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Bas_tet [7]
3 years ago
7

Can some one help me with number 2 urgent?!

Physics
1 answer:
babunello [35]3 years ago
6 0
A) I = 0
b) to the right side
c) to the left side
d) to the left side
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A 87.0 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. The astron
Ket [755]

Answer:

259.62521 seconds

Explanation:

m_1 = Mass of astronaut = 87 kg

m_2 = Mass of wrench = 0.57 kg

v_1 = Velocity of astronaut

v_2 = Velocity of wrench = 22.4 m/s

Here, the linear momentum is conserved

m_1v_1=m_2v_2\\\Rightarrow v_1=\frac{m_2v_2}{m_1}\\\Rightarrow v_1=\frac{0.57\times 22.4}{87}\\\Rightarrow v_1=0.14675\ m/s

Time = Distance / Speed

Time=\frac{38.1}{0.14675}=259.62521\ s

The time taken to reach the ship is 259.62521 seconds

4 0
3 years ago
What is the relationship between stream and drainage basin
Sergeeva-Olga [200]
Each stream in a drainage system drains into a certain area. In a drainage basin the water falling in the basin drain will fall into the same stream. A drainage divides drawing basin from other drainage basins
5 0
3 years ago
I NEED HELP ASAP!!!!!
Ratling [72]

Answer:

D) momentum of cannon + momentum of projectile= 0

Explanation:

The law of conservation of momentum states that the total momentum of an isolated system is constant.

In this case, the system cannon+projectile can be considered as isolated, because no external forces act on it (in fact, the surface is frictionless, so there is no friction acting on the cannon). Therefore, the total momentum of the two objects (cannon+projectile) must be equal before and after the firing:

p_i = p_f

But the initial momentum is zero, because at the beginning both the cannon and the projectile are at rest:

p_i = 0

So the final momentum, which is sum of the momentum of the cannon and of the projectile, must also be zero:

p_f = p_{cannon}+p_{projectile} =0

6 0
3 years ago
Which describes why people on earth can see light from the stars in the sky that are so far away?
Ugo [173]
Yes, C is correct. It self explains itself as we know light travels through a vacuum ( doesn't need a medium) and light is a type of electromagnetic wave.
7 0
3 years ago
A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

5 0
3 years ago
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