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2 years ago

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

Physics

2 answers:

SSSSS [86.1K]2 years 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]2 years 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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10. Scientists created models that show a rocky asteroid crashing into a moon (less mass) a

jolli1 [7]

Answer:

The diagram doesn’t tell you anything about the force on the asteroid. It only gives information about the force on the planet and the moon.

Explanation:

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2 years ago

Read 2 more answers

There are two forces on the 2 kg box in the overhead view of the following figure, but only one is shown. For F1=20N, a= 12 m/s2

maw [93]

Answer:

second force = 32.784

Magnitude = $\sqrt{32.784$

θ = -90°

Explanation:

Fnet = ma

F1 + F2 = ma

20N + F2 = 2(12 × cos30° + 12 ×sin30°)

F2 = 2 × 12 ( sin 30° + cos 30°)

= 24 × ( 1 + √3 )÷ 2

=12 (1 +√3 )

= 32.784

b) $\sqrt{12(1 +\sqrt{3}}$

= $\sqrt{12 ( 1+ 1.732)}$

= $\sqrt{12 (2.732)}$

= $\sqrt{32.784}$

θ = 30° + 180°

θ = 210°

210° - 300°

θ = -90°

8 0

2 years ago

Lincoln weighs 400 newtons. What’s his mass rounded to the nearest kilogram? Assume that acceleration due to gravity is 9.8 N/kg

Yanka [14]

Weight equals mass times gravitational acceleration=400N, so mass=400/9.8=41kg approx.

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2 years ago

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Two identical conducting spheres, A and B, sit atop insulating stands. When they are touched, 1.51 × 1013 electrons flow from sp

erma4kov [3.2K]

Answer:

A = -0.576 μC

B = 4.256 μC

Explanation:

Suppose a single electron charge is $1.6\times10^{-19}C$ . Then the total charge that is flowing from B to A is:

$1.6\times10^{-19} * 1.51 \times 10^{13} = 2.416\times10^{-6}C = 2.416 \mu C$

Let A and B be the initial charge of spheres A and B, respectively. Since the net charge is 3.68μC we have the following equation

$A + B = 3.68$ (1)

When they touch 2.416μC flows from B to A, then they are equal, so we have the following equation

$A + 2.416 = B - 2.416$

$-A + B = 2.416 + 2.416 = 4.832$ (2)

Add equation (1) to equation (2) we have

$2B = 3.68 + 4.832 = 8.512$

$B = 8.512 / 2 = 4.256 \mu C$

$A = 3.68 - B = 3.68 - 4.256 = -0.576 \mu C$

6 0

2 years ago

A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20.0ºC to 80.0ºC. (a) How much heatis required? Wh

Jet001 [13]

Answer:

heat used to rise temperature pan = 30.1%

heat used to rise temperature water = 69.9%

Explanation:

Given data

mass of water = 0.250 liter = 0.250 kg

aluminum pan mass = 0.500 kg

initial temperature = 20.0ºC

final temperature = 80.0ºC

to find out

heat used to rise temperature of pan and water

solution

we find here heat transferred to the water that is

heat transferred to the water = mass of water × specific heat of water × change in temperature ...........1

specific heat of water is 4186 J/kgºC

heat transferred to the water = 0.250 × 4186 × (80-20) kJ

heat transferred to the water = 62.8 kJ

and

heat transferred to the aluminum that is

heat transferred to the aluminum = mass of aluminum × specific heat of aluminum × change in temperature ...........2

here specific heat of aluminum is 900 J/kgºC

heat transferred to the aluminum = 0.500 × 900 × (80-20) kJ

heat transferred to the aluminum = 27 kJ

total heat = 62.8 + 27 = 89.8 kJ

heat used to rise temperature pan = 27/89.8 ×100% = 30.1%

heat used to rise temperature water = 62.8 / 89.8 ×100% = 69.9%

8 0

2 years ago