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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 10N force pulls to the right and friction opposes 2N. If the object is 20kg,find the acceleraton.
zmey [24]

Force = mass * acceleration

10 N - 2 N = 20 kg * acceleration

8 N = 20 kg * acceleration

8 / 20 = acceleration

2/5 m/s^2 = acceleration

8 0
1 year ago
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An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini
arsen [322]
Kinetic energy is calculated through the equation,

   KE = 0.5mv²

At initial conditions,

  m₁:  KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

  m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

   m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

   (0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

   V = 0.2977 m/s

The kinetic energy is,
   KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
   KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J. 
7 0
2 years ago
A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average
Kaylis [27]

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

3 0
2 years ago
A 250 Hz tuning fork is struck and the intensity at the source is I1 at a distance of one meter from the source. (a) What is the
Zina [86]

Answer:

a) 0.0625 I_1

b) 3.16 m

Explanation:

<u>Concepts and Principles  </u>

The intensity at a distance r from a point source that emits waves of power P is given as:  

I=P/4π*r^2                         (1)

<u>Given Data</u>

f (frequency of the tuning fork) = 250 Hz

I_1 is the intensity at the source a distance r_1 = I m from the source.  

<u>Required Data</u>

- In part (a), we are asked to determine the intensity I_2 a distance r_2 = 4 in from the source.

- In part (b), we are asked to determine the distance from the tuning fork at which the intensity is a tenth of the intensity at the source.  

<u>solution:</u>

(a)  

According to Equation (1), the intensity a distance r is inversely proportional to the distance from the source squared:

I∝1/r^2

Set the proportionality:  

I_1/I_2=(r_2/r_1)^2                                 (2)

Solve for I_2 :  

I_2=I_1(r_2/r_1)^2  

I_2=0.0625 I_1

(b)  

Solve Equation (2) for r_2:  

r_2=(√I_1/I_2)*r_1

where I_2 = (1/10)*I_1:

r_2=(√I_1/1/10*I_1)*r_1

     =3.16 m

3 0
2 years ago
Sharks are generally negatively buoyant; the upward buoyant force is less than the weight force. This is one reason sharks tend
Tresset [83]

Answer:

8.67807 N

34.7123 N

Explanation:

m = Mass of shark = 92 kg

\rho_{se} = Density of seawater = 1030 kg/m³

\rho_{f} = Density of freshwater = 1000 kg/m³

\rho_{sh} = Density of shark = 1040 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Net force on the fin is (seawater)

F_n=mg-V_s\rho_{se}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{se}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1030\times 9.81\\\Rightarrow F_n=8.67807\ N

The lift force required in seawater is 8.67807 N

Net force on the fin is (freshwater)

F_n=mg-V_s\rho_{f}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{f}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1000\times 9.81\\\Rightarrow F_n=34.7123\ N

The lift force required in a river is 34.7123 N

6 0
2 years ago
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