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sattari [20]
3 years ago
6

Assuming the acceleration due to gravity on the moon is exactly one-sixth of the acceleration due to gravity on Earth, what is t

he weight of the object on the moon?
Physics
1 answer:
uysha [10]3 years ago
5 0

<u>I have assumed a weight of 120 N on Earth.</u>

Answer:

<em>The object weighs 20 N on the moon</em>

Explanation:

Weight

The weight of an object depends on the mass m of the object and the acceleration of gravity g of the place they are in.

The formula to calculate the weight is:

W = m.g

If g_e is the acceleration of gravity on Earth, and g_m is the acceleration of gravity on the moon, we know:

g_m=1/6 g_e

Dividing by ge:

g_m/g_e = 1/6

An object of weight We=120 N on planet Earth has a mass of:

m = 120 / g_e

Multiplying by gm:

m.g_m=120 g_m/g_e

Substituting the ratio of accelerations of gravity:

m.g_m=120 * 1/6

Since m.gm is the weight on the Moon Wm:

W_m=20~N

The object weighs 20 N on the moon

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A toroidal solenoid has an inner radius of 12.0 cm and an outer radius of 15.0 cm . It carries a current of 1.50 A . Part A How
AleksAgata [21]

Answer:

The number of turns, N = 1750

Explanation:

It is given that,

The inner radius of a toroid, r = 12 cm

Outer radius, r' = 15 cm

The magnetic field at points within the coils 14 cm from its center is, B=3.75\ mT=3.75\times 10^{-3}\ T

R = 14 cm = 0.14 m

Current, I = 1.5 A

The formula for the magnetic field at some distance from its center is given by :

B=\dfrac{\mu_o NI}{2\pi R}

N=\dfrac{2\pi R B}{\mu_o I}

N=\dfrac{2\pi \times 0.14\times 3.75\times 10^{-3}}{4\pi \times 10^{-7}\times 1.5}

N = 1750

So, the number of turns must have in a toroidal solenoid is 1750. Hence, this is the required solution.

3 0
3 years ago
Read 2 more answers
What information did the movement of the bubble over a specified distance in the calibrated respirometer provide
emmasim [6.3K]

Answer:

The following explanatory section gives an explanation of this question.

Explanation:

  • This means that perhaps the bubble moves more than a certain duration throughout the calibration breath meter offers the rate as well as oxygenation consumed inside this cell.
  • Inside that respirometer, oscillation of something like the bubble gave a technique of multiplying the quantity of oxygenation that is used by the seedlings mostly through cell membrane breathing.
3 0
3 years ago
A toolbox of mass 3.2kg is lowered by a rope from the roof to the ground. Find the acceleration of the toolbox when the force of
Lisa [10]

Answer:

The answer to your question is:

a) 2.7 m/s²

b) -3.6 m/s²

Explanation:

Data

mass of the toolbox = 3.2 kg

a = ?

F = 40 N and F = 20 N

g = 9.81 m/s²

Formula

Second law of motion = F = ma

                              a + g = F / m

                              a = F/m - g

a)                            a = 40/3.2 - 9.81

                              a = 2.69 ≈ 2.7 m/s²   positive up

b)                            a = 20/ 3.2 - 9.81

                              a = 6.25 - 9.81

                                  = - 3.56 ≈ - 3.6 m/s²  negative down

8 0
3 years ago
A bicyclist is traveling at +25m/s when he begins to decelerate at -4m/s2. How fast is he traveling after 5 seconds
kotegsom [21]

Answer:

+5m/s

Explanation:

When doing the math we figure out that e is going to be slowing down at -4m/s² for 5 seconds. In total he is slowing down -20m/s which we take from the total speed of +25m/s to get his current new speed.

7 0
3 years ago
The sine of the incident angle is 0.217; the sine of the refracted angle is 0.173. calculate the index of refraction.
notka56 [123]
This can be solve using snell's law. snell's law equation is :

N1 / N2 = sin a2 / sin a1
where N1 is the index of  refraction of the air which is equal to 1
N2 is the index of refraction of the medium
a2 is the angle of refraction
a1 is the incident angle

subsitute the given values
1 / N2 = 0.173 / 0.217
N2 = 1 ( 0.217 / 0.173)
N2 = 1.25 is the index of refraction
5 0
4 years ago
Read 2 more answers
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