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Lena [83]
3 years ago
6

In Rutherford's experiment, which of the following proved that the

Physics
1 answer:
vampirchik [111]3 years ago
5 0

Answer:

b) the alpha particles were found to be attracted to the nucleus

Explanation:

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A copper wire 225m long must experience a voltage drop of less than 2.0v when a current of 3.5 a passes through it. compute the
MA_775_DIABLO [31]

Answer:

V = I * R

R = 2 / 3.5 = .571 ohms     maximum resistance of wire

R = ρ L / A  where R is proportional to L and inversely proportional to A

A = ρ L / R     minimum area of wire

ρ = 1 / μ  =     1.67E-8 ohm-m      resistivity inverse of conductivity

A = 1.67E-8 ohm-m * 225 m / .571 ohm = 6.68E-6 m^2

A = 6.68 mm^2       since 1 mm^2 = 10-6 m^2   or 1 mm = 10-3 m

A = Π r^2 = 6.68 mm^2

r = (6.68 / 3.14)^1/2 mm = 2.13 mm      radius of wire

d = 2 * r = 4.26 mm

5 0
2 years ago
1. What are the three ways an object can accelerate?
solong [7]

Answer:

increase speed, decrease speed, and change direction

Explanation:

4 0
3 years ago
What was the Law of Triads? A. Elements were arranged into three rows. B. Elements were arranged in groups of three. C. Elements
shtirl [24]

Answer:

its B

Explanation: i searched it up...

8 0
4 years ago
Classify the following situations as involving balanced or unbalanced forces.
tester [92]

Answer:

1) Unbalanced

2) Balance

3) Balanced

4) Unbalanced

5) Unbalanced

Explanation:

For 1 and 5, since the objects are not at a constant speed/velocity, their forces must be unbalanced.

For 2 and 3, their speeds are constant so that means the force is balanced. If the bycicle or box started accelerating, that would indicate an unbalanced force

For 4, the speed is constant, but it's direction is not indicating another unbalanced outside force causing it to turn. The force is unbalanced.

5 0
3 years ago
Block M = 7.50 kg is initially moving up the incline and is increasing speed with a = [09]____________________ m/s2 . The applie
Aliun [14]

Answer:

a) 73.2N

b) 66.6N

c) 20.28N

Explanation:

F= mg=7.5×9.8=73.5N

Force parallel to the plane Fp= 73.5sin25= 31.06N

b) Fv= 73.5 cos25= 66.6N

c) Ff= u×Fv= 0.312×66.6=20.28N

a) Normal ForceFn= F/(cos25) - Fp - Ff = ma

1.1F -31.06-20.28=7.5×3.82

1.1F -51.84=28.65

1.1F= 28.65+51.84

F= 80.49/1.1

F= 73.2N

4 0
3 years ago
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