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erastovalidia [21]
3 years ago
13

What two factors affect the rate of acceleration of an object?

Physics
1 answer:
Masja [62]3 years ago
5 0

For help with this answer, we look to Newton's second law of motion:

     Force = (mass) x (acceleration)

Since the question seems to focus on acceleration, let's get
'acceleration' all alone on one side of the equation, so we can
really see what's going on.

Here's the equation again:

                                                 Force = (mass) x (acceleration)

Divide each side by 'mass',
and we have:                            Acceleration = (force) / (mass) .

Now the answer jumps out at us:  The rate of acceleration of an object
is determined by the object's mass and by the strength of the net force
acting on the object.


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State which of the three materials would allow the thermometer to measure the
nignag [31]

Answer:

i dont know

Explanation:im sorry to do this to you but you dont have to watch ads if you answer questions

6 0
1 year ago
A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2,
Stella [2.4K]

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

Explanation:

The free-body  diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

  • the peripheral velocity that is directed downward (-V_y) along the y-axis
  • the linear velocity (V_x) that is directed along the x-axis

Now;

V_x = \frac{d}{dt}(12t^3+2) = 36 t^2

V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s

Also,

-V_y = R* \omega

where \omega(angular velocity) = \frac{d\theta}{dt} = \frac{d}{dt}(8t^4)

-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

3 0
3 years ago
Which state of matter would be described as a highly energized charge particles with moving extremely fast
Oksanka [162]
The correct answer is plasma
6 0
3 years ago
Read 2 more answers
Its not b...................
vladimir1956 [14]

sir what's the question you have all you wrote is it's not b

4 0
3 years ago
Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted
galina1969 [7]

Answer:

1) When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}

Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}

Where:

f = Focal length of the mirror = R/2

d_{i} = Image distance from the mirror

d_{o} = Object distance from the mirror

h_{i} = Image height

h_{o} = Object height

d_{o} is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

d_{i} is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When d_{o} < d_{i} (hence  d_{o} < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When d_{o} > d_{i} (and d_{o} < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0
3 years ago
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