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Harman [31]
2 years ago
15

Acceleration is best defined as the rate of change of __________ of an object. A. position B. force C. speed D. velocity 2. Whic

h of the following describes acceleration? A. s/m B. m/s C. m/s2 D. m2 3. Which of the following is the main difference between speed and velocity? A. Speed has direction and velocity B. Speed is measured over time C. Velocity has both speed and direction D. Speed is constant
Physics
2 answers:
Fittoniya [83]2 years ago
7 0
Velocity!
Hope this helps!
oksian1 [2.3K]2 years ago
3 0
Acceleration is the rate at which an object changes its' velocity. It's measured in m/s^2
D. 
C.
velocity is the rate of change of an objects position. 
C. velocity has speed and direction
(speed does not have direction)
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Veseljchak [2.6K]

Answer:

acceleration = v-u/ t

= 15-10/5

= 5/5

= 1 m/s2

Explanation:

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2 years ago
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The vector position of an object is given by r what is the torque acting on the object about the origin when a force f = (−12.5i
attashe74 [19]
Let the vector position of the object in the (x-y) plane be 
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3 years ago
What is helpful in calculating standard deviation?
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3 years ago
Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

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