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luda_lava [24]
2 years ago
10

On a box are the words 12 Volt : 50 mA What is the resistor in Kilo-Ohm?

Physics
1 answer:
katrin [286]2 years ago
6 0

Answer:

R = 0.24 kilo-ohm

Explanation:

Given that,

Voltage, V = 12 volt

Current, I = 50 mA = 0.05 A

We need to find the resistance of the resistor. We can find it using Ohm's law. So,

V = IR

R=\dfrac{V}{I}\\\\R=\dfrac{12}{0.05}\\\\R=240\ \Omega

or

R = 0.24 kohm

So, the value of resistance is 0.24 kilo-ohm.

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You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an
Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}

\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}

\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}

Substitute the given values to find "terminal speed"

\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}

\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}

\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}

\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}

The “terminal speed” of the ball bearing is 5.609 m/s

7 0
3 years ago
You use a lever to lift a heavy tree branch you apply a force of 50 n and the lever lifts the branch
valentinak56 [21]

1.8 is the mechanical advantage of the lever.

<h3>Definition of mechanical advantage</h3>

The theoretical mechanical advantage of a system is the ratio of the force that performs the useful work to the force applied, assuming there is no friction in the system.

The advantage gained by the use of a mechanism in transmitting force specifically the ratio of the force that performs the useful work of a machine to the force that is applied to the machine.

Mechanical advantage is given by the ratio of the load lifted to the force applied to lift the load.

In this case, Mechanical advantage=L/E where L is the load and E is the effort applied.

Mechanical advantage= 90/50 =1.8

Question-you use a lever to lift a heavy tree branch. you apply a force of 50 n and the lever lifts the branch with a force of 90 n. what is the mechanical advantage of the lever?

To learn more about the Mechanical advantage visit the link

brainly.com/question/16617083

#SPJ4

5 0
1 year ago
1. Bone has a Young’s modulus of about
Blababa [14]

#1

As we know that

Y = \frac{stress}{strain}

now plug in all data into this

1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}

strain = 9.33 \times 10^{-3}

now from the formula of strain

strain = \frac{\Delta L}{L}

9.33 \times 10^{-3} = \frac{\Delta L}{0.54}

\Delta L = 5.04 \times 10^{-3} m

\Delta L = 5.04 mm

#2

As we know that

pressure * area = Force

here we know that

Area = 3.53 \times 11.6 = 40.95 m^2

P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa

now force is given as

F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N

#3

As we know that density of water will vary with the height as given below

\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}

here we know that

\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa

B = 2.3 \times 10^9 N/m^2

now density is given as

\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}

\rho = 1185.3 kg/m^3

#4

as we know that pressure changes with depth as per following equation

P = P_o + \rho g h

here we know that

P = 3 P_0

now we will have

3P_0 = P_0 + \rho g h

2P_0 = \rho g h

2(1.01 \times 10^5) = 1025 (9.81)(h)

here we will have

h = 20.1 m

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

mg = \rho_1V_1g + \rho_2V_2g

A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)

46.6 = 43.89 - 922x + 1000x

x = 3.48 cm

so it is 3.48 cm below the interface

5 0
3 years ago
Convert 1erg into joule by dimensional method​
Valentin [98]

Answer:

1 * 10^-7 [J]

Explanation:

To solve this problem we must use dimensional analysis.

1 ergos [erg] is equal to 1 * 10^-7 Joules [J]

1[erg]*\frac{1*10^{-7} }{1}*[\frac{J}{erg} ] \\= 1*10^{-7}[J]

4 0
3 years ago
During a chemical reaction, increasing the temperature often
Nadya [2.5K]

Answer:c

Explanation:

the rate of reaction increases as temperature increases

5 0
3 years ago
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