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3 years ago

A car, moving along a straight stretch of highway,

1 answer:

5 0

<span>We know that an object in moving with acceleration follow the rule according that the distance covered will be : d = Vi*t + 1/2*a*t^2 where d is distance, Vi is initial speed, and a is acceleration Then after 1 km which is 1000 metres we have: 1000 = Vi *71.2 + 1/2*0.0499*(71.2)^2 Vi = (1000-1/2*0.0499*(71.2)^2)/71.2 = 1000/71.2 - 1/2*0.0499*71.2 = 12.27 m/s Then the car was going at 12.27 m/s when started to accelerate.</span>

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A 0.150 kg mass is attached to a spring with k = 18.9 N/m. At the equilibrium position, it moves 2.39 m/s. How much mechanical e

Nastasia [14]

Answer:0.428

Explanation: got it right on Acellus

5 0

2 years ago

A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (

natita [175]

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

5 0

2 years ago

If you jumped out of a plane, you would begin speeding up as you fall downward. Eventually, due to wind resistance, your velocit

MrRa [10]

Answer:

Mg or your weight.

Explanation:

When your velocity is constant, the net force acting on you is 0. That means the upwards force of air resistance must fully balance the downwards force of gravity on you, which is Mg.

5 0

2 years ago

In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera

Talja [164]

Answer:

η = 40 %

Explanation:

Given that

Qa ,Heat addition= 1000 J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

$\eta=\dfrac{W(net)}{Q(heat\ addition)}$

Now by putting the values in the above equation ,then we get

$\eta=\dfrac{400}{1000}$

η = 0.4

The efficiency in percentage is given as

η = 0.4 x 100 %

η = 40 %

Therefore the answer will be 40%.

4 0

3 years ago

During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±

Alla [95]

Answer:

P = 1 (14,045 ± 0.03 ) k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

Δ (Pₓ / Py) =?

Let's start by finding the momentum of each vehicle

car X

Pₓ = m vₓ

Pₓ = 2.34 2.5

Pₓ = 5.85 kg m

car Y

Py = 2,561 3.2

Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

ΔPₓ = m Δv + v Δm

ΔPₓ = 2.34 0.01 + 2.561 0.01

ΔPₓ = 0.05 kg m

Δ $P_{y}$ = m Δv + v Δm

ΔP_{y} = 2,561 0.01+ 3.2 0.001

ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

P = Pₓ / $P_{y}$

ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

ΔP = 0.006 + 0.0026

ΔP = 0.009 kg m

The result is

P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s

7 0

3 years ago