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2 years ago

A 30 ohm resistor and a 20 ohm resistor are

Physics

1 answer:

mariarad [96]2 years ago

8 0

Answer:

3.33 A

Explanation:

Equalent Resistance=30 ohms

I = V / R

I = 100 / 30

I = 3.33 A

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A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and

sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0

2 years ago

An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f

sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

$f=\frac{v_s}{2L}$ (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

$f=\frac{343m/s}{2(0.47228m)}=362.36Hz$

Closed tube:

$f'=\frac{v_s}{4L'}$

L': length of the closed tube = 0.702821m

$f'=\frac{343m/s}{4(0.702821m)}=122.00Hz$

Next, you use the following formula for the beat frequency:

$f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz$

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0

3 years ago

Transformer is used in rectifier step up and step down voltage?<br>

tensa zangetsu [6.8K]

Answer:

While the use of the type of transformer in a rectifier depends on the voltage requirement or to meet desired operating conditions, a step-down transformer is used mainly to reduce the voltage. It is used to bring the high AC voltage level to a reasonable value or the desired output voltage.

Explanation:

Hope it helps

Correct me if Im wrong

3 0

2 years ago

This speed is measured with respect to the space station the spacecraft was originally launched from. In interstellar space the

valentinak56 [21]

Answer:

15193.62 m/s

Explanation:

t = Time taken = 6.5 hours

u = Initial velocity = 0 (Assumed)

m = Mass of rocket = 1380 kg

F = Thrust force = 896 N

v = Final velocity

a = Acceleration of the rocket

Force

$F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{896}{1380}\\\Rightarrow a=0.6493\ m/s^2$

Equation of motion

$v=u+at\\\Rightarrow v=0+0.6493\times 6.5\times 60\times 60\\\Rightarrow v=15193.62\ m/s$

The velocity of the rocket after 6.5 hours of thrust is 15193.62 m/s

5 0

3 years ago

What will be the value of both charges if they are 5 cm apart and suffer a

Mrac [35]

Answer:

$\boxed{q = 1.2 \times {10}^{ - 6} C}$

Explanation:

$f_e = \frac{{q}^{2}k }{ {r}^{2} } \\ q = \sqrt{ \frac{f_e( {r}^{2} )}{k} } = \sqrt{ \frac{5.2(5 \times {10}^{ - 2} )^{2} }{9 \times {10}^{9} } } \\ q =\sqrt{ \frac{5.2(5 \times {10}^{ - 2} )^{2} }{9 \times {10}^{9} } } = \sqrt{ \frac{0.013}{9 \times {10}^{9} } } \\ q = 1.2 \times {10}^{ - 6}$

3 0

3 years ago