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3 years ago

Acar goes from 20m/s to 30m/s in 10 seconds what is its acceleration

Physics

1 answer:

prohojiy [21]3 years ago

8 0

Acceleration = change in velocity/change in time

= (30 - 20) / 10 - 0

= 10 / 10

Acceleration = 1 m/s²

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A car traveling at a speed of 30.0 m/s encounters an emergency and comes to a complete stop. how much time will it take for the

Vedmedyk [2.9K]

Answer:7.5seconds

Explanation:

V=0 u=30m/s a=-4m/s^2

t=(v-u)/a

t=(0-30)/-4=-30/-4

t=7.5second

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3 years ago

Technician A says that a power-assisted brake system reduces stopping distances compared to a nonpower-assisted brake system. Te

yanalaym [24]

Answer:

Technician B

Explanation:

here on analyzing both the statements from technician A and technician B. The Statement from Technician B is more logical and correct. That the power-assisted brake system reduces the force that the driver must exert on the brake pedal.

The power-assisted brake system does not reduce the distance of stopping. What it does is it reduces the force to be applied by the driver. Thus, making the drive more comfortable.

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3 years ago

During which segments is kinetic energy decreasing?<br> A)1,2,3<br> B) 4,5<br> C)1,3,5<br> D)2,4

Nutka1998 [239]

Answer:

Explanation:

kinetic energy is energy an object has due to its movement. for instance, if someone was riding down a hill, when the motion of the bike begins to decrease so does the kinetic energy

3 0

3 years ago

Read 2 more answers

Which two elements define a story's setting?

Archy [21]

Answer:

the plot structure defines a story's setting

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3 years ago

An Atwood's machine consists of two different masses, both hanging vertically and connected by an ideal string which passes over

Tasya [4]

Answer:

V₁ = √ (gy / 3)

Explanation:

For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2

Starting point

Em₀ = U₁ + U₂

Em₀ = m₁ g y₁ + m₂ g y₂

Let's place the reference system at the point where the mass m1 is

y₁ = 0

y₂ = y

Em₀ = m₂ g y = 2 m₁ g y

End point, at height yf = y / 2

$E_{mf}$ = K₁ + U₁ + K₂ + U₂

$E_{mf}$ = ½ m₁ v₁² + ½ m₂ v₂² + m₁ g $y_{f}$ + m₂ g $y_{f}$

Since the masses are joined by a rope, they must have the same speed

$E_{mf}$ = ½ (m₁ + m₂) v₁² + (m₁ + m₂) g $y_{f}$

$E_{mf}$ = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g $y_{f}$

How energy is conserved

Em₀ = $E_{mf}$

2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g $y_{f}$

2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2

3/2 v₁² = 2 g y -3/2 g y

3/2 v₁² = ½ g y

V₁ = √ (gy / 3)

5 0

3 years ago