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3 years ago

A gold bar contains 63.0 mol of gold, Au(s). How many atoms of gold are in the bar?

Physics

1 answer:

vredina [299]3 years ago

7 0

Answer:

63.0 x (6.02 x 10^23) = 3.79 x 10^25 atoms of gold

Explanation:

There are 6.02 x 10^23 representative particles in a mole of any substance. Multiply it by the number of moles, and you will get the number of representative particles (atoms, molecules, formula units)

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Circuit A

Licemer1 [7]

By V=IR
A: 24=I*20
I = 1.2A

B: 220 = I*250
I = 0.88A

C: 6= I*3
I = 2 A

C,A,B

3 0

4 years ago

Read 2 more answers

In a level parking lot, a man with a mass of 100 kg gives a shove to a boy with a mass of 50 kg who is on roller skates. The boy

natita [175]

Answer:

100N

Explanation:

Newton's third law states that whenever an object exerts a force on a second object, it exerts a force of equal magnitude and direction but in the opposite direction on the first. It is often stated as follows: Each action always opposes an equal but opposite reaction.

The subject 1 of 100kg is making a force F, to move an object from 50Kg to 2m / s ^ 2. This Force the object of 50Kg will reflect it in the opposite direction by Newton's third law.

Once the parameter of the force that both are experiencing is clarified, Newton's second law is applied to their respective calculation.

$F = ma = 50kg * 2m / s ^ 2 = 100N$

That is the force the boy exert on the man during the shove.

4 0

3 years ago

Flasher units are being discussed. Technician A says that only a DOT-approved flasher unit should be used for turn signals. Tech

zmey [24]

Answer: C

Both Technicians A and B

Explanation:

Only a DOT-approved flasher unit should be used for turn signals. And a parallel (variable-load) flasher will function for turn signal usage, although it will not warn the driver if a bulb burns out.

3 0

3 years ago

At a meeting of physics teacher in Montana, the teachers were asked to calculate where a flour sack would land if dropped from a

Harlamova29_29 [7]

At a distance of 469.2 m from the original point below the airplane.

Explanation:

First of all, we have to calculate the time it takes for the sack to reach the ground.

To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

s = 300 m is the vertical displacement

u = 0 is the initial vertical velocity

t is the time

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find the it takes for the sack to reach the ground:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(300)}{9.8}}=7.82 s$

Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:

$v_x = 60 m/s$