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Shalnov [3]
3 years ago
14

Which ball (if either) has the greatest speed at the moment of impact

Physics
2 answers:
Ierofanga [76]3 years ago
5 0
Are there any options??


I would have to say metal of course but without options I can't assume anything
Alenkinab [10]3 years ago
4 0
Whichever ball is placed higher, but if the balls are at the same level then both experience the same acceleration thus same speed.
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A 2.0 x 10^3-kilogram car travels at a constant speed of 12 meters per second around a circular curve of radius 30. meters. What
Vikentia [17]

The magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

<h3>Circular motion</h3>

From the question, we are to determine the magnitude of the centripetal acceleration.

Centripetal acceleration can be calculated by using the formula

a_{c} =\frac{v^{2} }{r}

Where a_{c} is the centripetal acceleration

v is the velocity

and r is the radius

From the given information

v = 12 \ m/s

and r = 30 \ m

Therefore,

a_{c} =\frac{12^{2} }{30}

a_{c} =\frac{144 }{30}

a_{c} = 4.8\ m/s^{2}

Hence, the magnitude of the centripetal acceleration of the car as it goes round the curve is 4.8 m/s²

Learn more on circular motion here: brainly.com/question/20905151

4 0
2 years ago
You are feeling like spaghetti. Although normally only about 2 meters tall, you are now about 25 meters long. (How fortunate, if
viva [34]

You are crossing the event horizon of a black hole

When you are feeling like spaghetti and you are normally only about 2 meters tall, you are now about 25 meters long, then look up over your head, you see things moving pretty quickly in the universe but that lasts only a brief instant, and then all contact with the universe is lost, you are crossing the event horizon of a black hole.

<h3>What happens when you are crossing the event horizon of a black hole?</h3>
  • The point of no return is the black hole's event horizon.
  • Anything that continues beyond this point will be absorbed by the black hole and disappear from the known universe forever.
  • The black hole's gravity is so strong at the event horizon that it cannot be overcome or resisted by any mechanical force.

<h3>Is it possible to endure inside an event horizon?</h3>
  • As a result, the individual would survive and gently float over the event horizon of the black hole without being harmed or stretched into a long, thin noodle.

<h3>What occurs beyond the horizon of the event?</h3>
  • A singularity is a truly tiny point that lies beyond the event horizon where gravity is so strong that space-time itself is infinitely bent.
  • The principles of physics as they exist presently break down at this point, making any hypotheses about what lies beyond mere conjecture.

To learn more about black hole visit:

brainly.com/question/27723143

#SPJ4

6 0
2 years ago
Two loudspeakers placed 8.0 m apart are driven in phase by an audio oscillator whose frequency range is 2.2 kHz to 2.9 kHz. A po
My name is Ann [436]

Answer:

The answer to the question is 2.2khz

Explanation:

<em>Let z₁ = 5.4m</em>

<em>Let z₂ = 4.6m</em>

<em>The path difference Δz = z₁-z₂ = 5.4 - 4.6 = 0.8m</em>

<em>For the interference= Δz λ, 2λ, 3λ......</em>

<em>The wavelength λ = 0.8m</em>

<em>The speed of sound v = 344m/s</em>

<em>The frequency f = v/λ = 344/0.8 = 430hz</em>

<em>Now,</em>

<em>f₁ =f, f₂= 2f, f₃ = 3f, f₄= 4f, f₅ =5f which is,</em>

<em>f₁ =f = 430Hz, f₂=2f =860Hz, f₃ =3f =1290Hz f₄ =4f =1720Hz and f₅=5f =2150Hz</em>

<em>f5 = 2120Hz = 2.200Hz </em>

<em>we will convert to two significant figures =2.2kHz</em>

<em> </em>

8 0
3 years ago
Utility poles are to be set every 30 meters. How many poles will be set in one mile if one was set at the beginning of the mile)
Zepler [3.9K]

Answer:

53

Explanation:

Because there are 1609.34 meters in a mile. 1609.34÷30=53.64 but because you put one at the beginning of the mile it will stay 53 and not round up to 54

4 0
3 years ago
Read 2 more answers
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.35 kg to a friend standing in front of him
nataly862011 [7]

Answer:

a) u_c=0\ m.s^{-1}       &        m_c.v_c=m_b.v_b\times \cos\theta

b) v_c=0.0566\ m.s^{-1}

c) p_e=2.9218\ kg.m.s^{-1}

Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

6 0
3 years ago
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