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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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eimsori [14]

Answer:B

Explanation:

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3 years ago
If a projectile travels in the air for 8 seconds when does a projectile reach its highest point
emmainna [20.7K]

Given

The projectile is in air for a time of t=8 sec

To find

The time it takes to reach the highest point

Explanation

A projectile moves up to the highest point and then again moves down following a parabolic path.

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The time taken to reach the highest point is 4 sec

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5 0
1 year ago
skateboarder, starting from rest, rolls down a 13.5 m ramp. When she arrives at the bottom of the ramp her speed is 7.37 m/s. If
scZoUnD [109]

Answer:

1.7 m/s²

Explanation:

d = length of the ramp = 13.5 m

v₀ = initial speed of the skateboarder = 0 m/s

v = final speed of the skateboarder = 7.37 m/s

a = acceleration

Using the equation

v² = v₀² + 2 a d

7.37² = 0² + 2 a (13.5)

a = 2.01 m/s²

θ = angle of the incline relative to ground = 29.9

a' = Component of acceleration parallel to the ground

Component of acceleration parallel to the ground is given as

a' = a Cosθ

a' = 2.01 Cos29.9

a' = 1.7 m/s²

7 0
3 years ago
The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the co
djyliett [7]

Answer:

<em>11.06m/s²</em>

Explanation:

According to Newtons second law of motion

\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

208 - 0.148(17)(9.8)cos 36 = 17a

208 - 24.6568cos36 = 17a

208 - 19.9478 = 17a

188.05 = 17a

a = 188.05/17

a = 11.06m/s²

<em>Hence the  the magnitude of the resulting acceleration is 11.06m/s²</em>

6 0
3 years ago
Which sphere is NOT a part of the cycling of oxygen through Earths systems ?
stepan [7]
The answer would be Exosphere because, there are 3 main regions that circulate oxygen through the Earths system, which are the Biosphere, Atmosphere, and the Lithosphere. 
5 0
3 years ago
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