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ElenaW [278]
3 years ago
6

According to the relationship between torque and angular acceleration, what happens when you have more torque (given a constant

rotational inertia)?
A. The angular acceleration increases.
B. The angular acceleration decreases.
C. The angular acceleration stays constant.
D. Not enough information to know.
Physics
1 answer:
gogolik [260]3 years ago
7 0

Answer:

The angular acceleration increases.

Explanation:

The relationship between the torque and angular acceleration is :

\tau=I\times \alpha

Where

I is the moment of inertia

\alpha is the angular acceleration

We can see that the torque is directly proportional to the angular acceleration. So, when we have more torque it means angular acceleration increases. Hence, the correct option is (A).

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What is free nitrogen
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Answer:

It is simply molecular nitrogen (N2). Nitrogen, in its molecular form, consists of two nitrogen atoms bound together with a tripple bond

Explanation:

4 0
3 years ago
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With a thunderstorm brewing, an electric field of magnitude 2.0 × 102 newtons/coulomb exists at a certain point in the earth’s a
hichkok12 [17]
(2.0 x 10² N/Coul) x (-1.6 x 10⁻¹⁹ Coul) = -3.2 x 10⁻¹⁷ N

The magnitude is 3.2 x 10⁻¹⁷ Newton.
3 0
3 years ago
A cylindrical, 0.500-m rod has a diameter of 0.02 m. The rod is stretched to a length of 0.501 m by a force of 3000 N. What is t
Salsk061 [2.6K]

Answer:

Y = 4.775 x 10⁹ Pa = 4.775 GPa

Explanation:

First, we calculate the stress on the rod:

stress = \frac{Force}{Area} = \frac{3000\ N}{\pi r^2}  \\\\stress = \frac{3000\ N}{\pi (0.01\ m)^2}\\\\stress = 9.55\ x\ 10^6\ Pa = 9.55 MPa\\

Now, we calculate the strain:

strain = \frac{Change\ in Length}{Original\ Length}\\\\strain = \frac{0.501\ m - 0.5\ m}{0.5\ m}\\\\strain =  0.002\\

Now, we will calculate the Young's Modulus (Y):

Y = \frac{stress}{strain}\\\\Y = \frac{9.55\ x\ 10^6\ Pa}{0.002} \\

<u>Y = 4.775 x 10⁹ Pa = 4.775 GPa</u>

6 0
3 years ago
If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the m
Xelga [282]

Answer:

The magnitude of the net force F₁₂₀ on the lid when the air inside the cooker has been heated to 120 °C is \frac{135.9}{A}N

Explanation:

Here we have

Initial temperature of air T₁ = 20 °C = ‪293.15 K

Final temperature of air T₁ = 120 °C = 393.15 K

Initial pressure P₁ = 1 atm = ‪101325 Pa

Final pressure P₂ = Required

Area = A

Therefore we have for the pressure cooker, the volume is constant that is does not change

By Chales law

P₁/T₁ = P₂/T₂

P₂ = T₂×P₁/T₁ = 393.15 K× (‪101325 Pa/‪293.15 K) = ‭135,889.22 Pa

∴ P₂ = 135.88922 KPa = 135.9 kPa

Where Force = \frac{Pressure}{Area} we have

Force = F_{120}=\frac{135.9}{A}N.

4 0
3 years ago
A block with a weight of 9.0 N is at rest on a horizontal surface. A 1.2 N upward force is applied to the block by means of an a
lesya [120]

Answer:

a)   9 - 1.2 = 7.8 N

b) Since the force exerted by the box is it's weight, it acts in a downward direction. So the box will exert a force downward (perpendicular to the horizontal surface).

Explanation:

A 9N block would exert 9N of normal force on the horizontal plane under normal conditions. But in this case, we have a spring taking away some of the force by applying an upward force on the box.

So the force exerted by the box on the surface would now be:

a)   9 - 1.2 = 7.8 N

b) Since the force exerted by the box is it's weight, it acts in a downward direction. So the box will exert a force downward (perpendicular to the horizontal surface).

8 0
3 years ago
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