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marusya05 [52]
3 years ago
15

The rate of spin of a disk is decreasing. A point on the disk that is near the rim of the disk will have...

Physics
1 answer:
NNADVOKAT [17]3 years ago
8 0
At a point near the rim of the disk, it will have a<span> non-zero radial acceleration and a zero tangential acceleration. Also known as centripetal acceleration, radial acceleration takes place along the radius of the disk. On the other hand, the tangential acceleration is along the path of disk's motion.</span>
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In what factors sound can be heard​
MariettaO [177]

Explanation:

Loudness of sound is a measure of response of sound to our ear. Loudness of sound is not simply the energy reaching the human ear, but it also tells about the sensitivity of human ear detecting this energy. Loudness of sound is measured in decibel (dB). As energy reaching the ear depends on square of amplitude, loudness of sound depends on various factors namely,

(i) Amplitude of sound waves

(ii) Sensitivity of ear

(iii) Distance from the source of the sound and the listener.

6 0
3 years ago
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What was the name of the voyages taken by Zheng he during the Ming dynasty on behalf of China
telo118 [61]

Answer:

Ming treasure voyages

Explanation:

6 0
3 years ago
An object accelerates at 32 m/s² when a force of 71 N is applied to it. What is the object’s mass? show your work
amid [387]

Answer:

Its  answer is 2.21 kg.

Explanation:

F =m × a

71 = m × 32

71 ÷ 32 = m

2.21 kg = m

6 0
2 years ago
A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple
aliina [53]

Answer:

T= 1 s

Explanation:

Given that

When x=  cm ,T= 1

we know that time period of spring mas system given as

T=2\pi \sqrt{\dfrac{m}{k}}

T= Time period

m= mass

k=spring constant

So from above equation we can say that time period of system does not depends on the value of x.

So when x= 10 cm ,still time period will be 1 s.

T= 1 s

5 0
3 years ago
A basketball player is 4.22 m from
max2010maxim [7]

Answer: The height above the release point is 2.96 meters.

Explanation:

The acceleration of the ball is the gravitational acceleration in the y axis.

A = (0, -9.8m/s^)

For the velocity we can integrate over time and get:

V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))

for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)

P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)

now, the time at wich the horizontal displacement is 4.22 m will be:

4.22m = 9.20*cos(69°)*t

t = (4.22/ 9.20*cos(69°)) = 1.28s

Now we evaluate the y-position in this time:

h =  -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m

The height above the release point is 2.96 meters.

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3 years ago
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