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aalyn [17]
3 years ago
8

What is a derived physical quantity? Name three derived physical quantities, and for each, give its S.I. units and its U.S. Cust

omary units.
Physics
1 answer:
anastassius [24]3 years ago
6 0

Answer:

Physical quantity is a physical property of an object or material that can be expressed by magnitude and unit.

The derived physical quantities are the type of physical quantities which can be expressed or defined by other physical quantities, called the base quantities. Example: Area, Volume, Velocity

Area- SI Unit: m², U.S. Customary unit: acre

Volume- SI Unit: m³, U.S. Customary unit: cubic inch

Velocity- SI Unit: m/s, U.S. Customary unit: ft/s

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How much money would be saved by turning off one 100.0-W lightbulb 3.0 h/day for 365 days if the
Pavlova-9 [17]

Answer:

the money that would be saved is $13.14.

Explanation:

Given;

power consumed by the light bulb, P = 100 W = 0.1 kW

time of running the bulb, t = 3 hours for 365 days = 1,095 hours

cost rate of power consumption, C = $0.12 per kWh

Energy consumed by the light bulb for the given days;

E = Pt

E = 0.1 kW  x 1,095 hr

E = 109.5 kWh

Cost of energy consumed = 109.5 kWh   x   $0.12 / kWh

                                            = $13.14

Therefore, the money that would be saved is $13.14.

3 0
2 years ago
To see why an MRI utilizes iron to increase the magnetic field created by a coil, calculate the current needed in a 400-loop-per
gtnhenbr [62]

Answer:

B = 4.059 x 10¹⁵ T

Explanation:

Given,

Number of loop, N = 400

radius of loop, r = 0.65 x 10⁻¹⁵ m

Current, I = 1.05 x 10⁴ A

Magnetic field at the center of the loop

B = \dfrac{\mu_0NI}{2R}

B = \dfrac{4\pi\times 10^{-7}\times 400 \times 1.05 \times 10^4}{2\times 0.65\times 10^{-15}}

B = 4.059 x 10¹⁵ T

7 0
3 years ago
A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th
Stolb23 [73]

Answer:

The current is  I_b  =  400 \ A

Explanation:

From the question we are told that

    The  length of the segment is  l  =  2.50  \  m

     The current is  I_a  =  1000 \ A

     The force felt is  F  =  4.0 \  N

        The distance of the second wire is  d =  5.0 \ cm  = 0.05 \  m

Generally the current on the second wire is mathematically represented as

        I_b  =  \frac{2 \pi * r * F }{ l *  \mu_o  *  I_a }

Here  \mu_o is the permeability of free space with value  \mu_o =  4 \pi * 10^{-7} \ N/A^2

=>      I_b  =  \frac{2 * 3.142  *  0.05 *  4 }{ 2.50  *  4\pi *10^{-7}  * 1000 }

=>      I_b  =  400 \ A

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