Properties of dot product

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

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(c) $A = 5.51 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

(c) For face centered cubic lattice:

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

6 0

2 years ago

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0

2 years ago

A 15.0-kg child descends a slide 2.40 m high and reaches the bottom with a speed of 1.10 m/s .

pickupchik [31]

The thermal energy that is generated due to friction is 344J.

<h3>What is the thermal energy?</h3>

Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.

Thus, the kinetic energy is given as;

KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J

PE = mgh = 15.0-kg * 9.8 m/s^2 * 2.40 m = 352.8 J

The thermal energy is; 352.8 J - 9.1 J = 344J

Learn more about thermal energy due to friction:brainly.com/question/7207509

#SPJ1

7 0

1 year ago

If the center of mass passes outside the area of support of an object, what will happen to it?

defon

Answer:

If a vertical line extending down from an object's CG extends outside its area of support, the object will topple

Explanation:

We can understand better this situation using a diagram with the forces acting on it.

In the attached image we can see that when the gravity center is bouncing outside from the area of the pedestal, the object will be out of balance and will fall.

6 0

2 years ago