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Tju [1.3M]
2 years ago
15

What are the phenomena that physics is applied in?

Physics
2 answers:
Ratling [72]2 years ago
5 0

Answer:

For empiricists like van Fraassen, the phenomena of physics are the appearances observed or perceived by sensory experience. Constructivists, however, regard the phenomena of physics as artificial structures generated by experimental and mathematical methods.

omeli [17]2 years ago
5 0

Answer:

Physics is applied every where? For example if you talking with you friends, that mean the sound that you produce oscillate through your friends' ear so he/ she can respond to you.

Explanation:

Physics is the fundamental unit about understanding the world around us.

You might be interested in
How energy and power are different ???
Naya [18.7K]

Answer:

Explanation:

Energy is what makes change happen and can be transferred form one object to another. ... Power is the rate at which energy is transferred. It is not energy but is often confused with energy. The watt is the most commonly used unit of measure for power.

6 0
2 years ago
Select the phrases that correctly describe density. Density is a chemical property. Density relates a mass to its volume. Densit
sp2606 [1]

Answer:

Density relates a mass to its volume.

Density varies with temperature

Density determines if a substance floats or sinks.

Density may have units of grams per milliliter (g/mL)

Explanation:

Density D is a characteristic property of a substance or material and is defined as the relationship between the mass m of a body or substance and the volume V it occupies:

D=\frac{m}{V}

This means the density is inversely proportional to the volume.

On the other hand, density is a scalar quantity and according to the International System of Units its unit is D=\frac{kg}{m^{3}} , although it can be also expressed in \frac{g}{ml}.

It should be noted that the density of a body is related to its buoyancy, a substance or body will float on another fluid if its density is lower. In addition, if the pressure of the substance remains constant, as the temperature increases, the density decreases; this means density varies with the temperature as well.

5 0
3 years ago
A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld
lesantik [10]
Answer: The magnitude of the proton's acceleration is 0.748 ×10^14 m/s²

Explanation:
the velocity ,v, of ththe proton = 5.02×10^6 m/s
Magnitude , B , of the magnetic field = 0.180 T

First , we need to find the magnitude of the Force on the proton. This is given by the relation :
F = q(v x B) = qvBsinθ

where 'q' is the charge of proton , q= 1.6×10^-19 C
θ is the angle the proton makes with the direction of the magnetic field

Putting the respective values of v, B ,θ in the above equation, we get:

F = (1.6×10^-19 C)(5.02×10^6 m/s)(0.180T) sin60°
∴ F = 1.25 ×10^-13 N

Now , from Newton's second law we know that ,
F=m×a

∴ a = F/m

Mass of a proton = 1.67×10^27 kg
a= 1.25 × 10^-13 N / 1.67 × 10^27 kg

a= 0.748 × 10^14 m/s² =acceleration of the proton

(To know more about Magnetic Fields : brainly.com/question/9095546)

5 0
2 years ago
A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the
Maurinko [17]

Answer:

(a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity R_{0}=10\ mCi

Time t_{1}=4\ hours

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

R=R_{0}e^{-\lambda t}

\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})

Put the value into the formula

\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})

\lambda=0.0000154\ s^{-1}

\lambda=1.55\times10^{-5}\ s^{-1}

We need to calculate the half life

Using formula of half life

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}

Put the value into the formula

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}

T_{\dfrac{1}{2}}=44.719\times10^{3}\ s

T_{\dfrac{1}{2}}=11.3\ hr

(b). We need to calculate the value of N₀

Using formula of N_{0}

N_{0}=\dfrac{3.70\times10^{6}}{\lambda}

Put the value into the formula

N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}

N_{0}=2.38\times10^{11}\ nuclei

(c). We need to calculate the sample's activity

Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

R=1.87\ mCi

Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

4 0
3 years ago
At which of the following angles will the sunlight received at a location on Earth spread out over the largest area?
AleksandrR [38]

The angle at which the sunlight received at a location on Earth spread out over the largest area is 10°. Last option is correct.

<h3>What is sunlight?</h3>

The light coming from the Sun reaching the Earth's surface is called as Sunlight.

When the sun is overhead, the intensity is high because sun's rays are perpendicular to the earth's surface, so the energy spreads over a small area and the heat is too high in that region.

When, the angle is smaller, the sunlight will spread out over a larger area.

Thus, at 10° the sunlight received at a location on Earth spread out over the largest area. Last option is correct.

Learn more about Sunlight.

brainly.com/question/23504828

#SPJ1

3 0
1 year ago
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