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amid [387]
3 years ago
12

the speed of a train is decreased in a uniform rate from 96 km/h to 48km/h through a distance of 800m. calculate the distance co

vered by the train from the moment of using the brakes till it stops if it was moving with the same acceleration​
Physics
1 answer:
GREYUIT [131]3 years ago
7 0

Answer:

1066.67 m

Explanation:

Given:

v₀ = 96 km/h = 26.67 m/s

v = 48 km/h = 13.33 m/s

Δx = 800 m

Find: a

v² = v₀² + 2aΔx

(13.33 m/s)² = (26.67 m/s)² + 2a (800 m)

a = -0.333 m/s²

Given:

v₀ = 26.67 m/s

v = 0 m/s

a = -0.333 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (26.67 m/s)² + 2 (-0.333 m/s²) Δx

Δx = 1066.67 m

Round as needed.

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A 750-newton person stands in an elevator that is
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When did agriculture first began?​
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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
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Answer:

Explanation:

All the displacement will be converted into vector, considering east as x axis and north as y axis.

5.3 km north

D = 5.3 j

8.3 km at 50 degree north of east

D₁= 8.3 cos 50 i + 8.3 sin 50 j.

= 5.33 i + 6.36 j

Let D₂ be the displacement which when added to D₁ gives the required displacement D

D₁ + D₂ = D

5.33 i + 6.36 j + D₂ = 5.3 j

D₂ = 5.3 j - 5.33i - 6.36j

= - 5.33i - 1.06 j

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D₂²= 5.33² + 1.06²

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4 0
3 years ago
At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0
3 years ago
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