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antiseptic1488 [7]
3 years ago
8

Which of the following is an accurate comparison of the weight of an astronaut on the moon and the Earth? The weight of the astr

onaut is the same on both. The weight of the astronaut on the moon is greater than the weight of the astronaut on the Earth. The weight of the astronaut on the moon is less than the weight of the astronaut on the Earth. The weight of the astronaut on the moon is half the weight of the astronaut on the Earth.
Physics
2 answers:
LiRa [457]3 years ago
7 0

The mass of the astronaut is the same on both, but weight is actually a force and it depends on the acceleration due to gravity.  On the moon, the acceleration due to gravity is 1/6 of the Earth’s so the astronaut’s weight will be 1/6 lighter on the moon.

Lorico [155]3 years ago
7 0

Answer:

The weight of the astronaut on the moon is less than the weight of the astronaut on the Earth.            

Explanation:

Matter contained in a body is known as mass. It remains constant even if one goes to another celestial body. Weight is the force due to gravity acting on a mass. Since, gravitational force varies at each celestial body, the weight also changes.

The acceleration due to gravity on the Moon is \frac{1}{6}^{th} the acceleration due to gravity on the Earth. Therefore, the astronaut would weigh less on moon than on the Earth. His weight would be \frac{1}{6}^{th} of that on Earth.

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Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e. the length and radius have twice the
saul85 [17]

Answer:

The wire now has less (the half resistance) than before.

Explanation:

The resistance in a wire is calculated as:

R=\alpha \frac{l}{s}

Were:

R is resistance

\alpha is the resistance coefficient

l is the length of the material

s is the area of the transversal wire, in the case of wire will be circular area (s=\pi r^{2}).

So if the lenght and radius are doubled, the equation goes as follows:

R=\alpha \frac{l}{\pi r^{2} } =\alpha \frac{2l}{\pi {(2r)}^{2} } =\alpha \frac{2l}{\pi 4 {r}^{2} }=\frac{1}{2} \alpha \frac{l}{\pi r^{2} }

So finally because the circular area is a square function, the resulting equation is half of the one before.

7 0
3 years ago
A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.
Zolol [24]
These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is 45^{\circ}.

Calling F the force of the wind, and d=15~km=15000~m the distance covered by the boat, the work done by the wind is:
W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J

The total time of the motion is t=1~h=3600~s and therefore the power of the wind is
P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55

Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
\epsilon =  \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
P= \frac{W}{t}
But we can rewrite the work as
W=Fdcos\theta
where F is the force applied, d the displacement of rock and \theta=60^{\circ] is the angle between the direction of the force and the displacement (3 m). 
Therefore we can rewrite the power as
P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta
where v=d/t=5~m/s is the velocity, Using the data of the exercise, we can then find the force, F:
F= \frac{P}{v cos\theta} =   \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N

and now we can also calculate the work, which is 
W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J
3 0
3 years ago
A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg.
rodikova [14]

Answer:

the moment of inertia of the merry go round is 38.04 kg.m²

Explanation:

We are given;

Initial angular velocity; ω_1 = 37 rpm

Final angular velocity; ω_2 = 19 rpm

mass of child; m = 15.5 kg

distance from the centre; r = 1.55 m

Now, let the moment of inertia of the merry go round be I.

Using the principle of conservation of angular momentum, we have;

I_1 = I_2

Thus,

Iω_1 = I'ω_2

where I' is the moment of inertia of the merry go round and child which is given as I' = mr²

Thus,

I x 37 = ( I + mr²)19

37I = ( I + (15.5 x 1.55²))19

37I = 19I + 684.7125

37I - 19 I = 684.7125

18I = 684.7125

I = 684.7125/18

I = 38.04 kg.m²

Thus, the moment of inertia of the merry go round is 38.04 kg.m²

7 0
3 years ago
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Scientists estimate the age of the universe to be (1 point)
Nesterboy [21]

12-15 billion years i think

4 0
3 years ago
chuck wagon travels with a constant velocity of 0.5 mile/minutes. determine the total distance traveled by Chuck Wagon during 12
n200080 [17]

Given:

Velocity: 0.5 mile/minute

Time: 12 minute

Now we know that speed and velocity have the same magnitude. Hence speed=velocity=0.5 mile/min

Substituting the given values in the above formula we get

Distance = 0.5 x 12= 6 miles

7 0
3 years ago
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