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ololo11 [35]
3 years ago
9

Q.1 Newton's second law gives the measure of--------------. 1)Acceleration 2) Force 3) Momentum 4)Angular momentum​

Physics
1 answer:
posledela3 years ago
4 0

Answer:

1. Acceleration

Explanation:

Newtons Second law gives the measure of acceleration

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Yes they are necessary 
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3 years ago
An electron is accelerated by a 3.6 kv potential difference. the charge on an electron is 1.60218 × 10−19 c and its mass is 9.10
katen-ka-za [31]
By definition, the potential energy is:
 U = qV
 Where,
 q: load
 V: voltage.
 Then, the kinetic energy is:
 K = mv ^ 2/2
 Where,
 m: mass
 v: speed.
 As the power energy is converted into kinetic energy, we have then:
 U = K
 Equating equations:
 qV = mv ^ 2/2
 From here, we clear the speed:
 v = root (2qV / m)
 Substituting values we have:
 v = root ((2 * (1.60218 × 10 ^ -19) * 3600) /9.10939×10^-31))
 v = 3.56 × 10 ^ 7 m / s
 Then, the centripetal force is:
 Fc = Fm
 mv ^ 2 / r = qvB
 By clearing the magnetic field we have:
 B = mv / qr
 Substituting values:
 B = (9.10939 × 10 ^ -31) * (3.56 × 10 ^ 7) / (1.60218 × 10 ^ -19) * 0.059
 B = 3.43 × 10 ^ -3 T
 Answer:
 
A magnetic field that must be experienced by the electron is:
 
B = 3.43 × 10 ^ -3 T
6 0
2 years ago
What are the consequences of sediment pollution?
Nonamiya [84]
Loss of habitats for fish, birds, and other wildlife. Sediment pollution is one of the leading causes of the loss of the wetlands, but it’s not just the wetlands. Changes in the nutrients in your water. The same problem that affects the fish in your area may also affect you. Other drinking water contamination.
6 0
3 years ago
A small bar of pure gold whose density is 19.3g/cm. Displaces 80 cm
nasty-shy [4]

Answer:

The mass of the gold bar is 1,544 g

Explanation:

3 0
2 years ago
A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan
Natasha2012 [34]

Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

6 0
2 years ago
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