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dezoksy [38]
3 years ago
5

Two objects of different mass rest on Earth. Which one experiences the greater acceleration due to gravity ?

Physics
1 answer:
Tamiku [17]3 years ago
3 0

the answer is a because if two objects of diffrent mass are on tge earth, the object that would recive greather acceleration on gravity, wouldbe none.

it would be none cuz since two objects are on the earth, it ha sthe same gravitational piull, from the other object.

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A rotating cylinder about 16 km long and 7.8km in diameter is designed to be used as a space colony.
Semmy [17]

Answer:

0.05 rad/s

Explanation:

7.8 km = 7800 m

For the residents inside the space cylinder to experience the same gravitation acceleration g = 9.81m/s2 on Earth, the centripetal acceleration must be the same as g

a = g

But centripetal acceleration is product of angular velocity squared and radius of rotation r

\omega^2r = 9.81

\omega^2 d/2 = 9.81

\omega^2 = \frac{2*9.81}{d} = \frac{19.62}{7800} = 0.0025

\omega = \sqrt{0.0025} = 0.05 rad/s

3 0
3 years ago
A force of 31 N south is used to accelerate an object uniformly from rest to 8 m/s in 22 s. What is the mass of the object?
harina [27]

this question is centered around newtons second law f=ma

First you need to find the acceleration

acceleration = \frac{8 - 0}{22}

that gives you 0.36 recurring(on both numbers) m/s²

mass= force ÷ acceleration

mass = \frac{31}{0.36recurring}

mass= 85.25 kg

if you want to check

f=ma

85.25 × 0.36 recurring (remember its on both numbers not just 6) = 31 N

4 0
3 years ago
Read 2 more answers
a sphere with a radius of 8cm carries a uniform volume charge density of 1.5 find the magnitude of the electric field
gladu [14]

Answer:

E = 5.65 x 10¹⁰ N/C

Explanation:

First we need to find the total charge on the sphere. So, we use the following formula for that purpose:

q = \sigma V\\

where,

q = total charge on sphere

V = Volume of Sphere = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.08\ m)^3 =  0.335\ m^3

σ = volume charge density = 1.5 C/m³

Therefore,

q = (0.335\ m^3)(1.5\ C/m^3) \\q = 0.502 C

Now, we use the following formula to find the electric field due to this charged sphere:

E = \frac{kq}{r^2}

where.

E  = Electric Field Magnitude = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

r = radius of sphere = 8 cm = 0.08 m

Therefore,

E = \frac{(9\ x\ 10^9\ Nm^2/C^2)(0.502 C)}{(0.08\ m)^2}\\\\

<u>E = 5.65 x 10¹⁰ N/C</u>

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%
Verdich [7]

Answer:

1.7 × 10^1^0

Explanation:

Knowing that, the volume of the sphere is given by, v=\frac{4}{3}\pi ^3

Thus, the fractional change in volume is given by,

=3 × \frac{0.02}{100}=\frac{0.06}{100}

Pressure at the bottom of the sea is,

Δp =pgh = 10^3 × 10 × 10^3=10^7 pa.

Knowing that,

Bulk modulus: \frac{10^7 * 100}{0.06}=\frac{10^9}{6*10^-^2}=\frac{10^1^1}{6}

B =\frac{10}{6}*10^1^0

B = 1.7*10^1^0N/M^2

Answer = 1.7 × 10^1^0

[RevyBreeze]

5 0
2 years ago
Read 2 more answers
An object on the surface of the earth weighs 90 lb. At two earth radii above the surface, It will welgh:
vredina [299]
I have no explanation, but saw this pop up on a test paper and the answe was 10lb
4 0
3 years ago
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