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3 years ago

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

Physics

2 answers:

MatroZZZ [7]3 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]3 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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A rope is being used to pull a mass of 10 kg vertically upward. Determine the tension on the rope, if starting from rest, the ma

Sergio [31]

Answer:

mgh= 10 x 8 x 10

= 800

but you can try 10 x 8 x 4^-1 x 10

6 0

3 years ago

Describe a ball's motion as it rolls up a slanted

emmasim [6.3K]

The ball will decelerate as it moves upwards.

The magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

The given parameters;

initial velocity of the ball, u = 1.25 m/s
time of motion of the ball, t = 4.22 s

As the ball rolls up the inclined plane, the velocity decreases and eventually becomes zero when the ball reaches the highest point of the plane.

Thus, the ball decelerate as it moves upwards.

The acceleration of the ball is calculate as;

$a = \frac{v_f -v_0}{t} \\\\$

at the highest point on the incline plane, the final velocity $v_f$ is zero

$a = \frac{0-1.25}{4.22} \\\\a = -0.3 \ m/s^2$

Thus, the magnitude of the ball's acceleration is 0.3 m/s² and it directed backwards.

Learn more here:brainly.com/question/23860763

4 0

3 years ago

The periodic table lists all the elements according to the number of __________ they have! You can’t change __________ without c

raketka [301]

Answer:

protons

atomic number

Note: The same atomic number can be associated with several different values of atomic mass, but an element can have only 1 atomic number,

5 0

2 years ago

Help pls i need this right now

pantera1 [17]

Answer:

The x-component of $F_{3}$ is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

$\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O$ (1)

Where:

$\vec F_{1}$ , $\vec F_{2}$ , $\vec F_{3}$ - External forces exerted on the ring, measured in newtons.

$\vec O$ - Vector zero, measured in newtons.

If we know that $\vec F_{1} = (70.711,70.711)\,[N]$ , $\vec F_{2} = (-126.859, 46.173)\,[N]$ , $F_{3} = (F_{3,x},F_{3,y})$ and $\vec O = (0,0)\,[N]$ , then we construct the following system of linear equations:

$\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N$ (2)

$\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N$ (3)

The solution of this system is:

$F_{3,x} = 56.148\,N$ , $F_{3,y} = -116.884\,N$

The x-component of $F_{3}$ is 56.148 newtons.

5 0

3 years ago

In a third class lever, the distance from the effort to the fulcrum is ____________ the distance from the load/resistance to the

padilas [110]

The complete sentence is:
In a third class lever, the distance from the effort to the fulcrum is SMALLER the distance from the load/resistance to the fulcrum.
In fact, in a third class lever, the fulcrum is on one side of the effort and the load/resistance is on the other side, so the effort is located somewhere between the two of them. This means that the distance effort-fulcrum is smaller than the distance load-fulcrum.

4 0

3 years ago