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igomit [66]
3 years ago
12

To counter the effects of centrifugal force and reduce vehicle traction it is important to to counter the effects of centrifugal

force and reduce vehicle traction it is important to

Physics
1 answer:
tatiyna3 years ago
5 0
Answer:  Add an incline or grade to the road track.

Explanation:
Refer to the figure shown below.

When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v =  linear (tangential) velocity to the circular path.

The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.

At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.

When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
 Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).


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The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 440g . The vibra
ANEK [815]

Answer:

Tension, T = 2038.09 N

Explanation:

Given that,

Frequency of the lowest note on a grand piano, f = 27.5 Hz

Length of the string, l = 2 m

Mass of the string, m = 440 g = 0.44 kg

Length of the vibrating section of the string is, L = 1.75 m

The frequency of the vibrating string in terms of tension is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

\mu=\dfrac{m}{l}

\mu=\dfrac{0.44}{2}=0.22\ kg/m

T=4L^2f\mu

T=4\times (1.75)^2\times (27.5)^2 \times 0.22

T = 2038.09 N

So, the tension in the string is 2038.09 N. Hence, this is the required solution.

6 0
4 years ago
How much higher is 0.25 meters than 0.0254 meters (with real objects)
sineoko [7]

Answer:

0.2246 meters

Explanation:

.25-0.0254 meters = 0.2246 meters

6 0
3 years ago
Consider the following situation, in which a compass is influenced by not only the force of Earth's magnetic field, but also an
DochEvi [55]

Answer:

1) The angle of deflection will be less than 45° ( C )

2)  The angle of deflection will be greater than 45° but less than 90° ( E )

Explanation:

1) Assuming that the force applied  has a direction which is perpendicular to the Earth's magnetic field

∴ Fearth > Fapplied   hence the angle of deflection will be < 45°

2) when the Fearth < Fapplied

the angle of deflection will be :   > 45°  but  < 90°

6 0
3 years ago
An investigator places a sample 1.0 cm from a wire carrying a large current; the strength of the magnetic field has a particular
Sonja [21]

Answer: <em>she will have to increase the factor of current by</em> 11

Explanation: The mathematical relationship between the strength of the magnetic field (B) created by a current carrying conductor with current (I) is given by the Bio-Savart law given below

B=\frac{u_{0}I }{2\pi r}

B=strength of magnetic field

I = current on conductor

r = distance on any point of the conductor from it center

u_{0} = permeability of magnetic field in space

from the question, the investigator is trying to keep a constant magnetic field meaning B has a fixed value such as the constants in the formulae, the only variables here are current (I) and distance (r). We can get this a mathematical function.

by cross multipying, we have

B* 2πr=u_{0}<em>I </em>

by dividing through to make <em>I </em>subject of formulae, we have that

<em>I </em>= \frac{B*2\pi r}{u_{0} }

B, 2π and u_{0} are all constants, thus

\frac{B*2\pi r}{u_{0} } = k(constant)

thus we have that

<em>I </em>=kr<em> (current is proportional to distance assuming magnetic field strength and other parameters are constant) </em>

thus we have that

\frac{I_{1} }{r_{1} }=\frac{I_{2} }{r_{2} }

r_{1}=1cm and r_{2}=11cm

\frac{1_{1} }{1}=\frac{I_{2} }{11}

thus I_{2}=11* I_{1}

which means the second current is 11 times the first current

8 0
3 years ago
Hello please help i’ll give brainliest
andriy [413]

Answer:

It should be (A few centimeters per year) About three to five centimeters

5 0
3 years ago
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