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nataly862011 [7]
3 years ago
7

A 2.0-kg object moving 3.0 m/s strikes a 1.0-kg object initially at rest. immediately after the collision, the 2.0-kg object has

a velocity of 1.5 m/s directed 30o from its initial direction of motion. what is the x-component of the velocity of the 1.0-kg object just after the collision?
Physics
1 answer:
Stells [14]3 years ago
5 0
The answer is 1.5 m/s
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Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of
morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner(v_1)=3.1 m/s

speed of second runner(v_2)=4.65 m/s

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Time required by second runner t=\frac{295.6}{4.65}=63.56 s

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3 0
3 years ago
20 Points available for physics help
Sonja [21]
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8 0
3 years ago
Read 2 more answers
A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes
Agata [3.3K]

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

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v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

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Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

8 0
3 years ago
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