1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Maru [420]
3 years ago
8

NEED IT ASAP PLEASE 3 examples of how we use physics in our everyday life. Please explain throughly.

Physics
1 answer:
Flauer [41]3 years ago
3 0

Answer:

  1. Alarm Clock. The buzzing sound of an alarm clock helps you wake up in the morning as per your schedule. The sound is something that you can’t see, but hear or experience.
  2. Cell Phones Cellphones have become like Oxygen gas in modern social life. Hardly, anyone would have been untouched by the effects of a cell phone. Whether conveying any urgent message or doing incessant gossips, cellphones are everywhere. But do you know how does a cell phone work? It works on the principle of electricity and the electromagnetic spectrum, undulating patterns of electricity and magnetism.
  3. Walking.Now, when you get ready for your office/school, whatever medium of commutation is, you certainly have to walk up to a certain distance. You can easily walk is just because of Physics. While you have a walk in a park or on a tar road, you have a good grip without slipping because of a sort of roughness or resistance between the soles of your shoes and the surface of the road.

Explanation:

physical is related to things perceived through the senses as opposed to the mind; tangible or concrete.

You might be interested in
You ride your bike for a distance of 30 km. You travel at a speed of 0.75 km/ minute. How many minutes does this take?
dolphi86 [110]
(30 km)/(0.75 km/min) = 40 min
4 0
3 years ago
Read 2 more answers
Four satellites are in orbit around the Earth. The heights of their four orbits
slavikrds [6]

The gravitational pull of Earth is stronger in satellite A

6 0
2 years ago
A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum
ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

\rho (r) \  \alpha  \  \delta (r -R)

\rho (r) = k \  \delta (r -R) \ \  at \ \  (r = R)

\rho (r) = 0\ \ since \ r< R  \ \ or  \ \ r>R---- (1)

To find the constant k, we  examine the total charge Q which is:

Q = \int \rho (r) \ dV = \int \sigma \times dA

Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2

∴

\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta  \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2

\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

Thus;

k * 4 \pi  \int ^{R}_{0}  \delta (r -R) * r^2dr = \sigma \times  R^2

k * \int ^{R}_{0}  \delta (r -R)  r^2dr = \sigma \times  R^2

k * R^2= \sigma \times  R^2

k  =   R^2 --- (2)

Hence, from equation (1), if k = \sigma

\mathbf{\rho (r) = \delta* \delta (r -R)  \ \  at   \ \  (r=R)}

\mathbf{\rho (r) =0 \ \  at   \ \  rR}

To verify the units:

\mathbf{\rho (r) =\sigma \ *  \ \delta (r-R)}

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

The integrated charge Q

Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \  sin \theta  \ dr \ d\theta \  d \phi  \\ \\  Q = \int ^{2 \pi}_{0} \  d \phi  \int ^{\pi}_{0} \ sin \theta  \int ^R_{0} \rho (r) r^2 \ dr

Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  \int ^R_0  * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  *R^2    since  ( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )

\mathbf{Q = 4 \pi R^2  \sigma  }

6 0
3 years ago
A car engine applies a force of 65,000 N, how much work is done by the engine as it pushed a car a distance of 75 m?
kupik [55]

Answer:

workdone = force \times distance \\  = 65000 \times 75 \\  = 4,875,000 \: J

3 0
3 years ago
What is it called when velocity changes over time
natka813 [3]
Well,

When an object's velocity changes, we call it acceleration.

Acceleration: The time rate of change in an object's velocity
3 0
3 years ago
Read 2 more answers
Other questions:
  • A block slides from rest with negligible friction down the track above, descending a vertical height of 5.0 m to point P at the
    5·1 answer
  • A heavy boy and a lightweight girl are balanced on a massless seesaw. The boy moves backward, increasing his distance from the p
    6·1 answer
  • Hw2-2 - show all your work, including the equation, express all answers in base units, (m, m/s, m/s2), unless stated otherwise i
    5·1 answer
  • A car is moving at 34mi/hr. If the car travels for 6hours how many miles (mi) did the car travel? Make sure you include the prop
    15·2 answers
  • Leta Stetter Hollingworth conducted pioneering work on __________. A. identity development in ethnic minorities B. cognitive pro
    10·2 answers
  • What is the energy (in J) stored in the 14.0 µF capacitor of a heart defibrillator charged to 6.90 ✕ 103 V?
    9·1 answer
  • Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, stra
    8·1 answer
  • Are forces present even when there is no movement?
    15·1 answer
  • A student determines that a sample has a mass of 157.2 g and a volume of 20 cm^3. Which
    15·1 answer
  • The table above lists the wavelength range, frequency range, possible uses, and hazards of the seven types of electromagnetic ra
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!