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1 year ago

An astronaut on a space walk floats a little too far away

Physics

2 answers:

SSSSS [86.1K]1 year ago

6 0

Answer:

The astronaut can throw the hammer in a direction away from the space station. While he is holding the hammer, the total momentum of the astronaut and hammer is 0 kg • m/s. According to the law of conservation of momentum, the total momentum after he throws the hammer must still be 0 kg • m/s. In order for momentum to be conserved, the astronaut will have to move in the opposite direction of the hammer, which will be toward the space station.

edge 2020

Lady bird [3.3K]1 year ago

3 0

Answer:

He can throw it away from himself.

Explanation:

Newtons Third Law says that everything has an equal, yet opposite reaction on other objects.

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A FBD of a rocket launching into space should include:

Vladimir [108]

Answer:

Explanation:

the force of the rocket engine pushing it up, the force of gravity pulling it down, maybe some force of air resistance as the rocket goes fast, hmmm Free Body Diagrams (FBD) should have any and all forces on the model, unless they are negligible . or so slight they really make little difference in the total outcome.

3 0

1 year ago

A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the

tankabanditka [31]

Answer

given,

$F_L= 1.52\ N$

$\theta_L= 31^0$

mass of book = 0.305 Kg

so, from the diagram attached below

$F_L cos {\theta_L} + F_b sin {\theta_b} = m g$

$1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8$

$F_b sin {\theta_b} = 2.989 -1.303$

$F_b sin {\theta_b} = 1.686$

computing horizontal component

$F_b cos {\theta_b} = F_L sin {\theta_L}$

$cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}$

$cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}$

$cos {\theta_b} = 0.464$

θ = 62.35°

5 0

2 years ago

A black, totally absorbing piece of cardboard of area A = 1.7 cm2 intercepts light with an intensity of 8.1 W/m2 from a camera s

Furkat [3]

Answer:

2.7x10⁻⁸ N/m²

Explanation:

Since the piece of cardboard absorbs totally the light, the radiation pressure can be found using the following equation:

$p_{rad} = \frac{I}{c}$

<u>Where:</u>

$p_{rad}$ : is the radiation pressure

I: is the intensity of the light = 8.1 W/m²

c: is the speed of light = 3.00x10⁸ m/s

Hence, the radiation pressure is:

$p_{rad} = \frac{I}{c} = \frac{8.1 W/m^{2}}{3.00 \cdot 10^{8} m/s} = 2.7 \cdot 10^{-8} N/m^{2}$

Therefore, the radiation pressure that is produced on the cardboard by the light is 2.7x10⁻⁸ N/m².

I hope it helps you!

3 0

2 years ago

Read 2 more answers

Here are the positions at three different times for a bee in flight (a bee's top speed is about 7 m/s). Time 6.6 s 6.9 s 7.2 s P

Ber [7]

Answer:

(A.) (- 4.33, 6.33 , 0); (B.) (- 3.66, 7.5, 0); (C.) average at (A) (- 4.33, 6.33 , 0) ; (D.) (- 0.2165, 0.3165, 0)

Explanation:

Given the following :

Time - - - - - - - 6.6s - - - - - - - - - 6.9s - - - - - 7.2s

Position - (1.8,5.0,0) - (0.5,6.9,0) - - (−0.4,9.5,0)

(a) Between 6.6 s and 6.9 s, what was the bee's average velocity?

Vavg = Distance / time

[(0.5,6.9,0) - (1.8,5.0,0)] / 6.9 - 6.6

Vavg = [(0.5 - 1.8), (6.9 - 5.0), (0 - 0)] / 0.3

Vavg = - 1.3 / 0.3, 1.9/0.3, 0/3

Vavg = (- 4.33, 6.33 , 0)

b) Between 6.6 s and 7.2 s, what was the bee's average velocity?

Vavg = [(−0.4,9.5,0) - (1.8,5.0,0)] / 7.2 - 6.6

Vavg = - 2. 2/0.6, 4.5/0.6, 0/0.6

Vavg = (- 3.66, 7.5, 0)

c.) Of the two averages (- 4.3, 6.3 , 0) is closer to the instantaneous Velocity at 6.6s

D.) (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s?

Displacement = Velocity * time

Vavg between 6.6 to 6.9 ; time = (6.65 - 6.6) = 0.05 s

= (- 4.33, 6.33 , 0) * 0.05

= (- 0.2165, 0.3165, 0)

5 0

2 years ago

Suppose Galileo dropped a lead ball (100 kilograms) and a glass ball (1 kilogram) from the Leaning Tower of Pisa. Which one hit

Reika [66]

The Answer is C Both at the same time

3 0

1 year ago