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Rudik [331]
3 years ago
11

While traveling on a horizontal road at speed vi, a driver sees a large rabbit ahead and slams on the brakes. The wheels lock an

d the car begins to slide against the road. The car collides with the rabbit at a final speed of vf, after T seconds of braking. What is the coefficient of kinetic friction of the tires against the road?
Physics
1 answer:
MA_775_DIABLO [31]3 years ago
4 0

Answer:

μk = (Vf - Vc)/(T×g)

Explanation:

Given

Vi = initial velocity of the car

Vf = final velocity of the car

T = Time of application of brakes

g = acceleration due to gravity (known constant)

Let the mass of the car be Mc

Assuming only kinetic frictional force acts on the car as the driver applies the brakes,

The n from Newtown's second law of motion.

Fk = Mc×a

Fk = μk×Mc×g

a = (Vf - Vc)/T

Equating both preceding equation.

μk×Mc×g = Mc × (Vf - Vc)/T

Mc cancels out.

μk = (Vf - Vc)/(T×g)

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What minimum speed does a 200 g puck need to make it to the top of a frictionless ramp that is 4.1 m long and inclined at 22 ∘?
myrzilka [38]

Answer:

5.5 m/ sec

Explanation:

Because the inclined surface is frictionless so we can assume that total change of energy is zero

i-e ΔE = 0

Or we can say that difference between final and initial energy is zero i-e

Ef- Ei =0

Where,

Ef= final energy at the top of the ramp= KEf+PEf

Ei= Initial energy at the bottom of the ramp=KEi+PEi

So we have

(KEf+PEf)-(KEi+PEi)=0

==>KEf-KEi+PEf-PEi=0            -------------(1)

KEf = mgh = 200×9.8×h

Where h= Sin 22 = h/d= h/4.1

or

0.375×4.1=h

or h= 1.54 m

So, PEf= 200×9.8×1.54=3018.4 j

and KEf= 1/2 mVf^{2}= 0.5×200×0=0 j

PEi= mgh = 200×9.8×0=0 j

KEi= 1/2 mVi^{2}=0.5×200×Vi^{2}=100Vi^{2} j

Put these values in eq 1, we get;

0-100 Vi^{2}+3018.4-0=0

-100 Vi^{2}=-3018.4

==> Vi^{2}= \frac{3018.4}{100} = 30.184

==>  Vi = \sqrt{30.184}  = 5.5 m.sec

7 0
3 years ago
What happens if a mid-ocean ridge occurs on land
Gnesinka [82]

Answer:

Despite being such prominent feature on our planet, much of the mid-ocean ridge system remains a mystery. While we have mapped about half of the global mid-ocean ridge in high resolution, less than one percent of the mid-ocean ridge has been explored in detail using submersibles or remotely operated vehicles. so therefore we do not have enough information about them to know what will happen

Explanation:

A mid-ocean ridge or mid-oceanic ridge is an underwater mountain range, formed by plate tectonics. This uplifting of the ocean floor occurs when convection currents rise in the mantle beneath the oceanic crust and create magma where two tectonic plates meet at a divergent boundary. Mid-ocean ridges occur along divergent plate boundaries, where new ocean floor is created as the Earth’s tectonic plates spread apart. As the plates separate, molten rock rises to the seafloor, producing enormous volcanic eruptions of basalt. The speed of spreading affects the shape of a ridge  slower spreading rates result in steep, irregular topography while faster spreading rates produce much wider profiles and more gentle slopes.

4 0
3 years ago
Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part
Fantom [35]

a) Angular acceleration: 17.0 rad/s^2

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

W=mg

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

W_p =mg sin \theta

where \theta is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta

where L is the length of the pencil.

The relationship between torque and angular acceleration \alpha is

\tau = I \alpha (1)

where

I=\frac{1}{3}mL^2

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for \alpha, we find:

\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when \theta=10^{\circ} is

\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude mg

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

\tau = Fd

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

\tau =mg\frac{L}{2} cos \theta

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

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MrRa [10]
I would say D) melting point.
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3 years ago
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Which electromagnetic has the shortest wavelength?.
Marina CMI [18]

Answer:

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Explanation:

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