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Ilya [14]
3 years ago
9

On the moon, what would be the force of gravity acting on an object that has a mass of 7 kg?

Physics
1 answer:
solniwko [45]3 years ago
5 0

Answer: 7 kg

Explanation:

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Two boats start together and race across a 48-km-wide lake and back. boat a goes across at 48 km/h and returns at 48 km/h. boat
jolli1 [7]

Answer:

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

Explanation:

Case 1: Boat 1

Speed of boat = \frac{distance of river}{time}

time = \frac{distance of river}{speed of boat}

While going to another end

time = \frac{distance of river}{speed of boat}

time = \frac{48}{48}

time = 1 hour

While going back,

time = \frac{distance of river}{speed of boat}

time = \frac{48}{48}

time = 1 hour

Total time taken by boat 1 is,

Total time by boat 1 = 1 hour + 1 hour = 2 hour

Total time by boat 1 = 2 hour

Total time taken by boat 1 for the round trip is 2 hour.

Case 2: Boat 2

Speed of boat = \frac{distance of river}{time}

time = \frac{distance of river}{speed of boat}

While going to another end

time = \frac{distance of river}{speed of boat}

time = \frac{48}{24}

time = 2 hour

While going back,

time = \frac{distance of river}{speed of boat}

time = \frac{48}{72}

time = 0.66 hour

Total time taken by boat 2 is,

Total time by boat 1 = 2 hour + 0.66 hour

Total time by boat 1 = 2.66 hour

Total time taken by boat 2 for the round trip is 2.66 hour.

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

5 0
3 years ago
What is a conversion factor?
rosijanka [135]

Answer:

A ratio of equivalent units

Explanation:

A conversion factor is a ratio of equivalent units and depends on which units are to be converted.

For example we want to convert 275 [mm] to inches, so we have to find the right conversion factor to allow us to work that conversion.

275 [mm] = inches = ?

275 [mm] * \frac{1in}{25.4mm} = 10.82 [in]

In this case the ratio is 1/25.4 = 0.039 [in/mm]

4 0
3 years ago
For high and low tides differences would they be caused by the moon and how?
Sunny_sXe [5.5K]
The moon has a small amount of gravity. Low tides mean the moon is not pulling on the water. High tides mean that the moon is pulling on the water.
8 0
3 years ago
HELP...<br> 3.00 amu = _____ Mev.<br><br> 3.22 x 10-3<br> 2.79 x 103<br> 3.10 x 102
DanielleElmas [232]
The correct answer is:
2.79 \cdot 10^3 MeV
Let's see why.

1 amu corresponds to the mass of the proton, which is:
m_p = 1.66 \cdot 10^{-27} kg
if we convert this into energy, using Einstein equivalence between mass and energy, we find:
E=mc^2 = (1.66 \cdot 10^{-27} kg)(3\cdot 10^8 m/s)^2 = 1.49 \cdot 10^{-10} J
Now we can convert it into electronvolts:
E= \frac{1.49 \cdot 10^{-10}kg}{1.6 \cdot 10^{-19} J/eV} =9.34 \cdot 10^9 eV = 934 MeV

So, 1 amu = 934 MeV. Therefore, 3 amu corresponds to 3 times this value:
3 amu = 3 \cdot 934 MeV  \sim 2790 MeV = 2.79 \cdot 10^3 MeV
5 0
3 years ago
Read 2 more answers
You are designing a delivery ramp for crates containing exercise equipment. The 1890 N crates will move at 1.8 m/s at the top of
Mashcka [7]

Answer:

K = 588.3 N/m

Explanation:

From a forces diagram, and knowing that for the maximum value of K, the crate will try to rebound back up (Friction force will point downward):

Fe - Ff - W*sin(22) = 0    Replacing Fe = K*X   and then solving for X:

X = \frac{Ff + W*sin(22)}{K}=\frac{1223}{K}

By conservation of energy:

\frac{K*X^{2}}{2}-mg*d*sin(22)-\frac{m*V^{2}}{2}=-Ff*d

Replacing our previous value for X and solving the equation for K, we get maximum value to prevent the crate from rebound:

K = 588.3 N/m

6 0
3 years ago
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