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den301095 [7]
3 years ago
7

There are some ways objects in motion are affected. _______________, or pushes and pulls, affect the motion of an object. ______

_________ is a force that pulls objects down to Earth. _______________ is a force that slows things down or can even make them stop. Did you know acceleration means more than the increase of an object’s speed? It also describes an object that is slowing down or changing _______________. Since acceleration is a change in speed or direction, that means, _______________ is a change in velocity
Physics
2 answers:
AysviL [449]3 years ago
8 0

Answer:

Forces, Gravity, Friction, Direction, Acceleration.

IgorC [24]3 years ago
4 0

Answer:

There are some ways objects in motion are affected. _Force_, or pushes and pulls, affect the motion of an object. _Gravity_ is a force that pulls objects down to Earth. _Friction_ is a force that slows things down or can even make them stop. Did you know acceleration means more than the increase of an object’s speed? It also describes an object that is slowing down or changing _velocity_. Since acceleration is a change in speed or direction, that means, _acceleration_ is a change in velocity.

Explanation:

This is kind of self explanatory, since these are just the definitions, so instead I'll give you examples so you can further understand these definitions.

FORCE: When you push a cart, and move a book from one desk to another.

GRAVITY: The force that causes a ball you throw in the air to come down again (notice how I mentioned force, which is why gravity is a type of force)

FRICTION: If you were walking in the park, you wouldn't be able to stop without friction.

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Give a example of frequency which an elephant can hear and a tiger cannot hear include a unit in ansawer
lisov135 [29]
P1.5.3 thats what i got........... hope its right not sure though
4 0
3 years ago
A car's acceleration is 3m/s2. If the car started at rest and it only took 10s for the car to reach this acceleration, what is t
Grace [21]

Answer:

30m/s

Explanation:

From law of motion equation

Vf= Vi + at

Where Vf= final velocity

Vi= initial velocity=0(the car started at rest)

a= acceleration= 3m/s2

t= time= 10s

Then substitute into the equation to get the final velocity.

Vf= 0+(10×3)

Vf= 30m/s

Hence, the car's final velocity is 30m/s

5 0
3 years ago
A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit
Amanda [17]

A) +1.67 rad/s^2

The angular acceleration of the wheel is given by

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

where

\omega_i = -6.00 rad/s is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

\omega_f = 4.00 rad/s is the final angular velocity (anticlockwise, so with a positive sign)

\Delta t= 6.00 s - 0=6.00 s is the time interval

Substituting into the equation, we find the angular acceleration:

\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant t_1 where the angular velocity becomes positive (so, \omega_i=0) and the time corresponding to the final instant t_2 = 6.0 s, where \omega_f = +6.00 rad/s. We can find this time interval by using

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

And solving for \Delta t we find

\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant t_1=0 when \omega_i=-4.00 rad/s) and the time corresponding to the instant in which the velovity becomes positive t_2, when \omega_f = 0 rad/s. We can find this time interval by using

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

And solving for \Delta t we find

\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{0-(-4.00 rad/s)}{+1.67 rad/s^2}=2.4 s

D) 5.6 rad

The angular displacement of the wheel is given by the equation

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where we have

\omega_i = -6.00 rad/s is the initial angular velocity of the wheel

\omega_f = 4.00 rad/s is the final angular velocity

\alpha=+1.67 rad/s^2 is the angular acceleration

Solving for \theta,

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{((+6.00 rad/s)^2-(-4.00 rad/s)^2}{2(+1.67 rad/s^2)}=5.6 rad

3 0
3 years ago
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Olin [163]
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4 0
3 years ago
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sattari [20]

momentum conservation

75x2 = 30 x her speed

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4 years ago
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