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AleksAgata [21]
3 years ago
12

If you mass 35kg on earth what will your mass be on the moon where gravity is 1/6 that of earths. PLZZZZ HElP NEED ASAP

Physics
1 answer:
abruzzese [7]3 years ago
4 0

Answer:

Mass will be same on moon as on Earth but weight will be one-sixth of Earth.

Explanation:

Mass of a body doesn't depend on gravity. Mass is a constant quantity. So, mass on moon will be same as mass on Earth.

But, the weight of a body depends on gravity as weight is given as:

\textrm{Weight}=\textrm{mass}\times \textrm{acceleration due to gravity}

Therefore, if g is acceleration due to gravity on Earth, then weight on Earth is, W_{E}=mg

Now, gravity on moon is one-sixth of Earth. So, g_{moon}=\frac{1}{6}g

Therefore, weight of the body on moon is, W_{moon}=mg_{moon}=m\times \frac{1}{6}g=\frac{1}{6}mg=\frac{1}{6}W_{E}

Therefore, a body has same mass both on moon and Earth but weight on moon is one-sixth of the weight on Earth.

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An iron block of density rhoFe and of volume l 3 is immersed in a fluid of density rhofluid. The block hangs from a scale which
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Answer:

R=m*g-∀fl*g*l3

Explanation:

<em>An iron block of density rhoFe and of volume l 3 is immersed in a fluid of density rhofluid. The block hangs from a scale which reads W as the weight. The top of the block is a height h below the surface of the fluid. The correct equation for the reading of the scale is</em>

From Archimedes' principle we know that a body when immersed in a fluid, fully or partially, experiences an the upward buoyant force equal to the weight of the fluid displaced. As the body is fully submerged in water, volume of water displaced  

density of iron =mass/ volume

rho=m/l3

mass=rhol3

weight fluid=rhofluid*g*Volume

weight of fluid=rhofluid*g*l3

F=∀fl*g*l3

Downward force is weight of iron

w=m*g

Reading on the spring scale

R=w-F

R=m*g-∀fl*g*l3

m=mass of iron

g=acceleration due to ravity

rhfld=density of fluid

l3=volume of fluid displaced

6 0
3 years ago
Need help asap pls
Zinaida [17]
B. third

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6 0
2 years ago
Scientists in a test lab are testing the hardness of a surface before constructing a building. Calculations indicate that the en
Over [174]
It depends on what that "certain amount" is.
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3 years ago
A source produces 20 crests and 20 troughs in 4 seconds. The second crest is 3 cm away from the first crest.Calculate :
sineoko [7]

Answer:

Solution given:

No of waves[N] =20crests & 20 troughs

=20waves

Time[T]=4seconds

distance[d]=3cm=0.03m

Now

<u>Wave</u><u> </u><u>length</u><u>=</u>3cm=3 × {10}^{ - 2}m

<u>Frequency</u>=\frac{No of waves}{time}

=\frac{20}{4}=5Hertz

and

Wave speed:wave length×frequency=3 × {10}^{ - 2}m×5=1.5 × {10}^{ - 1} \tt{ {ms}^{ - 1}}.

3 0
3 years ago
A 4 cm diameter ball is located 40 cm from a point source and 80 cm from a wall. What is the size of the shadow on the wall?
Slav-nsk [51]
<span>11.823 cm There is a slight ambiguity with this question in that I don't know if the measurements are from the surface of the ball, or the center of the ball. I will take this question literally and as such the point light source will be 124 cm from the wall. The key thing to remember is that ball won't be showing an effective diameter of 4 cm to the light source. Instead the shadow line is a tangent to the ball's surface. There is a right triangle where the hypotenuse is the distance from the center of the ball to the light source (42 cm), one leg of the triangle is the radius (2cm). That right triangle will define a chord that will be the effective diameter of the disk casting the shadow. The cosine of the half angle of the chord will be 2/42 = 1/21. The sine of the half angle then becomes sqrt(1-(1/21)^2) = sqrt(440/441) = 2sqrt(110) = 0.99886557. Now multiply that sine by 4 (radius of ball multiplied by 2 since it's the half angle and we want the full side of the chord) and we get an effective diameter of 3.995462279 cm. Now we need to calculate the effective distance that circle is from the wall. It will be slightly larger than 82 cm. The exact value will be 82 + cos(half angle) * radius. So 82 + 1/21 * 2 = 82 + 2/21 = 82.0952381 Now we have the following dimensions with a circle replacing the ball in the original problem. Distance from wall to effective circle = 82.0952381 cm Distance from effective circle to point source = 124 - 82.0952381 = 41.9047619 cm Effective diameter of circle = 3.995462279 cm And because the geometry makes similar triangles, the following ratio applies. 3.995462279/41.9047619 = X/124 Now solve for X 3.995462279/41.9047619 = X/124 124*3.995462279/41.9047619 = X 495.4373226/41.9047619 = X 11.82293611 = X The shadow cast on the wall will be a circle with a diameter of 11.823 cm</span>
7 0
4 years ago
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