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Mars2501 [29]
3 years ago
7

A 4000 kg rocket is launched, shooting 50 kg of burned fuel from its exhaust at a velocity of -625 m/s. What is the velocity of

the rocket after the fuel has burned? (Ignore the effects of gravity as this rocket is in the deepest darkest regions of interstellar space)

Physics
1 answer:
AnnZ [28]3 years ago
8 0
Hopefully this will help you.

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Gabriel is performing an experiment in which he is measuring the energy and work being done by a ball rolling down a hill.Which
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I think this one's B. energy and work are both measured in joules.
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a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w
nataly862011 [7]

Answer:

40 cm

Explanation:

We are given that

Load=800 N

Effort=200 N

Load  distance=10 cm

We have to find the effort distance.

We know that

load\times load\;distance=Effort\times effort\;distance

Using the formula

800\times 10=200\times effort\;distance

Effort distance=\frac{800\times 10}{200}

Effort distance=\frac{8000}{200}

Effort distance=40 cm

Hence,  the effort distance will be 40 cm.

7 0
3 years ago
66. Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown st
kozerog [31]

Answer:

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:  

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t  Equation 2

y: The vertical distance the ball moves at time t  

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t  

Known information

We know the following data:

Vi=15 m / s

g =9.8 \frac{m}{s^{2} }

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)  

a) t=1s

y = 15*1 - ½ 9.8*1^{2}=15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

y = 15*1.5 - ½ 9.8*1.5^{2}=22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

y = 15*2 - ½ 9.8*2^{2}= 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s  (The minus sign (-) indicates that the ball is already going down)

3 0
3 years ago
Anyone going to be my friend
Artemon [7]

Explanation:

I'd love to but we cant talk right now cause its 12:22 am here and I'm gonna sleep now lol.

but let's follow each other.

who knows we might be able to help each other.

whaddya say?

have a good day ♡

5 0
3 years ago
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