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WINSTONCH [101]
3 years ago
6

A curve of radius 166 m is banked at an angle of 11°. An 736-kg car negotiates the curve at 81 km/h without skidding. Neglect th

e effects of air drag and rolling friction. Find the following. (a) the normal force exerted by the pavement on the tires.

Physics
1 answer:
lianna [129]3 years ago
6 0

Answer:

F_n = 7509.33\ N

Explanation:

given,

radius of curve = 166 m

angle of the banked road = 11°

mass of car = 736 Kg

speed of the curve = 81 km/h

                                = 81 x 0.278 = 22.52 m/s

normal force acting on the tires

on tire there will be two force acting on it

first one will be force acting due to weight and the other force acting on the tire is due to centripetal force.

F_n = m g cos \theta+ \dfrac{mv^2}{r} sin \theta

F_n = 736 \times 9.8 \times cos 11^0+\dfrac{736 \times 22.52^2}{166} sin 11^0

F_n = 7080.28+429.047

F_n = 7509.33\ N

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Ivanshal [37]
<span>Answer: I'm pretty sure the SA / V ratio would get smaller. Assume that the cell is more or less spherical. SA = 4(pi)r^2, while V = (3/4)(pi)r^3. The ratio = (4(pi)r^2)/((3/4)(pi)r^3), which can be simplified to 3/r. Thus, the larger r gets, the smaller the ratio becomes.</span>
3 0
3 years ago
Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the
Murljashka [212]

Answer:

The net force on the box is 2 N to the left.

The box will move to the left.

The acceleration on the box is 0.5 m/s^2 to the left.

Explanation:

Let's say movement to the right is positive and left is negative.

Bob: +10 N

John: -12 N

Add those together and you get a net force of -2 N, and the negative sign means that the box is moving to the left.

For the acceleration:

Fnet = ma

-2 = (4 kg)a

a = -0.5 m/s^2

Again, the negative sign in this answer means the box is being accelerated to the left.

4 0
3 years ago
A physicist's right eye is presbyopic (i.e., farsighted). This eye can see clearly only beyond a distance of 97 cm, which makes
Ivan

Answer:

f = 19,877 cm   and  P = 5D

Explanation:

This is a lens focal length exercise, which must be solved with the optical constructor equation

        1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image.

In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye

let's calculate

        1 / f = 1/97 + 1/25

        1 / f = 0.05

        f = 19,877 cm

the power of a lens is defined by the inverse of the focal length in meters

         P = 1 / f

         P = 1 / 19,877 10-2

         P = 5D

5 0
3 years ago
A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t
Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=\sqrt{\frac{Gmm}{R+h} }         [(R+h) is the distance of the satellite   from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = h_{i} = 98 km = 98000 m

∴Initial Energy (E_{i})  = \frac{1}{2}mv^{2} + \frac{-GMm}{(R+h_{i} )}

Substituting v=\sqrt{\frac{GMm}{R+h_{i} } } in the above equation and simplifying we get,

E_{i} = \frac{-GMm}{2(R+h_{i}) }

Similarly for final condition,

h=h_{f} = 198km = 198000 m

∴Final Energy(E_{f}) = \frac{-GMm}{2(R+h_{f}) }

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = E_{f} - E_{i}

        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

Substituting ,

M = 6 × 10^{24} kg

m = 1036 kg

G = 6.67 × 10^{-11}

R = 6400000 m

h_{i} = 98000 m

h_{f} = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = \frac{1}{2}m[v_{f} ^{2} - v_{i} ^{2}]

                                                          = \frac{GMm}{2}[\frac{1} {R+h_{f} } - \frac{1} {R+h_{i} }]

                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} }]

                                                             = 2ΔE

                                                             = 968.907 MJ

3 0
3 years ago
Why might video games increase creativity while the use of cell phones, the internet, or computers do not?
taurus [48]

Explanation:

Video games are developed around a structure that is unique from your usual media, where there is a sense purpose in mind when playing a game, as decided by the player, and it allows them to explore creative options in order to solve scenarios, depending on the genre. In the use of phones, internet, or computers, this structure is more rare and diverse from video games, which does not lean more toward a creative purpose to build from. That idea gives people the inspiration and discover skills they never knew they even had.

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