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yaroslaw [1]
3 years ago
12

Which planet is small, with a rocky surface and an atmosphere?

Physics
2 answers:
Alex_Xolod [135]3 years ago
7 0

Answer:

Venus.

Explanation:

You can look up the third smallest planet with these too, and I took the unit test review and got it correct. So I hope it helps! also, if you ever have a question and need it quicker, you can just copy and paste it in google;-;

bija089 [108]3 years ago
3 0

Answer:

venus

Explanation:

its small and still its the hottest since it has carbon 4 oxide in its atmosphere

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Hich of the following equations is balanced correctly?
Alexus [3.1K]
A. the carbons are unbalanced B. the hydrogens are unbalanced. D. the chlorines are unbalanced. That leaves C. to be correctly balanced.
3 0
3 years ago
Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m
mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0
3 years ago
Consider the table below. Which of the following data sets depict an accelerating object? Mark all that apply.
Gre4nikov [31]

Answer:

A and B

Explanation:

The data sets that depict an accelerating object is Data Set A & Data Set B.

The both data sets show that the body is accelerating. Also, they show that the body started from rest (0m/s) at a 0sec.

Data Set A shows a non-constant acceleration which has changing amount of velocity with change in time. While Data Set B shows a constant acceleration which has constant amount of velocity with change in time.

7 0
3 years ago
The relationship between linear velocity and angular velocity
bonufazy [111]

In linear motion , when a body moves with uniform velocity , in time t , its linear displacement will be ;

S = r∅     S = vt

r∅ =  vt

r.∅ / t = v

As

v = rw

where ∅ = 90° is the angle between between radius vector r  and angular velocity w (omega )

In case ∅ ≠ 90° , we can write v = r w sin∅

It gives us v = w× r


7 0
3 years ago
What is the resistance of a 3.5 m copper wire (Rho= 1.7x10-8 Ohm·m) that 1 point
VikaD [51]

Answer:

(D)

Explanation:

Given :

l=3.5 m

A=5.26*10^{-6} m^{2}

p=1.7*10^{-8}  ohm.m

Resistance can be calculated as :

R=p\frac{l}{A} \\R=1.7*10^{-8} \frac{3.5}{5.26*10^{-6} }

R=\frac{5.95*10^{-2} }{5.26} \\R=1.13*10^{-2}

Resistance of the wire will be 1.1×10^{-2} ohms

Option D is correct

4 0
2 years ago
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