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2 years ago

How and why will my weight change if i took a trip to the moon

Physics

2 answers:

MatroZZZ [7]2 years ago

7 0

weight less on moon than on earth.

high on lift off - G force

low in orbit.

zero at a point between earth and moon

olganol [36]2 years ago

4 0

Answer:

Due to gravitational force

We know that,

weight (W) = m. g Newtons

Where m = mass , in Kg

g = gravitational force (m/s²)

= 9.8 m/s (it is constant on earth)

So, from equation we knew that Weight is proportional to the gravitational force assuming fixed mass.

On the moon the gravitational force (g) is very less compared to the earth gravitational force, i.e., 1.62 m/s²

Therefore from equation of weight W = m. g is very less by considering the gravitational force on moon.

So, obviously weight changes on moon due to gravitational pull

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In all measurements and calculations, magnitude (number) and units are always included.

san4es73 [151]

the equation of the line allows us to find the answer is

y = -27.8 t + 97.4

The equation of a line in a linear relationship between two variables, its general expression is

y = A x + B

in this case the slope is the quantity that the independent variable in this case A = -27.8 m / s

The cut-off point that is the value of the dependent variable for x = is b = 97.4 m

In this case we see that the slope has a unit of [m / s] and the dependent variable is a unit of length, therefore the independent variable must have a unit of time [s] so that the entire equation is in units of length

y = -27.8 t + 97.4

[m] = [m / s] [s] + [m]

[m] = [m]

The other two magnitudes with are necessary to write the equation r is the mean square root and gives an idea that the values also fit the line, the best value is 1

In conclusion, the equation of the line allows us to find the answer is

y = -27.8 t + 97.4

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2 years ago

what will happen to the fieldof view for each resultant magnification as you change objectives from 4 to 10 to 43

jonny [76]

Answer:

The field of view is reduced.

Explanation:

Given that,

The field of view for every resultant magnification like you change objectives from 4 to 10 to 43.

We know that,

Field of view :

When the view is observed at a point in a defined field then these field called field of view.

The normal angle of field of view is 90°.

The formula of field of view is define as,

$field\ of\ view = \dfrac{field\ number}{magnification}$

We can say that,

The field of view is inversely proportional to the magnification.

When magnification is low then field of view will be large.

When magnification is higher then field of view will be small .

According to question,

When the magnification adjust from 4 to 10 to 43, the field of view is reduced.

Hence, The field of view is reduced.

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Answer:

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Calculate the first and second velocities of the car with four washers attached to the pulley, using the formulas v1 = 0. 25 m /

castortr0y [4]

The first and second velocities of the car are 0.25/t1 and 0.50/t2 respectively.

Given data:

The value of the first distance is, d1 = 0.25 m.

The value of the second distance is, d2 = 0.50 m.

The ratio of distance covered by an object in a specific direction and the time taken to cover the distance is known as the velocity of the object. Mathematically, the expression for the velocity is,

v = d/t

Here,

d is the distance covered.

t is the average time taken to cover the distance.

Then the first velocity of the car at 0.25 m is,

v1 = d1/t1

v1 = 0.25 / t1

here, t1 is the average time for first distance.

And the second velocity of the car with four washers at the 0. 50 m mark is,

v2 = d2/t2

v2 = 0.50 /t2

here, t2 is the average time for the second distance.

Thus, we can conclude that the first and second velocities of the car are 0.25/t1 and 0.50/t2 respectively.

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