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3 years ago

Which statement(s) correctly compare the masses of protons, neutrons, and electrons?

1 answer:

3 0

Electron. the central part of an atom containing protons and neutrons ... which of the following is necessary to calculate the atomic mass of an element? ... which of the statements correctly comparesthe relative size of an ion to its neutral atom?

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The moon has a mass of 7.34 . 102 kg and a radius of 1.74 . 106 meters. If you have a mass of 66 kg,

Ierofanga [76]

Answer:

$F=1.06\times 10^{-18}\ N$

Explanation:

Given that,

Mass of the Moon, $M=7.34\times 10^2\ kg$

Mass of the person, m = 66 kg

The radius of Moon, $r=1.74\times 10^6\ m$

We need to find the force between the person and the Moon. The formula for the gravitational force is given by :

$F=G\dfrac{Mm}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{7.34\times 10^2\times 66}{(1.74\times 10^6)^2}\\\\=1.06\times 10^{-18}\ N$

So, the required force is $1.06\times 10^{-18}\ N$ .

3 0

3 years ago

Truck drivers approaching a steep hill that they must climb often increase their speed. What good does this do, if any?

Inga [223]

Um, this doesn't make any sense. By climbing a hill, you are decreasing your momentum and kinetic energy, so it slows you down. The only positive, is after you have climbed the hill, you have more potential energy, and it will be released once you go down the hill, but you will not be as fast as if you ignored the hill.

6 0

3 years ago

how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at

DiKsa [7]

Answer:

$|a|=2.83\ m/s^2$

$\theta=75^o$

Explanation:

Net Force And Acceleration

The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

$\vec F_n=m\vec a$

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

$\vec F_1=$

$\vec F_1=\ N$

The components of the second force are

$\vec F_2=$

$\vec F_2=\ N$

The net force is

$\vec F_n=$

$\vec F_n=\ N$

The magnitude of the net force is

$|F_n|=\sqrt{14.64^2+54.64^2}$

$|F_n|=\sqrt{3200}=56.57\ N$

The acceleration has a magnitude of

$\displaystyle |a|=\frac{|F_n|}{m}$

$\displaystyle |a|=\frac{56.57}{20}$

$|a|=2.83\ m/s^2$

The direction of the acceleration is the same as the net force:

$\displaystyle tan\theta=\frac{54.64}{14.64}$

$\theta=75^o$

5 0

3 years ago

What is the direction of the normal contact force of the road on the wheels?

Gekata [30.6K]

Answer:

The direction of the contact forces acting on a body is not necessarily perpendicular to the contact surface. The resolution of contact forces in two components i.e. perpendicular to contact surface and along surface. Perpendicular component is normal force and parallel component is friction.

Explanation:

3 0

2 years ago

Is it using energy when someone stands against a locker that doesn’t move

soldi70 [24.7K]

When someone stands against a locker and is does not moving at all, then there will be no displacement and since displacement = 0

Work done also becomes equal to zero.

Work done is usually defined as change in energy. Since the work done is zero there has been no energy used.

5 0

3 years ago