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1 month ago
7

When numbers are very small or very large, it is convenient to either express the value in scientific notation and/or by using a

prefix with the unit. A pain-relieving pill has a mass of 0.005 g. Express the pill’s mass in grams using scientific notation or in milligrams. Express your answer in scientific notation to one significant figure or use a prefix on the unit.
Physics
1 answer:
Maru [3.3K]1 month ago
4 0

Answer:

5 mg, 5\cdot 10^{-3}g

Explanation:

First, we should express the mass in grams using scientific notation.

so we find:

m = 0.005 g

To convert it into scientific notation, we observe that we have to shift the decimal three places to the right. Hence in scientific notation it becomes

m=5\cdot 10^{-3}g

Converting to milligrams, we must remember that

1 g = 1000 mg

Thus we can set up the ratio

1 g: 1000 mg = 0.005 g: x

and we conclude with

x=\frac{(1000 mg)(0.005 g)}{1 g}=5 mg

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If you secure a refrigerator magnet about 2mmfrom the metallic surface of a refrigerator door and then move the magnet sideways,
Yuliya22 [3333]

Response:

(A) 4* 6 ^ ⁻6 T m² (B) 2 * 10 ^ ⁻6 v

Clarification:

Solution

Given that:

A refrigerator magnet with a depth of approximately 2 mm

The estimated magnetic field strength of the magnet is = 5 m T

The Area = 8 cm²

Now,

(A) The magnetic flux ΦB = BA

Therefore,

ΦB = (5 * 10^⁻ 3) ( 4 * 10 ^⁻2) * ( 2 * 10^ ⁻2) Tm²

Thus,

ΦB = 4* 6 ^ ⁻6 T m²

(B) By employing Faraday's Law, the subsequent equation applies:

Ε = Bℓυ

Where,

ℓ = 2 cm equals 2 * 10 ^⁻2 m

B = 5 m T = 5 * 10 ^ ⁻3 T

υ = 2 cm/s = 2 * 10 ^ ⁻2 m/s

Therefore,

Ε = (5 * 10 ^ ⁻3 T) * (2 * 10 ^ ⁻2) (2 * 10 ^ ⁻2) v

E =2 * 10 ^ ⁻6 v

7 0
1 month ago
A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e
kicyunya [3294]

Answer:

Maximum emf = 5.32 V

Explanation:

Provided data includes:

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions each second

Magnetic field, B = 0.5 T

We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:

\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

For the maximum emf, \sin\omega t=1

Therefore,

\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V

Hence, the maximum emf generated in the loop is 5.32 V.

3 0
2 months ago
A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer
Keith_Richards [3271]

Response:

Clarification:

Provided

weight of disk m=12.5 kg

diameter of disc R=0.23 m

weight of ring m_r=7 kg

Force F=9.7 N

N=180 rpm

\omega =\frac{2\pi N}{60}

\omega =6\pi rad/s

Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2

=13.25\times 0.23^2=0.7009 kg-m^2

At this point, Torque is T=F\times R=I\cdot \alpha

9.7\times 0.23=0.7\times \alpha

\alpha =3.18 rad/s^2

Utilizing \omega _f=\omega +\alpha t

\omega _f=0 in this scenario

0=6\pi -3.18\times t

t=\frac{6\pi }{3.18}

t=5.92 s

7 0
2 months ago
A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the
kicyunya [3294]

Answer:

Induced EMF is 2 x 10⁻³ volts

Explanation:

B = strength of the magnetic field aligning with the loop's axis = 1 T

\frac{dA}{dt} = area change rate of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = the angle formed by the magnetic field and area vector = 0

E = the induced EMF across the loop

EMF can be calculated using the formula

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

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A paper in the journal Current Biology tells of some jellyfish-like animals that attack their prey by launching stinging cells i
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Given that, the starting speed of the cells is 0 since they were at rest. The cell's acceleration is specified, along with time t = 700 ns. We aim to calculate the peak speed achieved by the cells and the distance covered during the acceleration. Let v signify the final velocity. Let d represent the distance traversed. We'll apply the equations of motion to find the solution.
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