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dsp73
3 years ago
14

What is the amount of charge in the electrons of 5 grams of nickel?

Physics
1 answer:
sergiy2304 [10]3 years ago
6 0
1.) Two positive charges of 4.3 mC each are separated by 0.25 m.
What is the size and type of force between the two charges? Help

2.) A positive charge of 1.32 x 10-4 C and a negative charge of 9.8 mC
are 0.015 m apart. What is the size and type of force between them? H

3.) Two electrons in an atom are separated by 1.5 x 10-10 m , the typical size
of an atom. What is the size and type of force between them? H

4.) A negative charge of 4.7 mC exerts a repulsive force of 51.0 N on a 2nd charge
0.062 m away. What is the size and polarity (pos or neg) of the 2nd charge? H

5.) A negative charge of 9.3 x 10-5 C exerts an attractive force of 37.4 N when
placed 0.080 m away from a 2nd charge. Find size and polarity of 2nd charge. H

6.) Two positive charges of 3.0 mC exert a repulsive force of 2.0 N on each other.
By what distance are they separated? H

7.) Two charges q1 and q2 are separated by a distance, d and exert a force, F1
on each other. What new force (F2) will exist if:

a) q1 is doubled. H d) d is doubled. H
b) q1 is doubled and q2 is tripled. H e) d is tripled. H
c) q1 is cut in half. H f) q2 is doubled and d is cut in half. H
8.) a) How many electrons are there in 1 C ? H
b) How many excess electrons are on a ball with a charge of -5.26 x 10-17 C ? H
c) How many excess protons are on a ball with a charge of 7.29 x 10-12 C ? H

9.) How many coulombs of negative charge does a 5 gram nickel coin have?
[takes 3 steps; do a) , b) and c) to get answer, don't round off until part c]

a) Find the number of atoms in a nickel.
Its mass is 5 g and nickel has 6.02 x 1023 atoms / 58 g. H

b) Find the number of electrons in the coin. There are 28 electrons / atom. H

c) Find how many coulombs of negative charge in a nickel. 1.6 x 10-19 C / 1 e- H

10.) A lightning bolt transfers 35 C to Earth. How many electrons are transferred? H
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Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air
Nata [24]

The frequency of a sound is whatever frequency leaves the source. It doesn't change.

Voiced of swimmers at the pool don't change frequency in or out of the water. Only their speed and wavelength change.

5 0
3 years ago
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NEED HELP!!!!!!!!!!!!!
raketka [301]

it would be 'B' because it speeds up reactions

7 0
3 years ago
A 750 g air-track glider attached to a spring with spring constant 14.0 N/m is sitting at rest on a frictionless air track. A 20
alexandr402 [8]

Answer:

the amplitude of  the subsequent oscillations is 0.11  m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, m_{1} = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

the mass of glider, m_{2} = 200 g = 0.2 kg

the speed of glider,  v_{2} = 170 cm/s = 1.7 m/s

the amplitude of  the subsequent oscillations is A = 0.11  m

according to mechanical enery equation, we have

A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}

where

A is the amplitude and  v_{f} is the final speed.

to find v_{f}, we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

P_{f} = P_{i}

(m_{1} +m_{2} )v_{f} = m_{1} v_{1} +m_{2}v_{2}

v_{1} = 0, thus

(0.75+0.2)v_{f} = (0.75)(0)+(0.2)(1.7)

0.95 v_{f} = 0.34

v_{f} = 0.36 m/s

Now we can calculate the amplitude

A = \sqrt{\frac{0.75 +0.2 }{10} }0.36

A = 0.11  m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π\sqrt{\frac{m_{1}+m_{2}  }{k} }

T = 2π\sqrt{\frac{0.75+0.2  }{10} }

T = 1.94 s

7 0
3 years ago
What’s the answer for this
AlladinOne [14]

Answer:

52.5 m

Explanation:

The displacement is the area under the velocity vs time graph.

From 0 to 5 seconds, the area is the area of the triangle plus the area of the rectangle.

x = ½ (3 s) (15 m/s) + (2 s) (15 m/s)

x = 52.5 m

8 0
3 years ago
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A rock is thrown with a force of 500 N and an acceleration is 75 m/s^2. What is its mass?
artcher [175]

Answer:

We conclude that the mass of a rock with a force of 500 N and an acceleration of 75 m/s² is 6.7 kg.

Hence, option D is correct.

Explanation:

Given

  • Force F = 500 N
  • Acceleration a = 75 m/s²

To determine

Mass m = ?

Important Tip:

  • The mass of a rock can be found using the formula F = ma

Using the formula

F = ma

where

  • F is the force (N)
  • m is the mass (kg)
  • a is the acceleration (m/s²)

now substituting F = 500, and a = 75 m/s² in the formula

F = ma

500 = m(75)

switch sides

m\left(75\right)=500

Divide both sides by 75

\frac{m\cdot \:75}{75}=\frac{500}{75}

simplify

m=\frac{20}{3}

m=6.7 kg

Therefore, we conclude that the mass of a rock with a force of 500 N and an acceleration of 75 m/s² is 6.7 kg.

Hence, option D is correct.

7 0
3 years ago
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