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2 years ago

My mass is 65 kg and on Earth this equals a weight of 640 N, but on the moon where gravity is 1.7 m/s² my

Physics

1 answer:

Helen [10]2 years ago

4 0

Your weight on the moon given the data from the question is 110.5 N

<h3>Definition of mass and weight </h3>

Mass is simply defined as the quantity of matter present in an object. The mass of an object is constant irrespective of the location of the object.

Weight is simply defined as the gravitational pull on an object. The weight of an object varies from place to place due to gravity.

<h3>Relationship between mass and weight </h3>

Mass and weight are related according to the following equation

Weight (W) = mass (m) × Acceleration due to gravity (g)

<h3>How to determine the weight on the moon</h3>

Mass (m) = 65 Kg
Acceleration due to gravity on the moon (g) = 1.7 m/s²
Weight (W) =?

W = mg

W = 65 × 1.7

W = 110.5 N

Learn more about mass and weight:

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Can you please explain me how to do integration in physics in a easy way?

d1i1m1o1n [39]

The easiest way to explain it is roughly identical to the way that your teacher explained it in class. If there were any easier way ... like writing it here in a few paragraphs ... then that's what the teacher would have done. You would have been given the easy explanation on the first day of class, printed on one sheet of paper, and you would have had the rest of the year to practice it and get really good at it.
If the class spent a month teaching it, then that's about how long it takes. Sorry.

5 0

3 years ago

A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same m

Nina [5.8K]

Answer: $0.05\ mm$

Explanation:

Given

Cross-sectional area of wire $A_1=4\ mm^2$

Extension of wire $\delta l=0.1\ mm$

Extension in a wire is given by

$\Rightarrow \delta l=\dfrac{FL}{AE}$

where, $E=\text{Youngs modulus}$

$\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)$

for same force, length and material

$\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)$

Divide (i) and (ii)

$\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm$

5 0

2 years ago

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin

Zarrin [17]

Answer:

$\rm 9.186\times 10^{-7}\ C.$

Explanation:

<u>Given:</u>

Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
Distance of separation between the plates, d = 1.0 cm = 0.01 m.
Minimum value of electric field that produces spark, $\rm E=3\times 10^6\ N/C.$

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

$\rm E=\dfrac{\sigma}{\epsilon_o}.$

where,

$\rm \sigma$ = surface charge density of the plate of the capacitor = $\dfrac qA$ .

$\rm q$ = magnitude of the charge on each of the plate.
$\rm A$ = surface area of each of the plate = $\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.$
$\epsilon_o$ = electrical permittivity of free space, having value = $8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.$

For the minimum value of electric field that produces spark,

$\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.$