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Alika [10]
3 years ago
6

What is the mass of an object that has an acceleration of 2.63 m/s^2 when an unbalanced force of 112 N is applied to it?

Physics
1 answer:
shusha [124]3 years ago
6 0
112/2.63= 42.586
42.586 is your answer I need 20 characters
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Identify each process labeled in the diagram
marishachu [46]
I am going to need a picture for this question
3 0
4 years ago
A copper telephone wire has essentially
Lunna [17]

Answer:

128.21 m

Explanation:

The following data were obtained from the question:

Initial temperature (θ₁) = 4 °C

Final temperature (θ₂) = 43 °C

Change in length (ΔL) = 8.5 cm

Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)

Original length (L₁) =.?

The original length can be obtained as follow:

α = ΔL / L₁(θ₂ – θ₁)

17×10¯⁶ = 8.5 / L₁(43 – 4)

17×10¯⁶ = 8.5 / L₁(39)

17×10¯⁶ = 8.5 / 39L₁

Cross multiply

17×10¯⁶ × 39L₁ = 8.5

6.63×10¯⁴ L₁ = 8.5

Divide both side by 6.63×10¯⁴

L₁ = 8.5 / 6.63×10¯⁴

L₁ = 12820.51 cm

Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

12820.51 cm = 12820.51 cm × 1 m / 100 cm

12820.51 cm = 128.21 m

Thus, the original length of the wire is 128.21 m

5 0
3 years ago
Rainfall occurs on the earth due to effect of gravity​
Anna [14]

<em>Answer:</em>

<h3><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>True</em></h3>

  • <em>Because </em><em>Gravity is the force of attraction between two objects, and Earth's gravity pulls matter downward, toward its center. It pulls precipitation down from clouds and pulls water downhill. Gravity also moves air and ocean water. ... Gravity pulls denser air and water downward, forcing less dense air and water to move upward.</em>

<em>Carryonlearning</em>

3 0
3 years ago
A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.
Anestetic [448]

Answer:

\omega=0.37 [rad/s]  

Explanation:

We can use the conservation of the angular momentum.

L=mvR

I\omega=mvR

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

(I_{proffesor - stool}+mR^{2})\omega=mvR

Now, we just need to solve it for ω.

\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}

\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}      

\omega=0.37 [rad/s]  

I hope it helps you!

5 0
4 years ago
What mass of water must evaporate from the skin of a 70.0 kg man to cool his body 1.00 ∘C? The heat of vaporization of water at
romanna [79]

Answer:

100 cc

Explanation:

Heat released in cooling human body by t degree

= mass of the body x specific heat of the body x t

Substituting the data given

Heat released by the body

= 70 x 3480 x 1

= 243600 J

Mass of water to be evaporated

= 243600 / latent heat of vaporization of water

= 243600 / 2420000

= .1 kg

= 100 g

volume of water

= mass / density

= 100 / 1

100 cc

1 / 10 litres.

6 0
3 years ago
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