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RideAnS [48]
3 years ago
11

Are the units of the formula ma = mv2/2 dimensionally consistent? Select the single best answer.

Physics
1 answer:
Vesnalui [34]3 years ago
5 0

To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency

ma = \text{Mass}\times \text{Acceleration}

ma = kg \cdot \frac{m}{s^2}

At the same time we have that

\frac{1}{2}mv^2 = \text{Mass}\times \text{Velocity}^2

\frac{1}{2}mv^2 = kg ( \frac{m}{s})^2

\frac{1}{2}mv^2 = kg \cdot \frac{m^2}{s^2}

Therefore there is not have same units and both are not consistent and the correct answer is B.

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A series circuit has a current of 3 A. The circuit contains a 12 resistor. What is the voltage of the circuit?
muminat

Answer:

36 volts

If the circuit has a current of 3 amps, and a 12 ohm resistor.

12 x 3 = 36volts

6 0
2 years ago
Read 2 more answers
                            URGENT
Lena [83]
Just solve for a in this equation

Vf=Vi*A
7 0
3 years ago
Read 2 more answers
(a) Find the frequency of revolution of an electron with an energy of 114 eV in a uniform magnetic field of magnitude 46.7 µT. (
stira [4]

Answer:

(a)  1.3 x 10^6 Hz

(b) 76.73 cm

Explanation:

(a)

the formula for the frequency is given by

f = B q / 2 π m

where, B be the strength of magnetic field, q be the charge on one electron, m is the mass of one electron.

B = 46.7 micro tesla = 46.7 x 10^-6 T

q = 1.6 x 10^-19 C

m = 9.1 x 10^-31 kg

f = (46.7 x 10^-6 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31) = 1.3 x 10^6 Hz

(b) K = 114 eV = 114 x 1.6 x 10^-19 J = 182.4 x 10^-19 J

K = 1/2 mv^2

182.4 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v = 6.3 x 10^6 m/s

r = m v / B q

Where, r be the radius of circular path

r = (9.1 x 10^-31 x 6.3 x 10^6) / (46.7 x 10^-6 x 1.6 x 10^-19)

r = 0.7673 m = 76.73 cm  

5 0
3 years ago
Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage, by amplifying the input force
marusya05 [52]

Answer:

The solution and the explanation are in the Explanation section.

Explanation:

According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:

TA = W * (a * cosθ)

The torque due to effort at C point is:

TC = E * (b * cosθ)

The net torque is equal to 0, we have:

Tnet = 0

W * (a * cosθ) - E * (b * cosθ) = 0

E=W\frac{a}{b}

From the figure, you can observe that a/b < 1, thus E < W

8 0
3 years ago
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
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