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Sergeeva-Olga [200]

4 years ago

6

Please help with 3 questions about acceleration

1 answer:

Goryan [66]4 years ago

5 0

A great, helpful, useful definition of acceleration is

<em>A = (change in speed) / (time for the change)</em> . <== you should memorize this

This simple tool will directly solve all 3 problems.

The REASON for assigning these problems for homework is NOT to find the answers. It's to help YOU find out whether you know this definition, to let you go back and review it if you don't, and to give you a chance to practice using it if you do. Noticed that if you get the answers from somebody else, you lose all of these benefits.

The only wrinkle anywhere here is in #3, because when you use this definition, the unit of time has to be the same in both the numerator and the denominator.

So for #3, you have to EITHER change the km/hr to km/sec, OR change the 4sec to a fraction of an hour, before you plug anything into the definition.

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A book weighing 2 Newtons is at rest on a table. The net force on the book is zero.

natta225 [31]

Answer: I think its True

7 0

3 years ago

Case 1: A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t₁ = 11.9 seconds to get

olga_2 [115]

Answer:

Part a)

$\omega = 8.17 rad/s$

Part b)

$N = 7.74 rev$

Part c)

$\alpha = 0.69 rad/s^2$

Part d)

$\alpha = 0.48 rad/s^2$

Part e)

$t = 9.14 s$

Explanation:

Part a)

Angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi(\frac{78}{60})$

$\omega = 8.17 rad/s$

Part b)

Since turn table is accelerating uniformly

so we will have

$\theta = \frac{\omega_f + \omega_i}{2} t$

$\theta = \frac{8.17 + 0}{2}(11.9)$

$2N\pi = 48.6$

$N = 7.74 rev$

Part c)

angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{t}$

$\alpha = \frac{8.17 - 0}{11.9}$

$\alpha = 0.69 rad/s^2$

Part d)

When its angular speed changes to 120 rpm

then we will have

$\omega_2 = 2\pi (\frac{120}{60})$

$\omega_2 = 12.56 rad/s$

number of turns revolved is 15 times

so we have

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

$12.56^2 - 8.17^2 = 2\alpha (2\pi\times 15)$

$\alpha = 0.48 rad/s^2$

Part e)

now for uniform acceleration we have

$\omega_f - \omega_i = \alpha t$

$12.56 - 8.17 = 0.48 t$

$t = 9.14 s$

7 0

3 years ago

Which two options are forms of kinetic energy?

Georgia [21]

Answer:

A, C

Explanation:

SOUND ENERGY - energy from sound

MAGNETIC ENERGY - energy with magnetism

7 0

3 years ago

Read 2 more answers

A hockey puck sliding at 60.0m/s slows uniformly to 20.0m/s while travelling 800.0m.

julia-pushkina [17]

A) -3.75 meters/second
A=(20^2-80^2)/(2x800)
=(400-6400)/1600
=-6000/1600
=-3.75
B) 16 seconds
t=(20-80)/-3.75
=-60/-3.75
=16

3 0

3 years ago

A box of unknown mass is sliding with an initial speed vi = 5.10 m/s across a horizontal frictionless warehouse floor when it en

Mumz [18]

Answer:

5.2 m/s.

Explanation:

Let’s determine the initial kinetic energy of the box.

KE = ½ * m * 5.8^2 = m * 16.82

Let’s determine the friction force.

Ff = 0.1 * m * 9.8 = m * 0.98

To determine the work done by the friction force, multiply by 3.4 meters.

Work = m * 3.332

To determine the final kinetic energy of the box, subtract these two numbers.

KE = m * 13.448

½ * m * v^2 = m * 13.448

v = √26.956

Hope this helps :)

6 0

3 years ago