Answer:
See the answers below
Explanation:
This problem and its respective questions can be easily solved using Newton's law of universal gravitation. Which can be calculated by means of the following expression.
where:
G = it is the universal gravitation constant. = 6.673 x 10⁻¹¹ [N*m²/kg²]
m1 = mass of the first body [kg]
m2 = mass of the second body [kg]
r = distance among the bodies [m]
a. the mass of one is doubled?
When this happens we see that the force is increased twice as well, since the mass is in the numerator of the expression.
b. The masses of both are doubled?
If both masses are doubled the force is increased to four times its original value since the terms of the masses are in the numerator of the expression.
c. The distance between them is doubled?
In this case the force is decreased to half of its original value, since the distance is in the denominator of the expression of universal gravitation.
From Largest to Smallest:
Earth > Biosphere > Biome > Ecosystem > Community > Population > Organism > Organelles
Hope this helps!
Answer:
In summary, it is safe to handle this voltage with dry hands because the current value that you pass through the body is smaller than its underestimated sensitivity.
Explanation:
The current flowing through your system is described by Ohm's law
V = I R
where I is the current, V the voltage and R the resistance
in this case three barateras are taken in series giving a total voltage of V = 4.5 V the typical resistance values of dry skin is R = 1000 000Ohm and the resinification of wet skin is R = 100000 ohm
let's calculate the current flowing
I = V / R
I = 4.5 / 1000000
I = 4.5 10⁻⁶ A
this is the current with dry hands, we see that much less than the value that allows to feel a painful response by the body
If the skin is
I = 4,5 / 100,000
I = 4.5 10⁻⁵ A
This value is small, but it is close to the pain threshold, but it is in the range of slight discomfort.
In summary, it is safe to handle this voltage with dry hands because the current value that you pass through the body is smaller than its underestimated sensitivity.
Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor
Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.
<h3>What is the apparent weight of a body in a lift?</h3>
- Consider a body of mass m kept on a weighing machine in a lift.
- The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
- The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
- Here we have given with the actual weight of the body as 100lbs.
- This 100lb child is standing on the scale or the weighing machine, when it is riding .
- During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
- There is also<em> mg </em>downwards and a normal reaction in the upward direction.
- when we equate both the upward force and downward force, we get,
i.e. during riding the scale reads a weight less than that of actual weight.
- When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.
Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.
Learn more about the apparent weight of the body in a lift here:
brainly.com/question/28045397
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