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Yanka [14]
3 years ago
13

An object with a mass of 375g is moving with a constant velocity. It has a force of -20 N applied to it. Determine the accelerat

ion of the object. What is happening to the object?
Physics
1 answer:
olga55 [171]3 years ago
8 0

Answer:

Explanation:

Convert the mass to kg:

375g = 375/1000kg = 0.375kg

F = ma

-20 = 0.375a

a = -20/0.375

a = -53

The object is accelerating at 53m/s/s backwards assuming that the forward motion is positive.

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Yes. Acceleration means any change in speed or direction of motion. When an object coasts in a circular orbit at constant speed around the Earth, its direction is constantly changing. The acceleration is "CENTRIPETAL", which points toward the center of the circle.
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If the volume on your TV is low, turning the volume up one click of the remote control will make the TV seem louder than if the
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From what we know, we can confirm that this ratio (turning up the volume by one click relative to the TV's overall volume) can be quantified as the Weber fraction.

<h3>What is the Weber fraction?</h3>

This fraction describes the ratio needed for change to a stimulus in which the change is just barely noticeable. This question is a prime example in that it seeks to find out just how low of a difference is needed in TV volume in order for the difference to be noticeable.

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7 0
2 years ago
What is the difference between m/s and m.s.​
vfiekz [6]

Answer

m/s rate of change of dispalcement per sec. ie velocity

m/s^2 is (m/s)/s ie rate of change of velocity per sec. ie accelerationplanation:

5 0
3 years ago
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Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i
mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2}  =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}

\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad

w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}

The velocity at the contact point is equal to:

v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}

v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}

Matching both expressions:

1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s

b) The time during which the disks slip is:

t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s

a) The angular acceleration of each disk is

\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)

\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)

6 0
3 years ago
A shark travels with an average velocity of 12 m/s s. How long (time) would it take the shark to swim 42 m at that velocity
DerKrebs [107]

Answer:

Explanation:

d = 42 meters

v = 12 m/s

t = ?

t = d/v

t = 42 / 12

t = 3.5 seconds. That's awfully fast.

7 0
3 years ago
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