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Ilya [14]
3 years ago
14

A NASA explorer spacecraft with a mass of 1,000 kg takes off in a positive direction from a stationary asteroid. If the velocity

of the spacecraft is 250 m/s and the asteroid is pushed back –25 m/s, what is the mass of the asteroid? Assume there is no net force on the system.
A) 100 kg
B) 6,250 kg
C) 10,000 kg
D) 6,250,000 kg
Physics
2 answers:
Jet001 [13]3 years ago
3 0

Before the launch, the momentum of the (spacecraft + asteroid) was zero.  So after the launch, the momentum of the (spacecraft + asteroid) has to be zero.

Momentum = (mass) x (velocity)

Momentum after the launch:

Spacecraft:  (1,000 kg) x (250 m/s) = 250,000 kg-m/s

Asteroid: (mass) x (-25 m/s)

Their sum:  250,000 - 25(mass) .

Their sum must be zero, so  250,000 kg-m/s = (25 m/s) x (mass)

Divide each side by  25 :  10,000 kg-m/s = (1 m/s) x (mass)

Divide each side by (1 m/s) :  10,000 kg = mass


Deffense [45]3 years ago
3 0

Answer:

IT IS C EDGEINUTY

Explanation:

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spayn [35]

To solve letter a:

d1 = 85t1 = 16 km, 
85t1 = 16, 
t1 = 16 / 85 = 0.1882 h = 11.29 min. 

d2 = 115t2 = 16 km, 
115t2 = 16, 
t2 = 16 / 115 = 0.139 h = 8.35 min. 

t1 - t2 = 11.29 - 8.35 = 2.94 min. 
Car #2 arrives 2.94 minutes sooner.

To solve letter b:

15 min = 1/4 h = 0.25 h. 
d1 = d2, 
115t = 85(t + 0.25), 
115t = 85t + 21.25, 
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30t = 21.25, 
t = 21.25 / 30 = 0.71 h, 

d = 115 * 0.71 = 81.65 km.

8 0
3 years ago
What velocity must a 1210 kg car have in order to have the same momentum as a pickup truck
Mariulka [41]

Answer: 46.5 m/s

Explanation:

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1210 × v = 2250 × 25

v = 46.5 m/s

6 0
3 years ago
A centrifuge used in DNA extraction spins at a maximum rate of 7000rpm producing a "g-force" on the sample that is 6000 times th
defon

Answer:

A) a = 73.304 rad/s²

B) Δθ = 3665.2 rad

Explanation:

A) From Newton's first equation of motion, we can say that;

a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.

Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s

Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s

We are given; t = 10 s

Thus;

a = 733.04/10

a = 73.304 rad/s²

B) From Newton's third equation of motion, we can say that;

ω² = ω_o² + 2aΔθ

Where Δθ is angular displacement

Making Δθ the subject;

Δθ = (ω² - ω_o²)/2a

At this point, ω = 0 rad/s while ω_o = 733.04 rad/s

Thus;

Δθ = (0² - 733.04²)/(2 × 73.304)

Δθ = -537347.6416/146.608

Δθ = - 3665.2 rad

We will take the absolute value.

Thus, Δθ = 3665.2 rad

8 0
3 years ago
An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at
andrew-mc [135]

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

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m1u1 +m2u2 = m1v1+m2v2

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u1 = 15.4m/s

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u2 = 0m/s(at rest)

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v2.

Substitute

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123.2-31.2 = 10v2

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v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

6 0
3 years ago
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maks197457 [2]

Answer: Well first of all, for an Atom to be neutral, it shall have an equal number of the protons and the neutrons. So given that, the protons are able to be positively charged and since the electrons are negatively charged (Protons and Electrons are completely different from each other) then that means the Algebric are aggregated of charges because it equal to zero. So overall, the 'neutral atom' has to be neutral, and how? Protons must equal the number of electrons or else it wouldn't be considered neutral.

Hopefully this helps you very much!! :-) Have a great daayy!

6 0
3 years ago
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