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andreev551 [17]
4 years ago
9

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a

car can safely travel if the radius of the track is 79.0 m and the coefficient of friction is 0.39?
Physics
1 answer:
ad-work [718]4 years ago
6 0

Answer:

Maximum speed of the car is 17.37 m/s.

Explanation:

Given that,

Radius of the circular track, r = 79 m

The coefficient of friction, \mu=0.39

To find,

The maximum speed of car.

Solution,

Let v is the maximum speed of the car at which it can safely travel. It can be calculated by balancing the centripetal force and the gravitational force acting on it as :

v=\sqrt{\mu rg}

v=\sqrt{0.39\times 79\times 9.8}

v = 17.37 m/s

So, the maximum speed of the car is 17.37 m/s.

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Explanation:

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3 years ago
A 3.6kg mass is accelerated at 2.5m/s. Calculate the resultant force<br> acting on it.
AleksandrR [38]
Force = mass*acceleration so
3.6*2.5 =9 Newtons
5 0
3 years ago
Slope or velocit ime graph represent
Lubov Fominskaja [6]

Answer:

Please check the explanation.

Explanation:

The slope of the velocity-time graph illustrates the change in velocity with respect to change in time.

In other words, the acceleration of the object is defined by the slope of a velocity graph. The acceleration can be obtained by finding the slope at a particular time.

Hence, the slope of the velocity time graph represent represents acceleration.

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7 0
3 years ago
Jeff of the Jungle swings on a 7.6-m vine that initially makes an angle of 32 ∘ with the vertical.
shtirl [24]

To solve this problem we will use the trigonometric concepts to find the distance h, which will allow us to find the speed of Jeff and that will finally be the variable that will indicate the total tension, since it is the variable of the centrifugal Force given in the vine at the lowest poing of the swing.

From the image:

cos (32) = \frac{(7.6-h)}{7.6}

h = 1.1548m

When Jeff reaches his lowest point his potential energy is converted to kinetic energy

PE = KE

mgh = \frac{1}{2} mv^2

v = \sqrt{2gh}

v = \sqrt{2(9.8)(1.1548)}

v= 4.75m/s

Tension in the string at the lowest point is sum of weight of Jeff and the his centripetal force

T = W+F_c

T = mg + \frac{mv^2}{r}

T = (83)(9.8)+\frac{(9.8)( 4.75)^2}{7.6}

T = 842.49N

Therefore the tension in the vine at the lowest point of the swing is 842.49N

3 0
3 years ago
A particle with a charge of 5.13 μC has a velocity of 8.64 x106 m/s in a direction perpendicular to a magnetic fieldof 1.99 x 10
Svetach [21]

Answer:

F= 0.009 N

Explanation:

Given that

Charge ,q= 5.13 μC

Velocity ,V= 8.64 x 10⁶ m/s

Magnetic field , B = 1.99 x 10⁻⁴ T

The force on a charge q moving with velocity v is given as follows

F= q V B

Now by putting the values in the above equation we get

[tex]F= 5.13\times 10^{-6}\times 8.64\times 10^{6}\times 1.99\times 10^{-4}\ N [\tex]

F=0.00882 N

F= 0.009 N

Therefore the force on the particle will be 0.009 N.

5 0
4 years ago
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