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3 years ago

How long does a bullet take to travel 12.7cm when going 360 m/s

Physics

2 answers:

Vinil7 [7]3 years ago

8 0

Answer: 3.53 x 10^-4 s

Explanation:

12.7cm x 1m/100cm = 0.127m

V = d/t

t x V = d

t = d/v = 0.127m/(360m/s) = 0.000353s or 3.53 x 10^-4

Helen [10]3 years ago

7 0

Answer:

0.000352 seconds

Explanation:

$Speed = \frac{Distance}{Time} or \\Time = \frac{Distance}{Speed} = \frac{\frac{12.7}{100}} {360}$

Time = 0.000352 seconds

Given Distance in centimetres is converted to metres

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3 years ago

A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/

german

Answer:

<h2>128.61 Watts</h2>

Explanation:

Average power done by the torque is expressed as the ratio of the workdone by the toque to time.

Power = Workdone by torque/time

Workdone by the torque = $\tau \theta$ = $I\alpha * \theta$

I is the rotational inertia = 16kgm²

$\theta = angular\ displacement$

$\theta = 2 rev = 12.56 rad$

$\alpha \ is \ the\ angular\ acceleration$

To get the angular acceleration, we will use the formula;

$\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}$

$\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}$

Workdone by the torque = 16 * 1.28 * 12.56

Workdone by the torque = 257.23 Joules

Average power done by the torque = Workdone by torque/time

= 257.23/2.0

= 128.61 Watts

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3 years ago

A grindstone increases in angular speed from 6.00 rad/s to 12.20 rad/s in 16.00 s. Through what angle does it turn during that t

Akimi4 [234]

Answer:

Explanation:

Given that the grand stone has initial angular velocity of

w(ini)= 6rad/

And it has a final angular velocity of

w(fin)=12.20rad/sec

Time taken is t=16s

Using equation of angular motion

To get angular acceleration (α)

w(fin)=w(ini)+αt

12.20=6+16α

16α=12.20-6

16α=6.2

α=6.2/16

α=0.3875rad/sec²

The angular acceleration is 0.39rad/s²

Angle that he turn using

w(fin)²=w(ini)²+2αθ

12.2²=6²+2×0.3875θ

12.2²-6²=0.775θ

0.775θ=112.84

Then, θ=112.84/0.775

θ=145.6radian

The angular displacement is 145.6rad

6 0

3 years ago

A rotating merry-go-round makes one complete revolution in 4.0s a) what is the linear speed of a child seated 1.2 meters from th

lianna [129]

Answer:a) The linear speed of the rotating merry-go-round is 1.884 m/s.

b) The acceleration of the rotating merry-go-round device is $2.95m/s^2$ .

Explanation:

a) linear speed of the device:

Distance = $2\pi (r)$

Child seated 1.2 meters from the center, which means that radius ,r = 1.2 meters

Time taken to complete one revolution = 4.0 seconds

$\text{Linear speed}=v=\frac{distance}{time}=\frac{2\pi r}{t}=\frac{2\times 3.14\times 1.2 m}{4.0 s}=\frac{7.536 m}{4.0 s}=1.884 m/s$

The linear speed of the rotating merry-go-round is 1.884 m/s.

b) Acceleration

Linear velocity ,v = 1.884 m/s

$a_c=\frac{v^2}{r}$

$=\frac{1.884 m/s\times 1.884m/s}{1.2 m}=2.95m/s^2$

The acceleration of the rotating merry-go-round device is $2.95m/s^2$ .

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4 years ago

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Sedbober [7]

Answer:

This was my best estimation of the answers

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3 years ago