Answer:
Explanation:We should know that weight = mass * gravity.
That is weight equals mass times gravity.
Gravity is a force of attraction between any two bodies in the universe. It is directly proportional to product of their masses and inversely proportional to the square of the distance between them.
Gravity is generally measured in terms of acceleration due to gravity, denoted as g. For Earth it is, 9.8 m/s². And for moon, it is about 1.62 m/s².
On Earth, your weight is 70 kg = W
W = mass x 9.8
70 = mass x 9.8
Your mass is 70/ 9.8
i.e approximately 7.14
Weight at the Moon, W' = 7.14 x 1.62
Hence, your weight on the surface of the moon is just 11.56 kg.
Congratulations, you've lost about 58.14 kilograms without any hard exercise. And you're as light as a Sweedish Vallhund! Cheers!
Answer:
171.5 N
Explanation:
The gravitational force on an object due to the Earth is given by

where
m is the mass of the object
g is the acceleration due to gravity
The acceleration due to gravity at a certain height h above the Earth is given by

where:
G is the gravitational constant
is the Earth's mass
is the Earth's radius
Here,

So the acceleration due to gravity is

We know that the mass of the object is
m = 70 kg
So, the gravitational force on it is

Answer:
E = 31.329 N/C.
Explanation:
The differential electric field
at the center of curvature of the arc is
<em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>
where
is the radius of curvature.
Now
,
where
is the charge per unit length, and it has the value

Thus, the electric field at the center of the curvature of the arc is:


Now, we find
and
. To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

and this is
radians.
Therefore,

evaluating the integral, and putting in the numerical values we get:


Answer:
B
Explanation:
That's the answer. Hope it helped!
Answer:
25 seconds
Explanation:
Assuming the woman is accelerating at a constant rate of
from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.
Let she takes t seconds to cover the distance, s=115 m.
As acceleration, 

Now, from the equation of motion


[ from equation (i)]

seconds.
Hence, she takes 25 seconds to walk the distance.