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3 years ago

Rutherford's gold foil experiment revealed the atom has what subatomic particle

1 answer:

3 0

His gold foil experiment revealed the atom had a dense positive subatomic particle in the middle . He called it the NUCLEUS.

Hope this helps!

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Suppose that a person gets hit by a bus moving at 30 mi/h with a 58,000 lbs of force in the direction of motion. If the mass of

alexandr402 [8]

The impulse of a force is due to the change in the motion of an object

A. The persons speed after impact is approximately 59.38 mi/h

B. The expected speed is 29.89 mi/h which is less than the findings

Reason:

Known parameters are;

The speed of the bus, v = 30 mi/h

The force with which the person was hit, F = 58,000 lbs

Mass of the bus, M = 40,000 lbs

Mass of the person, m = 150 lbs

Duration of the impact, Δt = 0.007 seconds

A. The speed of the person at the end of the impact, v, is given as follows;

The impulse of the force = F × Δt = m × Δv

For the person, we get;

58,000 lbf ≈ 1866094.816 lb·ft./s²

58,000 lbf × 0.007 s = 150 lbs × Δv

1,866,094.816 lb·ft./s²

$\Delta v = \dfrac{1,866,094.816\ lbs \times 0.007 \, s}{150 \, lbs} \approx 87.084 \ ft./s$

Δv = v₂ - v₁

The initial speed of the person at the instant, can be as v₁ = 0

The final speed, v₂ = Δv - v₁

∴ v₂ ≈ 87.084 ft./s - 0 = 87.084 ft./s

≈ 87.084 ft./s

$v_2 \approx \dfrac{87.084 \ ft./s}{y} \times\dfrac{1 \ mi}{5280 \ ft.} \times \dfrac{3,600 \ s}{1 \, hour} \approx 59.38 \ mi/h$

The speed of the person at the end of the impact, v₂ ≈ 59.38 mi/h

B. Where the momentum is conserved, we have;

m₁·v₁ + m₂v₂ = (m₁ + m₂)·v

$v = \dfrac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_2 + m_1}$

$v = \dfrac{40,000 \times 30 + 150 \times 0}{40,000 + 150} \approx 29.89$

The expected speed of the person at the end of the impact is 29.89 mi/h, and therefore, the findings does not agree with the expectation

Learn more here:

brainly.com/question/18326789

3 0

3 years ago

I need to find the current resistance and voltage for each in this complicated circuit plz help

konstantin123 [22]

Explanation:

The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.

The net resistance is:

R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)

R = 20Ω

Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:

V = IR

120 V = I (20Ω)

I = 6 A

So the voltage drops are:

V = (4Ω) (6A) = 24 V

V = (10Ω) (6A) = 60 V

That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:

V = 120 V − 24 V − 60 V

V = 36 V

So the currents are:

I = 36 V / 11 Ω = 3.27 A

I = 36 V / 22 Ω = 1.64 A

I = 36 V / 33 Ω = 1.09 A

If we wanted to, we could also show this using Kirchhoff's laws.

7 0

3 years ago

The steps in the scientific process must be followed in order. True False

olya-2409 [2.1K]

Answer: True

Hope I helped :)

5 0

3 years ago

Read 2 more answers

In a collision between two unequal masses, which mass receives a greater magnitude impulse?

Contact [7]

The answer to this question depends upon Newton's third law of motion. For every action, there's an equal and opposite reaction. Because of this law, during the collision between two unequal masses, the impulse that each mass receives will be of equal magnitude and and opposite sign.

8 0

3 years ago

A spring with a mass on the end of it hangs in equilibrium a distance of 0.4200 m above the floor. The mass is pulled down a dis

den301095 [7]

Answer:

0.48 m

Explanation:

I'm assuming that this takes place in an ideal situation, where we neglect a host of factors such as friction, weight of the spring and others

If the mass is hanging from equilibrium at 0.42 m above the floor, from the question, and it is then pulled 0.06 m below that particular position. This pulling is a means of adding more energy into the spring, when it is released, the weight compresses the spring and equals its distance (i.e, 0.06 m) above the height.

0.42 m + 0.06 m = 0.48 m

At the highest point thus, the height is 0.48 m above the ground.

3 0

3 years ago