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Alina [70]
3 years ago
6

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]3 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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Explanation:

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Unknown:

Speed of kangaroo  = ?

Velocity of kangaroo  = ?

Solution:

To solve this problem,

    Speed  = \frac{distance}{time }   = \frac{84}{7}  = 12m/s

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Answer:

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Explanation:

At the peak, the velocity is 0.

Given:

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Find:

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Rounded to 3 sig-figs, the astronaut halloweener reaches a maximum height of 2.78 meters.

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Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a
malfutka [58]

Answer:

a) p_i=1.568\hat{i}+0.752 \hat{j}

b) v_{fx}=1.668\ m.s^{-1}

c) v_{fy}=0.7999\ m.s^{-1}

Explanation:

Given masses:

m_1=0.49\ kg

m_2=0.47\ kg

Velocity of mass 1, v_1=3.2 \hat{i}\ m.s^{-1}

Velocity of mass 2, v_2=1.6 \hat{j}\ m.s^{-1}

a)

Initial momentum:

p_i=m_1.v_1+m_2.v_2

p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}

p_i=1.568\hat{i}+0.752 \hat{j}

b)

magnitude of initial momentum:

p_i=\sqrt{1.568^2+0.752 ^2}

p_i=1.739\ kg.m.s^{-1}

From the conservation of momentum:

p_f=p_i

m_f.v_f=1.739

v_f=\frac{1.739}{0.49+0.47}

v_f=1.85\ m.s^{-1} is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

tan\theta=\frac{0.752 }{1.568}

\theta=25.62^{\circ}

\therefore v_{fx}=1.85\ cos25.62^{\circ}

v_{fx}=1.668\ m.s^{-1}

c)

Vertical component of final velocity:

v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

6 0
4 years ago
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