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MrMuchimi
2 years ago
8

Show that the units N/kg can be written using only units of meters (m) and second (s). Is this a unit of mass ,acceleration or f

orce.
Physics
1 answer:
nikdorinn [45]2 years ago
3 0
<span>When the difference between two results is larger than the estimates error, the result is</span>
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What is a partial and total lunar eclipse?
Minchanka [31]

Partial Lunar Eclipse:

A partial lunar eclipse is when the earth gets between the Sun and Moon. However, all three bodies are not in alignment meaning we are able to see some more like part of the moon's surface as it moves in route of the Earth's shadow.

Total Lunar Eclipse:

The three celestial bodies are perfectly aligned which allows for the earth to completely block the sun's rays from hitting/reaching the moon. The sun is positions is in back of the Earth which then causes the shadow of the earth to be cast on the Moon covering the moon completely. When that happens you get the phenomenon called a total lunar eclipse.

Hopefully this helped and good luck.



7 0
2 years ago
A 10.0-g bullet moving at 300 m/s is fired into a 1.00-kg block at rest. the bullet emerges (the bullet does not get embedded in
sesenic [268]
Momentum before the hit:
p = mv = 0.01 * 300 + 1 * 0 

Momentum after the hit:
p = 0.01 * 150 + 1 * v

Momentum is conserved:
0.01 * 300 = 0.01 * 150 + v 
3 = 1.5 + v
v = 1.5

The velocity of the block after the collision is 1.5 m/s.
7 0
3 years ago
A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif
Serggg [28]

Answer:

Height of the rocket be one minute after liftoff is 40.1382 km.

Explanation:

v(t)=-gt-v_e\times \ln \frac{m-rt}{m}

v = velocity of rocket at time t

g = Acceleration due to gravity =9.8 m/s^2

v_e = Constant velocity relative to the rocket = 2,900m/s.

m = Initial mass of the rocket at liftoff = 29000 kg

r = Rate at which fuel is consumed = 170 kg/s

Velocity of the rocket after 1 minute of the liftoff =v

t = 1 minute = 60 seconds'

Substituting all the given values in in the given equation:

v(60)=-9.8 m/s^2\times 60 s-2,900m/s\times \ln (\frac{29,000 kg-170 kg/s\times 60 s}{2,9000 kg})

v(60) = 668.97 m/s

Height of the rocket = h

Velocity=\frac{Displacement}{time}

668.97 m/s=\frac{h}{60 s}

h=668.97 m/s\times 60 s=40,138.2 m = 40.1382 km

Height of the rocket be one minute after liftoff is 40.1382 km.

4 0
3 years ago
Which of the following is true about radiation from the sun?
Lelu [443]
><span>It can travel through vacuum. 

The rays must travel in the vacuum of space between Earth's atmosphere and the sun.</span>
3 0
3 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0
3 years ago
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