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crimeas [40]
3 years ago
5

Two identical blocks are connected to opposite ends of a compressed spring. the blocks initially slide together on a frictionles

s surface with a velocity v to the right. at some instant, the left block is moving at v/2 to the left and the other block is moving to the right. what is the speed of the center of the mass of they system at that instant?
Physics
1 answer:
lana66690 [7]3 years ago
7 0

Answer:

v to the right.

Explanation:

<u>If there is no external force on the system, then the velocity of the center of mass is </u><u>constant</u><u>. </u>

The initial velocity of center of mass is v to the right. Therefore, the velocity of the center of mass at any point is v to the right.

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A ____________ is a condition or behavior that may threaten an individual's well-being
pickupchik [31]
It would be risk factor
7 0
3 years ago
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S the work done on Amanda's car while speeding up (i) greater than, (ii) less than, or (iii) the same as the work done on Bertha
ankoles [38]

Answer:

Explanation:

Mass of amanda and bertha car = Ma = Mb

Initial velocity of amanda, ua = 10 m/s

Final velocity of amanda, va = 20 m/s

Initial velocity of bertha, ub = 20 m/s

Final velocity of bertha, vb = 30 m/s

Workdone = change in Energy = 1/2 mv^2 - 1/2 mu^2

Ea = Ma × 1/2 × (20^2 - 10^2)

= Ma × 150

= 150 Ma J

Workdone = change in Energy = 1/2 mv^2 - 1/2 mu^2

Eb = Mb × 1/2 × (30^2 - 20^2)

= Mb × 250

= 250 Mb J

Option ii. Less than. Thus this is because the workdone by amanda car (150Ma) is less than the workdone in berthas car (250Mb).

B.

Impulse, p = force × time

p = Mass × change in velocity

pa = Ma × va - Ma × ua

= Ma × (20 - 10)

= 10 × Ma

= 10 Ma

p = Mass × change in velocity

pa = Mb × vb - Mb × ub

= Mb × (30 - 20)

= 10 × Mb

= 10 Mb

Option iii. The same as.

The impulse as seen above is the same in amanda car (10Ma) as the same with bertha (10Mb)

7 0
4 years ago
A basketball player shoots toward a basket 4.9 m away and 3.0 m above the floor. If the ball is released 1.8 m above the floor a
Zolol [24]

Answer:

  v₀ = 6.64 m / s

Explanation:

This is a projectile throwing exercise

          x = v₀ₓ t

          y = y₀ + v_{oy} t - ½ g t²

In this case they indicate that y₀ = 1.8 m and the point of the basket is x=4.9m y = 3.0 m

         

the time to reach the basket is

        t = x / v₀ₓ

we substitute

        y- y₀ = \frac{ v_o \ x \ sin \theta  }{ v_o \ cos \theta} - \frac{1}{2} g \ \frac{x^2 }{v_o^2 \ cos^2 \theta }

        y - y₀ = x tan θ - \frac{ g \ x^2 }{ 2 \ cos^2 \theta } \ \frac{1}{v_o^2 }

         

we substitute the values

        3 -1.8 = 3.0 tan 60 - \frac{ 9.8 \ 3^2 }{2 \ cos^2 60 } \ \frac{1}{v_o^2}

        1.2 = 5.196 - 176.4 1 / v₀²

        176.4 1 / v₀² = 3.996

        v₀ = \sqrt{ \frac{ 176.4}{3.996} }

        v₀ = 6.64 m / s

6 0
3 years ago
Two harmonic sound waves reach an overseveer simulatenouslt. the obsever hears the sound intensity rise and fall with a time of
irakobra [83]

Answer:

dF=2.5Hz

Explanation:

From the question we are told that:

Time T=0.2sec

Generally the Period is given as

 T= 2 * 0.2 = 0.4

Therefore difference in frequency dF

 dF=\frac{1}{T}

 dF=\frac{1}{0.4}

 dF=2.5Hz

4 0
3 years ago
Struggling on this, really need help
Bad White [126]

Answer:

Guessing you just need help with the definition but if it's the question I can still help you.

7 0
3 years ago
Read 2 more answers
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