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A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0

3 years ago

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

7 0

3 years ago

The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point

Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth, R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

w = 2π / T

w = 2π / 24*3600

w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

v1 = R*w

v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

π/2 ........... s

x ............ 1/5 s

x = π/2*5 = 18°

- The radius of the earth R' at point where θ = 18° from the equator is:

R' = R*cos(18)

R' = (6.37 * 10 ^6)*cos(18)

R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

v2 = R'*w

v2 = (6058230.0088)*(7.27 * 10^-5)

v2 = 440.433 m/s

5 0

3 years ago

Learning Task 2: Write the words that can be associated with the ''Music During Classical Era''. You may ask help from the membe

Natasha_Volkova [10]

Answer:

classical

concert

exposition

orchesta

devolopment

recapitulation

sympathy

sonata

sinfonia

soloist

Explanation:

5 0

3 years ago

7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N

artcher [175]

Answer:

20.4 $m/s^{2}$

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 $m/s^{2}$

7 0

3 years ago