sin 30^{\circ}. Due to the presence of the acceleration g on the vertical direction (pointing in the opposite direction of the initial vertical velocity), the vertical velocity of the ball decreases as it goes higher, up to a point where it becomes zero and it reverses its direction: when the vertical velocity becomes zero, the ball has reached its maximum height. 
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Through which of the following can magnetic force pass
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All of the above hope this helps :)
8 0
4 years ago
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A car travels 24 m East in 3 s. what is the velocity of the car?
Bad White [126]

Well, since you noted the direction of travel is “due east”, that clearly means the question is asking about a frame of reference that included planetary rotation into the mix. At the equator, that’s about 460m/s in the eastward direction. Clearly, at the north pole, that’s zero—you’re effectively stationary. So, let’s assume you’re somewhere between the equator and the pole, so we’ll take the average of the two, and say the earth is contributing 230m/sec of eastward velocity. So, at time t=0s, your velocity is 230m/sec; at time t=30s, your velocity is 254m/sec. Thus, you plug those into the formula, and you get (254–230)/(30–0), or 0.8m/sec^2.

Now, if the question was to consider this from a larger frame of reference, we’d also have to take the rotation of the earth around the sun into consideration, which is about 107,000km/hr, or about 29,722m/sec. The problem is that we don’t know if we need to *add* that to the rotational velocity of the earth, and motion of the car, or *subtract* it; that all depends on whether the side of the earth that the car is on is facing the sun, or away from the sun. If we assume that sane people do math experiments on their cars only when the sun is shining, then we need to add the velocity in as well; so we get 29,722+230+24, or (29,976–29,952)(30–0). However, that only works if you do your vehicular calculations in the daytime. If, on the other hand, you’re a dark, brooding vigilante who only comes out after darkness falls to drive around, then we need to adjust our calculations to account for the fact that you’re now going retrograde with respect to the sun. Thus, the calculation would become 29,722–230–24, or (29468–29492)/(30–0), or -0.8m/sec^2.

HOPE THIS HELPS :)

8 0
3 years ago
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