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stich3 [128]
3 years ago
6

When a force is acting at axis of rotation,why torque is zero?

Physics
1 answer:
Vikki [24]3 years ago
4 0

Explanation:

see, torque=force × perpendicular distance

...that perpendicular distance is between axis of rotation and the point where force acts... so in above's case perpendicular distance is zero... so the torque is zero!

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During which process does layer upon layer of sediment build up, exerting pressure on the layers below?
irinina [24]
I think the answer is b. compaction
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3 years ago
The x vector component of a displacement vector has a magnitude of 86.2 m and points along the negative x axis. The y vector com
Flauer [41]

Answer:

(a) Magnitude of Vector = 207.73 m

(b) Direction = 65.48°

Explanation:

(a)

The formula to find out the magnitude of a resultant vector with the help of its x and y components is given as follows:

Magnitude\ of\ Vector = \sqrt{d_{x}^{2} + d_{y}^{2}} = \sqrt{(86.2\ m)^{2} + (189\ m)^{2}}\\\\

<u>Magnitude of Vector = 207.73 m</u>

(b)

For the direction of the vector we have the formula:

Direction = tan^{-1}(\frac{y}{x})\\\\Direction = tan^{-1}(\frac{189\ m}{86.2\ m})\\\\

<u>Direction = 65.48°</u>

4 0
3 years ago
For thermometers to read temperatures correctly they must be
madam [21]
For thermometers to read temperatures correctly, they must be immersed in the substance being measured while reading the temperature. When they are taken out the substance, there is a possibility that the fluid inside would change its level due to the sudden change in the temperature. Thus, giving you an erroneous reading of the temperature. 
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Read 2 more answers
I need to know which is represented as a element
Tema [17]

Answer:

the yellow one

Explanation:

2 of the same elements resolute as the same element

6 0
3 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
3 years ago
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