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rjkz [21]
3 years ago
6

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

Physics
2 answers:
Burka [1]3 years ago
7 0
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time  t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is  7/3 .              v1 = 7/3 m/s .

At time  t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
                                     v2 = zero .

At time  t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is  -3/5 .            v3 = -0.6 m/s .              
wel3 years ago
7 0
The slope or gradient of a line is a number that describes both the direction and the steepness of the line. For a distance-time graph, it signifies the velocity.

at t1=2.0s
v =  7/3 m/s 

At time  t2 = 4.0 s
v = zero

At time  t3 = 13 s
v = -0.6 m/s

Hope this answers the question. Have a nice day.
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When a condenser discharges electricity, the instantaneous rate of change of the voltage is proportional to the voltage in the c
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Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = e^{kt}

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = e^{-0.01t}

V = V₀e^{-0.01t}

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀e^{-0.01t}

0.1V₀ = V₀e^{-0.01t}

0.1V₀/V₀ = e^{-0.01t}

0.1 = e^{-0.01t}

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

3 0
3 years ago
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