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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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If a 2 kg object is falling at 3 m/s at what rate is gravity working on the object
777dan777 [17]

Answer:

+9.8m/s^2

Explanation:

The rate of gravity of the object is constant thriughout the surface of the earth.

For falling object, the rate of gravity is positive since the body is coming down (falling)

The rate of gravity is negative if the body is going up

The constant value for acceleration due to gravity is 9.8m.s^2

Since the object is falling, hence the acceleration due to gravity is positive.

Rate of gravity working on the object will be +9.8m/s^2

4 0
3 years ago
A crate resting on a rough horizontal floor is to be moved horizontally. The coefficient of static friction is 0.36. To start th
luda_lava [24]

Answer:

Explanation:

The direction of force will be in upward direction making an angle of θ with the vertical .

Reaction force R = mg - F cosθ

Friction force = μR

= .36 (mg - F cosθ )

Horizontal component of applied force

= F sinθ

For equilibrium

F sinθ = .36 (mg - F cosθ)

F sinθ + .36 F cosθ =.36 mg

F (sinθ + .36 cosθ) = .36 mg

F R( cosδsinθ +sinδ cosθ) = .36 mg ( Rcosδ = 1 . Rsinδ= .36 )

F R sin( θ+δ )  = . 36 mg

F = .36 mg / Rsin( θ+δ )

For minimum F , sin( θ+δ ) should be maximum

sin( θ+ δ ) = sin 90

θ+ δ  = 90

Rsinδ / Rcosδ  = .36

δ = 20⁰

θ = 70⁰ Ans

5 0
3 years ago
I tried looking at formulas, if you know the formula can you list it?
serg [7]

Answer:y=mx+b

Explanation:

y=mx+b

5 0
3 years ago
Read 2 more answers
Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r
cricket20 [7]

Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

                                         E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

                                 E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

                                  E_net = E_2 + E_4

                                  E_2 = E_4

                                  E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

                                  r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

                                  r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

                                  E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

                                 E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

Hence,

                                  E_net = 2*E_part(a)*cos(Q)

- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:

                                  Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

- Now, compute the net electric field E_net:

                                  E_net = 2*(99.518)*cos(27.12)

                                  E_net = 177.151 N / C

               

5 0
3 years ago
Most of the mass of the milky way exists in the form of.
Andrews [41]

Answer: Dark matter.

Explanation: Hope it helps :)

7 0
2 years ago
Read 2 more answers
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