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Nana76 [90]
3 years ago
10

Find the GCF of each set of numbers. 12, 21, 30 Math

Physics
2 answers:
svetlana [45]3 years ago
8 0

Answer:

3 is the GCF for all these numbers if thats what you're asking

natima [27]3 years ago
5 0
3 because
12=2’2x3
21=3x7
30=2x3x5
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Calculate the phase angle (in radians) for a circuit with a maximum voltage of 12 V and w-50 Hz. The voltage source is connected
Vinvika [58]

Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance C= 20\times10^{-2}\ F

Inductance L=20\times10^{-3}\ H

Resistance R=  50\ Omega

We need to calculate the impedance

Using formula of impedance

z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}

z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}

z=50.00

We need to calculate the phase angle

Using formula of phase angle

\theta=\cos^{-1}(\dfrac{R}{z})

\theta=\cos^{-1}(\dfrac{50}{50.00})

\theta=0.0180\ rad

Hence, The phase angle is 0.0180 rad.

3 0
3 years ago
A point charge q1q1 is held stationary at the origin. A second charge q2q2 is placed at point aa, and the electric potential ene
Yuri [45]

Explanation:

The given data is as follows.

            U_{a} = 5.4 \times 10^{-8} J

        W_{/text{a to b}} = -1.9 \times 10^{-8} J

        Electric potential energy (U_{b}) = ?

Formula to calculate electric potential energy is as follows.

            U_{b} = U_{a} - W_{/text{a to b}}

                        = 5.4 \times 10^{-8} J - (-1.9 \times 10^{-8} J)

                        = 7.3 \times 10^{-8} J

Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is 7.3 \times 10^{-8} J.

6 0
3 years ago
Methods to determine the specific heat capacity of a substance​
Gre4nikov [31]

The heat capacity and the specific heat

5 0
3 years ago
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A 4.00-m-long, 470 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building
adell [148]

Answer:

12164.4 Nm

Explanation:

CHECK THE ATTACHMENT

Given values are;

m1= 470 kg

x= 4m

m2= 75kg

Cm = center of mass

g= acceleration due to gravity= 9.82 m/s^2

The distance of centre of mass is x/2

Center of mass(1) = x/2

But x= 4 m

Then substitute, we have,

Center of mass(1) = 4/2 = 2m

We can find the total torque, through the summation of moments that comes from both the man and the beam.

τ = τ(1) + τ(2)

But

τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)

= 9221.4Nm

τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm

τ = τ(1) + τ(2)

= 9221.4Nm + 2943Nm

= 12164.4 Nm

Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm

6 0
3 years ago
When lamps with wattages greater than the rating of the luminaire are installed, a fire could occur because the luminaire is bei
sergejj [24]

Answer:

true

Explanation:

Yes, it is true.

As the wattage is more than the prescribed wattage, it becomes overheated.

6 0
3 years ago
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