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2 years ago

What is a law in physics?

Physics

1 answer:

Rina8888 [55]2 years ago

6 0

Answer:

By nature, laws of Physics are stated facts which have been deduced and derived based on empirical observations. Simply put, the world around us works in a certain way, and physical laws are a way of classifying that “working.”

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A student sits on a pivoted stool while holding a pair of weights. The stool is free to rotate about a vertical axis with neglig

Blababa [14]

Answer:

Please Mark As Brainliest!!

a) 4.99 rad/sec b) 6.24 rad/sec c) 7.03 J

Explanation:

a) If the student completes one turn in 1.26 sec, this is called the period of the movement.

If we take into account that the angle rotated during one turn is 2π rads, by definition of angular velocity, we can get this value as follows:

ω = Δθ / Δt = 2*π rad / 1.26 seg = 4.99 rad/sec.

b) As no external torques are acting on the system, the total angular momentum must be conserved, so we can write the following equation:

Li = Lf ⇒ I₁ * ω₁ = I₂* ω₂

So, we can solve for ω₂, as follows:

ω₂ = (I₁ * ω₁) / I₂ = 6.24 rad/sec

c) Appying the work-energy theorem, we know that the work done by the student, must be equal to the change in the kinetic energy, which in this case is only rotational, so we can write:

W = 1/2 I₂* ω₂² - 1/2 I₁ ω₁²

W =1/2 ((2.25 kg.m² * (6.24)²) (rad/sec)² - (1.8 kg.m²* (4.99)²) (rad/sec)²)

W = 7.03 J

4 0

2 years ago

A bottle of water with mass 0.9 kg is left out in the sun, the radiation from the sun warms up the water bottle. If the water bo

natita [175]

Answer:

Final temperature, T2 = 314.9 Kelvin

Explanation:

Given the following data:

Mass = 0.9kg

Initial temperature, T1 = 10°C to Kelvin = 10 + 273 = 283K

Quantity of heat = 120,000 J

Specific heat capacity = 4182 j/kgK

To find the final temperature;

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

Making dt the subject of formula, we have;

$dt = \frac {Q}{mc}$

Substituting into the equation, we have;

$dt = \frac {120000}{0.9*4182}$

$dt = \frac {120000}{3763.8}$

dt = 31.9K

Now, the final temperature T2 is;

But, dt = T2 - T1

T2 = dt + T1

T2 = 31.9 + 283

T2 = 314.9 Kelvin

8 0

2 years ago

What are the uses of a magnet?

Doss [256]

Answer:

A magnet is used in a compass to show the direction.

Magnets are used in medical equipment.

Powerful magnets are used to lift objects

they are used in refrigerator, televisions, earphones etc

5 0

1 year ago

Read 2 more answers

A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.

alisha [4.7K]

A)

It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

$S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s$

Through Definition of Velocity, comes:

$\Delta v= \frac{\Delta S}{\Delta t} \\ v_x= \frac{36}{2} \\ \boxed {v_{x}=18m/s}$

B)

Using the Velocity Hourly Equation in vertical direction, we have:

$v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}$

The angle of impact is given by:

$cos(\theta) =\frac{v_{x}}{v_{y}} \\ cos(\theta) = \frac{18}{20} \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}$

If you notice any mistake in my english, please let me know, because i am not native.

7 0

2 years ago

Read 2 more answers

A concert loudspeaker suspended high off the ground emits 34 W of sound power. A small microphone with a 1.0 cm2 area is 44 m fr

rjkz [21]

Answer:

Part A

I = 1.4 mW/m²

Part B

β = 91.46 dB

Explanation:

Part A

Sound intensity is the power per unit area of sound waves in a direction perpendicular to that area. Sound intensity is also called acoustic intensity.

For a spherical sound wave, the sound intensity is given by;

$I = \frac{P}{A}$

$I = \frac{P}{4\pi r^{2}}$

Where;

P is the source of power in watts (W)

I is the intensity of the sound in watt per square meter (W/m2)

r is the distance r away

Given:

P = 34 W,

A = 1.0 cm²

r = 44 m

The sound intensity at the position of the microphone is calculated to be;

$I = \frac{34}{4\pi (44)^{2}}$

I = 0.0013975 W/m²

≈ I = 0.0014 W/m² = 1.4 × 10⁻³ W/m²

I = 1.4 mW/m²

The sound intensity at the position of the microphone is 1.4 mW/m².

Part B

Sound intensity level or acoustic intensity level is the level of the intensity of a sound relative to a reference value. It is a a logarithmic quantity. It is denoted by β and expressed in nepers, bels, or decibels.

Sound intensity level is calculated as;

β $= 10log_{10}\frac{I}{I_{0}}$ dB

Where,

β is the Sound intensity level in decibels (dB)

I is the sound intensity;

I₀ is the reference sound intensity;

By pluging-in, I₀ is 1.0 × 10⁻¹² W/m²

∴ β $= 10log_{10}\frac{1.4 * 10^{-3} W/m^{2}}{1.0 * 10^{-12} W/m^{2}}$

β $= 10log_{10} (1.4 * 10^{9})$

β = 91.46 dB

The sound intensity level at the position of the microphone is 91.46 dB.

4 0

2 years ago

What is a law in physics?​

What is a law in physics?