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
Lelu [443]
2 years ago
10

What is the function of physics education

Physics
1 answer:
Fudgin [204]2 years ago
8 0

Answer:

Physics is the natural science that studies matter its fundamental constituents, its motion and behavior through space and time, and the related entities of energy and force

Explanation:

You might be interested in
Which of the following would increase​
bazaltina [42]

Answer:

1 and 3

Explanation:

<u>1 and 3  </u>

Increasing coils increases strength

   COOLING the wire would increase current flow and strength of magnet

Adding an iron core will definitely increase the strength of the electromagnet

7 0
2 years ago
Read 2 more answers
A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2
Ahat [919]

Answer:

Approximately 0.608 (assuming that g = 9.81\; \rm N\cdot kg^{-1}.)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

  • Let m represent the mass of this car.
  • Let r represent the radius of the circular track.

This answer will approach this question in two steps:

  • Step one: determine the centripetal force when the car is about to skid.
  • Step two: calculate the coefficient of static friction.

For simplicity, let a_{T} represent the tangential acceleration (1.90\; \rm m \cdot s^{-2}) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to 90^\circ or \displaystyle \frac{\pi}{2} radians.

The angular acceleration of this car can be found as \displaystyle \alpha = \frac{a_{T}}{r}. (a_T is the tangential acceleration of the car, and r is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity (u and v) to (tangential) acceleration a_{T} and displacement x:

v^2 - u^2 = 2\, a_{T}\cdot x.

The idea is to solve for the final angular velocity using the angular analogy of that equation:

\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta.

In this equation, \theta represents angular displacement. For this motion in particular:

  • \omega(\text{initial}) = 0 since the car was initially not moving.
  • \theta = \displaystyle \frac{\pi}{2} since the car travelled one-quarter of the circle.

Solve this equation for \omega(\text{final}) in terms of a_T and r:

\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}.

Let m represent the mass of this car. The centripetal force at this moment would be:

\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}.

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, m\, g.

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let \mu_s denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g.

The size of this force should be equal to that of the centripetal force when the car is about to skid:

\mu_s\, m\, g = \pi\, m\, a_{T}.

Solve this equation for \mu_s:

\mu_s = \displaystyle \frac{\pi\, a_T}{g}.

Indeed, the expression for \mu_s does not include any unknown letter. Let g = 9.81\; \rm N\cdot kg^{-1}. Evaluate this expression for a_T = 1.90\;\rm m \cdot s^{-2}:

\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608.

(Three significant figures.)

7 0
3 years ago
An 800-kHz radio signal is detected at a point 2.7 km distant from a transmitter tower. The electric field amplitude of the sign
dimaraw [331]

Answer:

Option D is correct: 170 µW/m²

Explanation:

Given that,

Frequency f = 800kHz

Distance d = 2.7km = 2700m

Electric field Eo = 0.36V/m

Intensity of radio signal

The intensity of radial signal is given as

I = c•εo•Eo²/2

Where c is speed of light

c = 3×10^8m/s

εo = 8.85 × 10^-12 C²/Nm²

I = 3×10^8 × 8.85×10^-12 × 0.36²/2

I = 1.72 × 10^-4W/m²

I = 172 × 10^-6 W/m²

I = 172 µW/m²

Then, the intensity of the radio wave at that point is approximately 170 µW/m²

7 0
3 years ago
Scuba divers are warned that if they must make a rapid ascent, they should exhale on the way up. If a diver rapidly ascends to t
Whitepunk [10]

Air for a diver comes out of a high pressure tank at - Same- pressure compared to the water around the diver (metered by the regulator).

This means the lungs are inflated with - Highly pressurized- gas.

This does not adversely affect the diver when deep underwater, because the entire environment around the diver is at -Same - pressure.

If the diver suddenly surface, the air in the alveoli in the lungs will still be at - a higher - pressure compared to the air around the diver, which will be at - a lower - pressure.

The gas in the diver's lungs will - expand - and can damage the alveoli.

5 0
3 years ago
After an initial race George determines that his car loses 35 percent of its acceleration due to air resistance travelling at 38
ddd [48]

Answer:

64.2 m/s

Explanation:

We are given that

Speed ,v=38 m/s

We have to find the maximum speed when his car reach on flat ground.

Using dimensional analysis

F_{res}\propto v^2

If 35% acceleration reduced by F(res) at 38 m/s

Then, 100% acceleration  can be reduced  by F(res) at v' m/s

\frac{F_1}{F_2}=\frac{v^2}{v'^2}

v'^2=\frac{F_2}{F_1}v^2

v'=v\sqrt{\frac{F_2}{F_1}}

Substitute the values

v'=38\times \sqrt{\frac{100}{35}}

v'=64.2 m/s

Hence, the maximum speed when his car can reach on flat ground=64.2 m/s

3 0
3 years ago
Other questions:
  • Theo mixes cake batter and places it in the oven at 350°F for 45 minutes. Explain what type of reaction baking a cake represents
    15·2 answers
  • Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the
    12·1 answer
  • Consider the case of the car starting at rest and accelerating forward. A. Since the air inside the car is not leaking out, it m
    7·1 answer
  • What’s an example of contact friction
    7·1 answer
  • Jimmy held the end of a metal bar over a fire while holding on to the opposite end. After a few minutes, the end he was holding
    6·1 answer
  • In a building with 10.000 cubic feet where the air changes every two hours, what the rate of air change? A. 167.7 cfm B. 83.3 cf
    15·2 answers
  • If a ball is 10m high with what velocity will it fall?
    13·1 answer
  • A clarinet sounds as a closed pipe. if a clarinet sounds a note with a pitch of 375 hz, what are the frequencies of the lowest t
    6·1 answer
  • How quickly can the nervous system relay messages?
    13·2 answers
  • Does charged battery have energy
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!