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Natali5045456 [20]

3 years ago

11

What is the term to describe the rate of flow of electricity?

1 answer:

SVEN [57.7K]3 years ago

7 0

The rate of flow of electric CHARGE past any point is described in the unit of electric CURRENT ... the Ampere.

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The terms sanity and insanity:

IgorLugansk [536]

Whats the question exactly?

3 0

2 years ago

Read 2 more answers

A wheel is rotating at 30.0 rpm. The wheel then accelerates uniformly to 50.0 rpm in 10.0 seconds. Determine the – a) rate of an

Mademuasel [1]

Answer:

The angular acceleration is $0.209\ rad/s^2$

Explanation:

Given that,

Angular velocity, $\omega_{i} = 30.0\ rpm$

Angular velocity, $\omega_{f} = 50.0\ rpm$

Time t = 10.0 sec

We need to calculate the angular acceleration

Using formula of angular acceleration

$\alpha=\dfrac{\Delta \omega}{\Delta t}$

$\alpha=\dfrac{\omega_{f}-\omega_{i}}{\Delta t}$

$\alpha=\dfrac{50.0-30.0}{10.0}$

Now, we change the angular velocity in rad/s.

$\omega=20\times\dfrac{2\pi}{60}$

$\omega=2.09\ rad/s$

$\alpha=\dfrac{2.09}{10.0}$

$\alpha=0.209\ rad/s^2$

Hence, The angular acceleration is $0.209\ rad/s^2$

5 0

3 years ago

Read 2 more answers

Question 2 The gravitational force between two objects with identical masses that are 10 m apart, is 2.67 x10-10 N. To the neare

xxMikexx [17]

Answer 888990,0 kg

Explanation:

3 0

3 years ago

Using -10 m/s2 for acceleration due to gravity, what would be the total displacement of the object if it took 8 seconds before h

ASHA 777 [7]

If it starts from 0m/s...
s=?
u=0
a=-10
t=8
s=ut +1/2at^2
so s=(0×8)+ (0.5×-10×64)
s=0+(32×-10)
s=32×-10
s=-320metres

6 0

3 years ago

The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho= rho0[1+α(T−T0)] , where

Stolb23 [73]

Answer:

At 81. 52 Deg C its resistance will be 0.31 Ω.

Explanation:

The resistance of wire = $R_T =\frac{\rho_T \ l}{A}$

Where $R_T$ =Resistance of wire at Temperature T

$\rho_T$ = Resistivity at temperature T $=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]$

Where $T_0 =20\ Deg\ C , \ \rho_0 = Constant, \alpha =3.9 \times 10^-^3 DegC^-1 \ (Given)$

l=Length of the wire

& A = Area of cross section of wire

For long and thin wire the resistance & resistivity relation will be as follows

$\frac{R_T_1}{R_T_2}=\frac{\rho_0(1+\alpha \cdot(T_1-2 0 )}{\rho_0(1+\alpha \cdot (T_2 -20 )}$

$\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}$

$1.24=1+\alpha (T-20)$

$0.24=\alpha(\ T -20 )$

$Putting\ the\ value\ of \alpha = 3.9 \times 10^-^3 DegC^-1$

T = 81.52 Deg C

4 0

2 years ago