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Musya8 [376]
3 years ago
15

how does inertia explain the movement of your body when a car in which you are riding comes to a sudden stop? (8 sentences)

Physics
1 answer:
nydimaria [60]3 years ago
5 0
Your body continues to move unless stopped by the seatbelt. An object in motion will remain in motion. Since your body was already moving it will continue to.
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g A ball thrown straight up into the air is found to be moving at 7.94 m/s after falling 2.72 m below its release point. Find th
kati45 [8]

The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

<em>y</em> = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

and

<em>v</em> = <em>v</em>₀ - <em>g t</em>

where <em>v</em>₀ is its initial speed and <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

-2.72 m = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

Substitute this into the first equation and solve for <em>v</em>₀ :

-2.72 m = <em>v</em>₀ (<em>v</em>₀ + 7.94 m/s) /<em>g</em> - 1/2 <em>g</em> ((<em>v</em>₀ + 7.94 m/s)/<em>g</em>)²

-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>

2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)

-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²

<em>v</em>₀² = 9.73 m²/s²

<em>v</em>₀ = 3.12 m/s

3 0
3 years ago
If you break a magnet in two you get
notka56 [123]
Two separate magnets.
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3 years ago
Which of these results in kinetic energy of an object? (1 point)
Aleksandr [31]
Motion I’m pretty sure
7 0
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The bulb

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5 0
4 years ago
The wheel having a mass of 100 kg and a radius of gyration about the z axis of kz=300mm, rests on the smooth horizontal plane.a.
pickupchik [31]

Answer:

a) 20 rad/s

b) 6 m/s

Explanation:

b) Force acting on the wheel is 200 N

mass of the wheel is 100 kg

From Newton's second law of motion, F = m × a

Where F is the net force acting on the body

m is mass of the body

a is the acceleration of the body

By substituting the values we get, a = 2 m/s²

As acceleration is constant, we can use the below formula for calculating the final velocity of the object

v = u + a × t

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

u = 0 (∵ it starts from rest)

By substituting the values we get

v = 0 + 2 × 3 = 6 m/s

∴ Speed of center of mass after 3 seconds = 6 m/s

a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel

∴ Radius of the wheel = 300 mm

Torque acting on the wheel about axis of rotation = 300 mm × 200 N =

60 N·m

Torque = (Moment of inertia) × (angular acceleration)

Assuming that the mass of spokes of the wheel to be negligible,

Moment of inertia of the wheel about axis of rotation = 100 × 300² × 10^{-6} = 9 kg·m²

Then,

60 = 9 × (angular acceleration)

∴ angular acceleration ≈ 6·67 rad/s²

As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed

w_{f} = w_{i} + α × t

Where

w_{f} is the final angular velocity

w_{i} is the initial angular velocity

α is the angular acceleration

t is the time taken

w_{i} is 0 (∵ initially it starts from rest)

By substituting the values we get

w_{f} = 6·67 × 3 = 20 rad/s

∴ Angular velocity of the wheel after three seconds = 20 rad/s

3 0
4 years ago
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