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Ksju [112]
3 years ago
14

If you can't test an explanation through experiment,

Physics
1 answer:
Gala2k [10]3 years ago
6 0

Answer:

B

Explanation:

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jack JACKKK

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3. What is the main difference between a physical change and a chemical
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Answer:

B and D could both be right as they are quit similar.

Consider two rods of the same length and diameter,

Increasing the diameter of one would change the expansion qualities of that rod even though there would be no chemical changes,

However, leaving the physical appearance of both rods the same while applying a reactive substance (acid or something) to one of the rods would not necessarily change the physical appearance of that rod but could make a considerable change in the  physical properties of that rod.

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3 years ago
An electrical motor spins at a constant 1975.0 rpm. If the armature radius is 7.112 cm, what is the acceleration of the edge of
stealth61 [152]

Answer:

Option D is the correct answer.

Explanation:

Since value of angular acceleration is constant, the body has only centripetal acceleration.

Centripetal acceleration

               a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2

We have radius = 7.112 cm = 0.07112 m

Frequency, f = 1975 rpm = 32.92 rps

Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s

Substituting in centripetal acceleration equation,

              a=r\omega ^2=0.07112\times 206.82^2=3042.17m/s^2

Option D is the correct answer.                

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Which cells in the immune system identify pathogens and distinguish one pathogen from another?
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Answer:

T cell

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7 0
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To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug
Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

\tau = Fr

Where,

F = Force

r = Radius

Replacing we have that,

\tau = Fr

\tau = 21cm (\frac{1m}{100cm})* 550N

\tau = 11.55Nm

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2

I = 0.394kg\cdot m^2

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}

\alpha = \frac{11.55}{0.394}

\alpha = 29.3 rad/s^2

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians (\pi / 3), therefore

W = \tau \theta

W = 11.5* \frac{\pi}{3}

W = 12.09J

4 0
3 years ago
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