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1 year ago

7

what are the speeds of (a) a proton that is accelerated from rest through a potential difference of −1000 v−1000 v and (b) an el

ectron that is accelerated from rest through a potential difference of 1000 v?

1 answer:

Evgen [1.6K]1 year ago

5 0

Answer:

This is the answer: The speed of a proton is about 5.0 × 10⁵ m/s

Explanation:

Because of the speeds of protons! :D

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Monochromatic light is incident on a metal surface, and electrons are ejected. If the intensity of the light increases, what wil

drek231 [11]

Answer:The rate of ejection of photoelectrons will increase

Explanation:

If the frequency of incident monochromatic light is held constant and its intensity is increased, the rate of ejection of photoelectrons from the metal surface increases with increase in intensity of the monochromatic light. More current flows due to more ejection of photoelectrons.

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3 years ago

The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-

nikdorinn [45]

Answer:

The answer is " $60.74^{\circ}$ ".

Explanation:

Cavity and benzene should be extended in equal quantities.

$\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\$

$\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\$

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3 years ago

A 1300 kg car traveling at 35 mph rear-ends a 1000 kg car traveling 25 mph. Just after the collision (but before the driver’s sl

guajiro [1.7K]

Answer

given,

before collision

mass of car A = m_a = 1300 kg

velocity of car A = v_a = 35 mph

mass of car B = m_b= 1000 kg

velocity of car B = v_b = 25 mph

after collision

V_a = 30 mph

V_b = 31.5 mph

Initial momentum

$P_1 = m_av_a + m_b v_b$

$P_1 = 1300 \times 35+ 1000 \times 25$

$P_1 =70500 Kg.m/s$

final momentum

$P_2 = m_aV_a + m_b V_b$

$P_2 = 1300 \times 30+ 1000 \times 31.5$

$P_2 =70500 Kg.m/s$

here initial momentum is equal to the final momentum of the car.

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3 years ago

Which part(s) of the electromagnetic spectrum are visible to humans?

jarptica [38.1K]

Color aka the visible light spectrum

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3 years ago

There are 11 steps one should follow when shooting <br>true or false<br>(Archery)

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Answer:

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1. Stance

2. Nock

3. Set Draw Hand

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8. Aim

9. Shot Set Up

10. Release

11. Follow-Through & Reflect

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2 years ago