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AVprozaik [17]
3 years ago
8

A horizontal disk rotates about a vertical axis through its center. Point P is midway between the center and the rim of the disk

, and point Q is on the rim. If the disk turns with constant angular velocity, which of the following statements about it are true? (There may be more than one correct choice.)
A. P and Q have the same linear acceleration.
B. The angular velocity of Q is twice as great as the angular velocity of P.
C. The linear acceleration of P is twice as great as the linear acceleration of Q.
D. The linear acceleration of Q is twice as great as the linear acceleration of P.
E. is moving twice Q as fast as P.
Physics
1 answer:
hjlf3 years ago
6 0

Answer:

The answer is explained below.

Explanation:

All the point on the disk has same angular acceleration. Here, the point P is at the midway between the center and the rim of the disk and the point Q is at rim of the disk.

So, the distance of the point Q from the axis is twicee the distance of the point P from the axis.

<em>Rp - R</em>

<em>Rq - 2R</em>

The linear acceleration is

α2 - Rα

So, the linear acceleration of Q is twice as great as the linear acceleration of P.

The speed of the particle when it is in the circular motion depends on the radius of the particle.

In this case, the speed of point Q is twice the speed of point P.

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A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.
kow [346]

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, h_{max}, is given by the following equation;

h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}

Therefore, substituting the known values for the pebble, we have;

h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439

Therefore, the maximum height of the pebble projectile, h_{max} ≈ 6.4 m.

3 0
3 years ago
The relative humidity would be ________% if the actual water vapor in the air was 4 grams per cubic meter, the air's capacity to
AveGali [126]

Answer:

50%

Explanation:

Humidity is the amount water vapor present in the atmosphere.

Relative humidity is defined as the ratio of partial water vapor present in air to the actual water vapor at a particular temperature. It is expressed in percentage and the higher the percentage RH, the more the saturated water vapor present in the atmosphere and vice versa.

It is expressed mathematically as shown;

RH = actual water vapor in air/saturated water vapor × 100%

If the actual water vapor in the air was 4 grams per cubic meter and the air's capacity to hold water vapor was 8 grams per cubic meter

Actual water vapor = 4g/cm³

Air's water capacity (saturated water vapor) = 8g/cm³

RH = 4/8×100

RH = 50%

3 0
3 years ago
What happens to an object after being transported to three planets in the solar system?
alukav5142 [94]
The weight changes but the mass will stay the same.
6 0
3 years ago
An electric vehicle starts from rest and accelerates at a rate of 2.4 m/s2 in a straight line until it reaches a speed of 27 m/s
DENIUS [597]
A) use v=u+at for both

First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.

b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.

First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.
6 0
3 years ago
Which of the following would decrease in size during the contraction of a sarcomere? The width of the I-bands The width of the A
ANEK [815]

Hi!


The correct answer would be: the width of I-bands


The sacromere is the smallest contractile unit of striated muscles. These units comprise of filaments (fibrous proteins) that, upon muscle contraction or relaxation, slide past each other. The sacromere consists of thick filaments (myosin) and thin filaments (actin).


<em>Refer to the attached picture to clearly see the structure of a sacromere.</em>


<u>When a sacromere contracts, a series of changes take place which include:</u>

<em>- Shortening of I band, and consequently the H zone</em>

<em>- The A line remains unchanged</em>

<em>- Z lines come closer to each other (and this is due to the shortening of the I bands) </em>

The only changes that take place occur in the zones/areas in the sacromere (as mentioned), not in the filaments (actin and myosin) that make the up the sacromere; hence all other options are wrong.


Hope this helps!

8 0
3 years ago
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