The gravitational potential energy of an object is due to

The chart shows data for a moving object.

yuradex [85]

Answer:

Explanation:

Let’s begin with a particle with an acceleration a(t) is a known function of time. Since the time derivative of the velocity function is acceleration,

\[\frac{d}{dt}v(t)=a(t),\]

we can take the indefinite integral of both sides, finding

\[\int \frac{d}{dt}v(t)dt=\int a(t)dt+{C}_{1},\]

where C1 is a constant of integration. Since

\[\int \frac{d}{dt}v(t)dt=v(t)\]

, the velocity is given by

\[v(t)=\int a(t)dt+{C}_{1}.\]

Similarly, the time derivative of the position function is the velocity function,

\[\frac{d}{dt}x(t)=v(t).\]

Thus, we can use the same mathematical manipulations we just used and find

\[x(t)=\int v(t)dt+{C}_{2},\]

where C2 is a second constant of integration.

We can derive the kinematic equations for a constant acceleration using these integrals. With a(t) = a a constant, and doing the integration in (Figure), we find

\[v(t)=\int adt+{C}_{1}=at+{C}_{1}.\]

If the initial velocity is v(0) = v0, then

\[{v}_{0}=0+{C}_{1}.\]

Then, C1 = v0 and

\[v(t)={v}_{0}+at,\]

which is (Equation). Substituting this expression into (Figure) gives

\[x(t)=\int ({v}_{0}+at)dt+{C}_{2}.\]

Doing the integration, we find

\[x(t)={v}_{0}t+\frac{1}{2}a{t}^{2}+{C}_{2}.\]

If x(0) = x0, we have

\[{x}_{0}=0+0+{C}_{2};\]

so, C2 = x0. Substituting back into the equation for x(t), we finally have

\[x(t)={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2},\]

3 0

4 years ago

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Click on the reset button, and stack one 50kg. crate on top of the other, so that the total mass is 100kg. The Friction should b

Helen [10]

Answer:

490.5 N

Explanation:

Coefficient of friction is 0.5 since friction force is set to halfway between none and lots. Minimum force is given by multiplying the weight and coefficient of friction

F= kN where k is coefficient of friction while N is weight. Also, N=mg where m is mass and g is acceleration due to gravity.

F=kmg=0.5*100*9.81=490.5 N

6 0

4 years ago

Is it possible for a nitrogen atom to change into an oxygen atom? If so, how?

Sliva [168]

Answer:

The nitrogen atom of an amine may be contained in a ring, a common feature of Nitrogen or oxygen atoms are the most common targets for protein methylation.

Explanation:

8 0

4 years ago

The creation and use of GM plants can reduce genetic diversity.<br><br>A.True<br>B.False

Bad White [126]

Technically, this is true...

I'm a bit rusty at this, tho

5 0

3 years ago

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1) [25 pts] A 90-kg merry-go-round of radius 2.0 m is spinning at a constant speed of 20 revolutions per minute. A kid standing

Ray Of Light [21]

Answer:

(A) 180 kg·m²

(B) 0.111 rad/s²

(C) The number of revolutions the merry-go-round will complete until it finally stops is 3.142 or π rev

Explanation:

The equation of the moment of inertia of a solid cylinder is presented as follows;

$I = \frac{1}{2}MR^2$

Where:

I = Moment of inertia of the merry-go-round

M = Mass of the merry-go-round

R = Radius of the merry-go-round

Therefore, I = 1/2×90×2² = 180 kg·m²

(B) For the angular acceleration we have;

Therefore, since the force × radius = The torque, we have, angular acceleration is found as follows

F × R = τ

10.0 × 2.0 = 20 = I×α = 180×α

α = 20/180 = 0.111 rad/s².

angular acceleration = 0.111 rad/s².

(C) Here we have ω₀ = 20 rev/ min = 20×2×π rad/min = π·40/60 rad/s

2/3·π rad/s

ω = ω₀ - α×t

∴ t = ω₀/α = (2/3·π rad/s)/(0.111 rad/s²) = 18.85 s

Hence we have

θ = ω₀·t + 1/2·α·t², plugging in the values, we have;

θ = 2/3·π×18.85 - 1/2·0.111·18.85²

θ = 19.74 rad

Therefore, since 2·π radian = 1 revolution

The number of revolutions the merry-go-round will complete until it stops is 19.74/(2·π) = 3.142 or π revolutions.

5 0

3 years ago