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
storchak [24]
3 years ago
7

A car goes round a curve of radius 48m, the road is banked at an angle of 15 with the horizontal,at what maximum speed may the c

ar travel if there is to be low tendency to skid even on very slippery pavement​
Physics
1 answer:
Marizza181 [45]3 years ago
3 0

Answer:

11 m/s

Explanation:

Draw a free body diagram.  There are two forces acting on the car:

Weigh force mg pulling down

Normal force N pushing perpendicular to the incline

Sum the forces in the +y direction:

∑F = ma

N cos θ − mg = 0

N = mg / cos θ

Sum the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Substitute and solve for v:

(mg / cos θ) sin θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Plug in values:

v = √(9.8 m/s² × 48 m × tan 15°)

v = 11.2 m/s

Rounded to 2 significant figures, the maximum speed is 11 m/s.

You might be interested in
What’s the velocity of a sound wave traveling through air at a temperature of 18°C (64.4°F)?
Anna11 [10]

Answer:

342 m/s

Explanation:

The velocity of sound in air is approximated as:

v ≈ 331.4 + 0.6 T

where v is the velocity in m/s and T is the temperature in Celsius.

At T = 18:

v ≈ 331.4 + 0.6 (18)

v ≈ 342.2

The velocity is approximately 342 m/s.

4 0
4 years ago
Read 2 more answers
An object on Earth weighs 150 N. What is its mass?
Alex777 [14]

Answer:

\tt \: mass \:  =  \frac{weight}{acceleration \: due \: to \: gravity}

\longrightarrow \tt \:  \frac{150}{10}

\longrightarrow \boxed{ \tt{15 \: kg}}

  • Our final answer is 15 kg .

----------- HappY LearninG<3 ------------

3 0
2 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
3 years ago
Wat is the meaning of inference?
VladimirAG [237]

The meaning of inference is the process of inferring  something and it is a noun


5 0
3 years ago
Read 2 more answers
If you hear thunder 5 seconds after you see a lightning bolt how far is the storm
NikAS [45]
I think the answer is 3 miles because its storming now where I live

4 0
3 years ago
Read 2 more answers
Other questions:
  • Your pencil is 11 cm long . How long is it in millimeters?
    7·1 answer
  • A stone was dropped off a cliff and hit the ground with a speed of 136 ft/s. what is the height of the cliff? (use 32 ft/s2 for
    6·1 answer
  • The graph shows a wave that oscillates with a frequency of 60 Hz. Based on the information given in the diagram, what is the spe
    6·1 answer
  • An 19-cm-long bicycle crank arm, with a pedal at one end, is attached to a 23-cm-diameter sprocket, the toothed disk around whic
    15·1 answer
  • A 2.55-m-long rod, as measured in its rest frame, speeds by you longitudinally at 6.11Ã10^7. You measure its length as it passes
    5·1 answer
  • A person trapped outside during a thunderstorm should
    7·2 answers
  • I NEED HELP WITH THE LAST QUESTION ASAP D':
    5·1 answer
  • To test the strength of a retainment wall designed to protect a nuclear reactor,
    14·1 answer
  • Objectives
    10·1 answer
  • How long will it take a person walking at 3.2 m/s to travel 16 m? *
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!