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Katyanochek1 [597]
3 years ago
12

A speed-time graph shows a car moving at 10 m/s for 10 s. The cars speed constantly decreases until it comes to a stop at 30 s.

Which describes the slope of the graph from 10 s to 30 s
Physics
2 answers:
egoroff_w [7]3 years ago
7 0

Actually the correct answer is going to be Linear, sloping downward.

Please mark Brainliest if this answer helped you.

Ray Of Light [21]3 years ago
5 0

speed of the car at t = 10 s is given as 10 m/s

now at t = 30 s car comes to rest so v = 0

now in this velocity time graph we can say slope will show the acceleration of the car as we know that acceleration is rate of change in velocity

a = \frac{v_f - v_i}{\Delta t}

as we know that

v_f = 0

v_i = 10 m/s

\Delta t = (30 - 10) = 20 s

now from above formula

a = \frac{0 - 10}{20} = -0.5 m/s^2

So slope of the graph will represent the acceleration of car which is 0.5 m/s^2

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A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the obje
kobusy [5.1K]

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

3 0
3 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
aivan3 [116]

Answer:

Explanation:

Given that,

Initial angular velocity is 0

ωo=0rad/s

It has angular velocity of 11rev/sec

ωi=11rev/sec

1rev=2πrad

Then, wi=11rev/sec ×2πrad

wi=22πrad/sec

And after 30 revolution

θ=30revolution

θ=30×2πrad

θ=60πrad

Final angular velocity is

ωf=18rev/sec

ωf=18×2πrad/sec

ωf=36πrad/sec

a. Angular acceleration(α)

Then, angular acceleration is given as

wf²=wi²+2αθ

(36π)²=(22π)²+2α×60π

(36π)²-(22π)²=120πα

Then, 120πα = 8014.119

α=8014.119/120π

α=21.26 rad/s²

Let. convert to revolution /sec²

α=21.26/2π

α=3.38rev/sec

b. Time Taken to complete 30revolution

θ=60πrad

∆θ= ½(wf+wi)•t

60π=½(36π+22π)t

60π×2=58πt

Then, t=120π/58π

t=2.07seconds

c. Time to reach 11rev/sec

wf=wo+αt

22π=0+21.26t

22π=21.26t

Then, t=22π/21.26

t=3.251seconds

d. Number of revolution to get to 11rev/s

∆θ= ½(wf+wo)•t

∆θ= ½(0+11)•3.251

∆θ= ½(11)•3.251

∆θ= 17.88rev.

5 0
3 years ago
Read 2 more answers
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