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3 years ago

6

A 100 kg object and a 10 kg object are dropped simultaneously in a vacuum. Which of the following statements is true?

2 answers:

ella [17]3 years ago

7 0

In a vacuum, all objects fall at the same rate. Meaning that the 100 kg ball will fall at the same speed as the 10 kg ball. Assuming that both objects share the same starting acceleration, they will keep that acceleration until the fall is stopped.
In other words, your answer is the first one, Both objects will accelerate at 9.8 m/s

Zepler [3.9K]3 years ago

3 0

Answer:

Both objects will accelerate at 9.8 m/s².

Explanation:

According to given condition, a 100 kg object and a 10 kg object are dropped simultaneously in a vacuum. If there is no air resistance, when two objects of same or different masses will reach the ground at same time. The rate of descent does not depend on the amount of matter contained inside the object.

In the vacuum, no air resistance is present. Both of the objects will accelerate at 9.8 m/s² i.e. under the action of gravity.

So, the correct option is (a). Hence, this is the required solution.

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3 years ago

The potential difference between two parallel conducting plates in vacuum is 165 V. An alpha particle with mass of 6.50×10-27 kg

shepuryov [24]

Answer:

kinetic energy (K.E) = 5.28 ×10⁻¹⁷

Explanation:

Given:

Mass of α particle (m) = 6.50 × 10⁻²⁷ kg

Charge of α particle (q) = 3.20 × 10⁻¹⁹ C

Potential difference ΔV = 165 V

Find:

kinetic energy (K.E)

Computation:

kinetic energy (K.E) = (ΔV)(q)

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4 0

3 years ago

You throw an 18.0 N rock into the air from ground level and observe that, when it is 15.0 m high, it is traveling upward at 17.0

stellarik [79]

Answer:

Vi = 24.14 m/s

Explanation:

If we apply Law of Conservation of Energy or Work-Energy Principle here, we get: (neglecting friction)

Loss in K.E of the Rock = Gain in P.E of the Rock

(1/2)(m)(Vi² - Vf²) = mgh

Vi² - Vf² = 2gh

Vi² = Vf² + 2gh

Vi = √(Vf² + 2gh)

where,

Vi = Rock's Speed as it left the ground = ?

Vf = Final Speed = 17 m/s

g = 9.8 m/s²

h = height of rock = 15 m

Therefore,

Vi = √[(17 m/s)² + 2(9.8 m/s²)(15 m)]

Vi = √583 m²/s²

<u>Vi = 24.14 m/s</u>

4 0

4 years ago

What is the magnetic field strength required to make a proton with a speed of 5.0 x 10^(5) m/s follow a circular path of radius

Nuetrik [128]

Answer:

Magnetic field strength required for this is 0.25 T

Explanation:

As we know that the proton moves in circular path in uniform magnetic field

so the radius of the path of the circle is given as

$R = \frac{mv}{qB}$

here we know that

$q = 1.6 \times 10^{-19}C$

$m = 1.6 \times 10^{-27} kg$

$R = 0.0200 m$

$v = 5 \times 10^5 m/s$

now we have

$B = \frac{1.6 \times 10^{-27} (5 \times 10^5)}{(1.6 \times 10^{-19})(0.02)}$

so we have

$B = 0.25 T$

8 0

4 years ago