All categories

3 years ago

Where would a bowling ball and a napkin fall with the same acceleration?

2 answers:

5 0

That would happen at any place where they don't have to
fall through air or anything else.

Examples:

-- on the moon
-- on an asteroid
-- on a comet
-- on Mercury
-- on Earth, in a vacuum chamber with all the air pumped out of it

deff fn [24]3 years ago

4 0

Answer: In absence of any drag force- in space/in vacuum

Explanation:

A Bowling ball and a napkin would fall with the same acceleration in space where there is vacuum. Basically where there is no drag force and the two objects are freely falling under gravity. The two objects would fall with same acceleration due to gravity. The drag force depends on mass of the object and surface area and it acts in the opposite direction of motion thus changing the net acceleration of the body.

You might be interested in

2. Determine the current in a 120-watt bulb plugged into a 120-volt outlet.

Sergio039 [100]

Answer:

1 ampere

Explanation:

Power = VI

V = voltage

I = current

Given

Power = 120 watts

V = 120 volts

Therefore

120 = 120 x I

Divide both sides by 120

120/120 = 120/120 x I

1 = I

I = 1 ampere

8 0

3 years ago

Use your periodic table to answer the following question.

Pavel [41]

Helium has only 2 electrons in it's valence shell

So, option D is your answer.

Hope this helps!

6 0

3 years ago

What is the best summary for this cartoon?

Paha777 [63]

Answer:

C: World War I resulted from a chain reaction involving many nations.

Explanation:

Just did the questions and got it right

6 0

3 years ago

Read 2 more answers

Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. Take T = 1.3 kN·m. De

My name is Ann [436]

Answer:

τ (bc. max) =25.37 MPa

Explanation:

From the question, T = 1.3Kn.m;

(G = 77.2 GPa) and from the image of this solid shaft system i attached;

d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m

So ΣT = 0 → Ta + Tc = 1.3Kn.m

So the system is statically indeterminate.

Let's check at the equation that makes it compatible ;

ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0

[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =

ΣT = 0 and T(bc) = T(c)

ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m

Now,

[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0

So, T(c) = 273374 Nmm = 273.374Nm

T(a) = 1300Nm - 273.374Nm = 1026. 63Nm

From the beginning we saw that;

T(bc) = T(c) = 273.374Nm

Now let's find the maximum shear stress in shaft BC;

τ (bc. max) = {τ (bc) x r(bc)} / J(bc)

τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707

= 25.37 MPa

3 0

3 years ago

As energy moves out from the sun’s core toward its surface, it first travels by _______, then by _______, and then by _______.

7nadin3 [17]

Answer:

neutrinos and seismic vibrations

Explanation:

6 0

3 years ago