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2 years ago

DESPERATEA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Physics

1 answer:

77julia77 [94]2 years ago

5 0

Answer:

i=1250000000Nm

ii=50000m/s

Explanation:

energy=force ×distance

500000N×(2.5×1000)m

=1250000000Nm

ii. Force=mass ×acceleration

500000N =40000kg ×a

500000 ÷40000=12.5a

s=1/2at^2

2500m=1/2×12.5×t^2

2500m=6.25×t^s

2500÷6.25=400

t^2=400

√t^2=√400

t=20seconds

speed=distance ×time taken

=2500m × 20s

=50000m/s

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2 years ago

At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci

kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

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3 0

2 years ago

3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-di

irinina [24]

Answer:

The initial velocity of the water from the tank is 5.42 m/s

Explanation:

By applying Bernoulli equation between point 1 and 2

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At the point 1

P₁=0 ( Gauge pressure)

V₁= 0 m/s

Z₁=3 m

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P₂=0 ( Gauge pressure)

Z₂= 0 m/s

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Now by putting the values

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3 years ago

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Scorpion4ik [409]

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Pick choice<em> (a)</em>.

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1 year ago

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Answer:

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2 years ago