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____ [38]
2 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]2 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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A Labrador retriever breeder needs to determine if three male pups who were born to Sadie, his black Labrador retriever, are the
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Answer: The things we know are:

Sadie, the mother, is a black labrador.

Sam, the possible father, is a yellow labrador.

Discarding things such as a scientific indagation, the breeder could look for traits in the pups.

For example, yellow is recessive against black, so if the pups are yellow, there is probable that the father is Sam, none the less, if the pups are black there is also the possibility where Sam is the father, just in this case the color genes that the pups received are the ones of the mother.

A good starting point is looking at the color, the posture, and things like that.

In the end, these things are only statistics, there is not any determination here, the only real (scientific) determinative process is a paternity test.

7 0
2 years ago
A proton moves along the x-axis with vx=1.0×107m/s. As it passes the origin, what are the strength and direction of the magnetic
Sunny_sXe [5.5K]

Answer:

Magnetic field will be ZERO at the given position

Explanation:

As we know that the magnetic field due to moving charge is given as

B = \frac{\mu_0 qv sin\theta}{4\pi r^2}

so here we know that for the direction of magnetic field we will use

\hat B = \hat v \times \hat r

so we have

\hat B = \hat i \times (\hat i + 0\hat j + 0\hat k)

so magnetic field must be ZERO

So whenever charge is moving along the same direction where the position vector is given then magnetic field will be zero

3 0
2 years ago
For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir
abruzzese [7]

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s




Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


hence closest is in case C at 5.1 m/s




7 0
2 years ago
Read 2 more answers
Jess kicks a soccer ball, and it rolls across the ground. The force diagram shows all the forces acting on the ball at the momen
juin [17]
Arrow at the left side pointing towards right side represents the frictional force as it always acts opposite to motion
7 0
2 years ago
Read 2 more answers
A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

4 0
2 years ago
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