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Roman55 [17]
3 years ago
13

A disc initially at rest experiences an angular acceleration of 3.11 rad/s for a time of 15.0 s. What will the angular speed of

the disc be at this time, in units of rad/s?
Physics
1 answer:
kolezko [41]3 years ago
6 0

Answer:

The angular speed of the disc be at this time is 46.65 rad/s.

Explanation:

Given that,

Angular acceleration \alpha= 3.11\ rad/s^2

Time t =15.0 s

We will calculate the angular speed of the disc

A disc initially at rest.

So, \omega=0

Using rotational kinematics equation

\omega'=\omega+\alpha\ t

Where, \omega = initial angular speed

\omega' =final angular speed

\alpha = angular acceleration

Put the value in the equation

\omega'=0+3.11\times15

\omega'=46.65\ rad/s

Hence, The angular speed of the disc be at this time is 46.65 rad/s.

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Extend the life of your __________ by avoiding fast starts, stops and sharp turns.
skelet666 [1.2K]

Answer:

Tires.

Explanation:

There are the few steps which are discussed below should be taken to increase or extend the life of tires.

(1) Avoid fast starts: Fast start of the vehicle will increase the pressure on the tires due to the friction between the tires and the road will decrease the life of tires.

(2)  Avoid fast stop: Fast stop of the vehicle will also increase the pressure on the tires due to the friction between the tires and the road will decrease the life of tires.

(3) Avoid sharp turns: The alignment of the wheels and tires are in such a way that they work properly when vehicle is drive in a straight path but sharp turn will increase the uneven pressure on the tires will lead to decrease the life of tires.

Therefore, the life of tires can be extend by avoiding all the above mention actions such as fast stop, start and sharp turns.

5 0
3 years ago
How long does light take to travel from the sun to earth? heres the exact question: light travels at 300,000 km.s. The sun is a
Artist 52 [7]
I've heard it takes around 8 min
4 0
3 years ago
A loudspeaker located in air generates sound waves of frequency 1,000 Hz. Some of these sound waves enter a pool of water, where
prisoha [69]

Answer:

1.4 m

Explanation:

v = Speed of sound in water = 1400 m/s

f = Frequency of sound = 1000 Hz

\lambda = Wavelength

When we multiply the frequency and the wavelength of a wave we get the velocity of sound in that medium

v=f\lambda\\\Rightarrow \lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{1400}{1000}\\\Rightarrow \lambda=1.4\ m

The wavelength of the sound waves in water is 1.4 m

8 0
3 years ago
A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate
Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is \bf{2.81 \times 10^{4}~n~C^{-1}}.

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and '\epsilon_{0}' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm

Part(c):

If we apply Gauss' law of electrostatics, then

&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}

3 0
3 years ago
A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it
Savatey [412]

The charge of the object must be 1.11 \times e^{-5} \text { coulomb }

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

         Electric field, E=\frac{\text { Force }(F)}{q}

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

      q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }

6 0
3 years ago
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