1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
nikitadnepr [17]
3 years ago
14

An astronaut with mass 84kg is at rest in space, before firing her water pack to move toward the shuttle. If the amount of water

shot out is
2kg and it is fired at a speed of 10m/s, what will the speed of the astronaut be?
Physics
1 answer:
Travka [436]3 years ago
4 0

Answer:

The astronaut is moving at a speed of 0.238 m/s in a direction opposite the direction of the water shot out.

Explanation:

We are given;

Mass of astronaut; m1 = 84 kg

Mass of water shoot out; m2 = 2 kg

Initial speed of astronaut; u1 = 0 m/s

Initial speed of water shoot out; u2 = 0 m/s

Final speed of shoot out; v2 = 10 m/s

From law of conservation of momentum, we can say that;

Initial momentum = final momentum

Thus;

m1•u1 + m2•u2 = m1•v1 + m2•v2

Where v1 is the final speed of the astronaut

Plugging in the relevant values, we get;

(84 × 0) + (2 × 0) = (84 × v1) + (2 × 10)

0 = 84v1 + 20

-20 = 84v1

v1 = -20/84

v1 = -0.238 m/s

The negative sign indicates that the astronaut is moving 0.238 m/s in a direction opposite the direction of the water shot out.

You might be interested in
The total momentum of two marbles before a collision is 0.06 kg-m / s. no outside forces act on the marbles. what is the total m
SpyIntel [72]
Hi my friend, since momentum is always conserved without external forces, the momentum after the collosion will still be 0.06 kg*m/s. Hope it helps☺
4 0
3 years ago
If the distance between objects is the same for each pair, which of the following have the greatest gravitational force between
tamaranim1 [39]

Answer:

B

Explanation:

3 0
4 years ago
What cell processes occour during interphase
Feliz [49]
The cell cycle has two main phases, interphase and mitosis. Mitosis is the process during which one cell divides into two. Interphase is the time during which preparations for mitosis are made. Interphase itself is made up of three phases -- G1 phase, S phase, and G2 phase -- along with a special phase called G0.
7 0
3 years ago
True or false communication is important in healthy relationship​
zvonat [6]

Answer:

True

Explanation:

If two partners don't communicate they'll grow apart and it may lead to one or the other cheating.

5 0
3 years ago
Read 2 more answers
A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20
statuscvo [17]

Answer:

A) v₁ = 10.1 m/s t₁= 4.0 s

B) x₂= 17.2 m

C) v₂=7.1 m/s

D) x₂=7.5 m

Explanation:

A)

  • Assuming no friction, total mechanical energy must keep constant, so the following is always true:

       \Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0  (1)

  • Choosing the ground level as our zero reference level, Uf =0.
  • Since the child starts from rest, K₀ = 0.
  • From (1), ΔU becomes:
  • \Delta U = 0- m*g*h = -m*g*h (2)  
  • In the same way, ΔK becomes:
  • \Delta K = \frac{1}{2}*m*v_{1}^{2}  (3)      
  • Replacing (2) and (3) in (1), and simplifying, we get:

       \frac{1}{2}*v_{1}^{2}  = g*h  (4)

  • In order to find v₁, we need first to find h, the height of the slide.
  • From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:

       h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)

       Replacing (5) in (4) and solving for v₁, we get:

      v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s  (6)

  • As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.
  • Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.
  • We can do this in more than one way, but a very simple one is using kinematic equations.
  • If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:

       v_{1}^{2} - v_{o}^{2} = 2*a* x_{1}  (7)

  • Since v₀ = 0 (the child starts from rest) we can solve for a:

       a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2  (8)

  • Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:

       t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s  (9)

B)

  • Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:

       x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2}  (10)

  • Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):

       x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2}  = 17.2 m (11)

C)  

  • From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:

       v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)

D)

  • Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:

       \frac{1}{2}*v_{2}^{2}  = g*h_{2}   (13)

  • Replacing from (12) in (13), we can solve for h₂:

       h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m  (14)

  • Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:

       x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)

4 0
3 years ago
Other questions:
  • A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth
    14·2 answers
  • A student discovers that sound waves travel 1,687.5 meters in 5 seconds through air at a temperature of 10°C. Based on this info
    11·1 answer
  • A Newton is a unit of...
    6·2 answers
  • Which of the following is true for a parallel circuit?
    10·1 answer
  • a man hits a gold ball (0.205kg) which accelerates at a rate of 20.0 m/s squared. what is the amount of force acted on the ball
    14·1 answer
  • A car drives 400 meters in 5 seconds. What is the cars speed in m/s?
    15·1 answer
  • Describe an experiment to show that air support burning​
    13·1 answer
  • 2.1 Define Resultant vector​
    5·1 answer
  • Which layer lies between the stratosphere and the
    13·2 answers
  • How do you find the oscillation period in seconds for different pendulum lengths?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!