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2 years ago

What is known physical quantity

Physics

1 answer:

jeka942 years ago

5 0

The quantities which are measured is called physical quantity.

Ex - Time, Length, Density. etc.

I hope it's help you..

Thanks♥♥

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The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$

Thus, wf=12.68rad/s

C) We can use our result in B)

$\alpha=3.52\frac{rad}{s^2}$

I hope this is useful for you

regards

3 0

2 years ago

Read 2 more answers

2 * 1.5 * (.850/2)^2A small ball with mass 1.50 kg is mounted on one end of a rod 0.850 m long and of negligible mass. The syste

grin007 [14]

Answer

given,

mass of the rod = 1.50 Kg

length of rod = 0.85 m

rotational velocity = 5060 rev/min

now calculating the rotational inertia of the system.

$I = m L^2$

where L is the length of road, we will take whole length of rod because mass is at the end of it.

$I = 1.5 \times 0.85^2$

I = 1.084 kg.m²

hence, the rotational inertia the system is equal to I = 1.084 kg.m²

8 0

2 years ago

A 1.2 KG rubber ball is being thrown in the air if the ball is traveling at 2.0 M/S when it is 3.0 M off the ground what is the

Vitek1552 [10]

Answer:

37.7 J

Hope this helps! (see pictures)

6 0

2 years ago

A current of 16.0 mA is maintained in a single circular loop of 1.90 m circumference. A magnetic field of 0.790 T is directed pa

pav-90 [236]

Answer

given,

current (I) = 16 mA

circumference of the circular loop (S)= 1.90 m

Magnetic field (B)= 0.790 T

S = 2 π r

1.9 = 2 π r

r = 0.3024 m

a) magnetic moment of loop

M= I A

M= $16 \times 10^{-3} \times \pi \times r^2$

M= $16 \times 10^{-3} \times \pi \times 0.3024^2$

M= $4.59 \times 10^{-3}\ A m^2$

b) torque exerted in the loop

$\tau = M\ B$

$\tau = 4.59 \times 10^{-3}\times 0.79$

$\tau = 3.63 \times 10^{-3} N.m$

8 0

2 years ago

A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with

beks73 [17]

Answer:

4.6 kHz

Explanation:

The formula for the Doppler effect allows us to find the frequency of the reflected wave:

$f'=(\frac{v}{v-v_s})f$

where

f is the original frequency of the sound

v is the speed of sound

vs is the speed of the wave source

In this problem, we have

f = 41.2 kHz

v = 330 m/s

vs = 33.0 m/s

Therefore, if we substitute in the equation we find the frequency of the reflected wave:

$f'=(\frac{330 m/s}{330 m/s-33.0 m/s})(41.2 kHz)=45.8 kHz$

And the frequency of the beats is equal to the difference between the frequency of the reflected wave and the original frequency:

$f_B = |f'-f|=|45.8 kHz-41.2 kHz|=4.6 kHz$

6 0

2 years ago

Read 2 more answers

What is known physical quantity​

What is known physical quantity