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____ [38]
3 years ago
7

A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction

al force required to keep the faster car on the road without skidding is F, then the frictional force required to keep the slower car on the road without skidding is __-
Physics
1 answer:
yawa3891 [41]3 years ago
8 0

Answer:

F'=\dfrac{F}{4}

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

For the slower car on the road,

F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

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Object 1 of mass m moves with speed v in the positive direction. Object 2 of mass 3 m moves with speed 4 v in the negative x-dir
Klio2033 [76]

Answer:

This means that the kinetic energy of second object is 48times that of the first object

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion e.g motion of an accelerating car. Mathematically,

Kinetic energy = 1/2mv² where;

m is the mass of the object

v is the velocity of the object

If Object 1 of mass m moves with speed v in the positive direction, its kinetic energy will be expressed as;

K1 = 1/2mv²

For Object 2 of mass 3m moving with speed 4v in the negative x-direction, its kinetic energy can be expressed as;

K2 = 1/2(3m)(4v)²

K2 = 1/2(3m)(16v²)

K2 = (3m)(8v²)

K2 = 24mv²

To compare the kinetic energy of both bodies, we will take the ratio of K2:K1 to have;

K2/K1 = 24mv²/(1/2)mv²

K2/K1 = 24/(1/2)

K2/K1 = 48

K2 = 48K1

This means that the kinetic energy of second object is 48times that of the first object and moving in the negative x direction since the body of mass 3m initially moves in the negative x direction.

3 0
3 years ago
If you can throw a stone straight up to height h, what’s the maximum horizontal distance you could throw it over level ground?
Mariana [72]
To answer this question, first we take note that the maximum height that can be reached by an object thrown straight up at a certain speed is calculated through the equation,
                            Hmax = v²sin²θ/2g
where v is the velocity, θ is the angle (in this case, 90°) and g is the gravitational constant. Since all are known except for v, we can then solve for v whichi s the initial velocity of the projectile. 

Once we have the value of v, we multiply this by the total time traveled by the projectile to solve for the value of the range (that is the total horizontal distance). 
8 0
3 years ago
Read 2 more answers
A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the
11Alexandr11 [23.1K]

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\

<u>v_f = 10.85 m/s</u>

4 0
2 years ago
What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified
oee [108]

Answer:

hello your question is not properly arranged attached below is the arranged table and solution

answer : attached table below

Explanation:

Given data:

02 molecules size = 10^-10m

smoke particles size = 0.3 mm

cloud droplets size = 20 mm

Rain droplets size = 3 mm

Attached below is a table showing the kind of scattering that is expected to occur at various wave lengths

Note : For Rayleigh scattering the wave particle is smaller than the wave length while for Non-selective scattering the wave particle is greater than the wavelength.

and  For Mie scattering the wavelength is the same as the wavelength.

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