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2 years ago

HOW TO ANSWER THIS QUESTION PLEASE

Physics

2 answers:

Over [174]2 years ago

5 0

Answer:

Radioactive decay is a long proxy
Use long tongs

↔ for safety precautions the scientist should used this upon working with radioactive materials in a laboratory.

Zigmanuir [339]2 years ago

4 0

Answer:

radioactive decay is a random process
use long tongs

Explanation:

Quantum theory has taught us that virtually everything to do with the nature of matter in the universe is a random process. The decay of atomic nuclei is something that makes that randomness "visible" with appropriate instrumentation. As such, the rate of decay will vary over the "short" term. Consequently, it is not surprising to see a variation of a few counts in an activity measurement over a period of a few minutes.

Safety with respect to radioactive materials involves staying as far away from them as possible. The use of long tongs for handling would be appropriate.

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A student rapidly rubs the palms of both hands

Lapatulllka [165]

Answer:

Option a

Explanation:

The action of the rapid rubbing of the hands are a result of chemical reaction and responses that takes place inside our body.

When we rub our hands together, that shows the mechanical motion of the body or we can say that while rubbing hands chemical energy gets converted to mechanical energy (here, Kinetic energy).

While rubbing, due to friction, heat is produced and hence we can say that this mechanical energy has transformed to heat energy.

Thus the conversion here is from chemical to mechanical to heat energy.

3 0

3 years ago

The magnitude of the gravitational field strength near Earth's surface is represented by

Zanzabum

Answer:

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

Explanation:

Let be M and m the masses of the planet and a person standing on the surface of the planet, so that M >> m. The attraction force between the planet and the person is represented by the Newton's Law of Gravitation:

$F = G\cdot \frac{M\cdot m}{r^{2}}$

Where:

$M$ - Mass of the planet Earth, measured in kilograms.

$m$ - Mass of the person, measured in kilograms.

$r$ - Radius of the Earth, measured in meters.

$G$ - Gravitational constant, measured in $\frac{m^{3}}{kg\cdot s^{2}}$ .

But also, the magnitude of the gravitational field is given by the definition of weight, that is:

$F = m \cdot g$

Where:

$m$ - Mass of the person, measured in kilograms.

$g$ - Gravity constant, measured in meters per square second.

After comparing this expression with the first one, the following equivalence is found:

$g = \frac{G\cdot M}{r^{2}}$

Given that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972 \times 10^{24}\,kg$ and $r = 6.371 \times 10^{6}\,m$ , the magnitude of the gravitational field near Earth's surface is:

$g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(6.371\times 10^{6}\,m)^{2}}$

$g \approx 9.82\,\frac{m}{s^{2}}$

The magnitude of the gravitational field strength near Earth's surface is represented by approximately $9.82\,\frac{m}{s^{2}}$ .

4 0

2 years ago

If Scoobie could drive a Jetson's flying car at a constant speed of 450.0 km/hr across oceans and space, approximately how long

vladimir2022 [97]

Answer:

The value is $t = 3.6 \ days$

Explanation:

From the question we are told that

The speed is $v = 450.0 km/h$

The radius of the earth is $R = 6200 \ km$

Generally the circumfernce of the earth is mathematically evaluated as

$C = 2\pi R$

=> $C = 2 * 3.142 * 6200$

=> $C = 38960.8 \ km$

Generally the time taken is mathematically represented as

$t = \frac{38960.8}{450.0}$

$t = 86.6 \ hr$

Converting to days

$t = \frac{86.6}{24}$

=> $t = 3.6 \ days$

7 0

2 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an

sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The work is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the two tugboats "pull the tanker a distance 0.83km toward the north", that is the displacement, and you have to calculate the net force toward the north.

3. Tugboat #1.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N

4. Tugboat #2:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle: = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N =

5. Total net force, Fn:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. Work, W:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

7 0

3 years ago

The movement of electrons through a wire is called.....

Elenna [48]

Answer:

Electric Current

Explanation:

5 0

2 years ago

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HOW TO ANSWER THIS QUESTION PLEASE