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Volgvan
3 years ago
11

Which is an example of qualitative data?

Physics
1 answer:
LiRa [457]3 years ago
6 0
The answer is C)the rating that the golfers give Callaway clubs
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A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a
blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

E=7*10^{9}N/C

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field  is mathematically given as

E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\

E=7*10^{9}N/C

For more information on this visit

brainly.com/question/21811998

4 0
3 years ago
Motion and force of physics. Plz answer.​
Whitepunk [10]

Explanation:

a. " for every action there is an equal and opposite reaction".

b. The electric fan does not stop moving just after the switch turns off because of rational inertia force.

c. The force applied to first vehicle is 120N.

d. In my view it doesn't support the law of conservation of momentum. Momentum of 1 and 2 object before the collision is equal to the total momentum of two object after collision.

7 0
3 years ago
In Wagner’s opera Das Rheingold, the goddess Freia is ransomed for a pile of gold just tall enough and wide enough to hide her f
Vesna [10]

Answer:

Price=$7×10⁷

Explanation:

Step 1: Estimate the volume of the pile,

Step 2: Multiply it by the density to get its mass

Step 2: Then multiply the mass by the price per gram to get the total price

So

The average pile dimensions are=45.7×45×172.7

V=3.6*10^{5}cm^{3}\\  m_{g}=V_{p}=3.6*10^{5}*19.3=7*10^{6}g\\

Price=m×$10

Price=(7×10⁶)×$10

Price=$7×10⁷

5 0
3 years ago
What are the Standard System of unit​
Natalija [7]

Answer:

SI

Explanation:

6 0
3 years ago
A cosmic ray travels 60.0 km through the earth's atmosphere in 500 μs , as measured by experimenters on the ground. You may want
zavuch27 [327]

Answer:

t' = 458.26 μs

Explanation:

given,

distance traveled by the cosmic ray = 60 Km

time = 500 μs

time taken for the journey according to cosmic ray = ?

now,

 v =\dfrac{distance}{time}

 v =\dfrac{60\times 10^3}{500\times 10^{-6}}

      v = 1.2 x 10⁸ m/s

according to time dilation formula

t' = t\sqrt{1-\dfrac{v^2}{c^2}}

t' = 500\sqrt{1-\dfrac{(1.2\times 10^8)^2}{(3\times 10^8)^2}}

     t' = 500 x 0.9165

    t' = 458.26 μs

6 0
3 years ago
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