PHYSICS COLLEGE

How fast can the car take this curve this curve without skidding to the outside of the curve?

Answers

Answer 1

Answer:

Lets write the data down. That will help us solve the problem later:

R = 36 m

θ = 18º

m = 1492 kg

μ = 0.67

g = 9.8 m/s²

Lets draw all the forces that act on the car:

In order to the car won't skidding to the outside of the curve we must have the centripetal force equals the friction force:

$F_(cp)=f_a$

$(m.v^(2))/(R)=\mu.F_N$

Related Questions

A stone is thrown with an initial speed of 11.5 m/s at an angle of 50.0 above the horizontal from the top of a 30.0-m-tall building. Assume air resistance is negligible, and g = 9.8 m/s2. What is the magnitude of the horizontal displacement of the rock?
A vibrating tuning fork makes 500 vibrations in one second. What is the wavelength of the sound produced if the air temperature is 20°C? Group of answer choices0.500 m0.686 m0.343 m1.46 m1.87 m
On a day when the water is flowing relatively gently, water in the Niagara River is moving horizontally at 4.5 m/sm/s before shooting over Niagara Falls. After moving over the edge, the water drops 53 mm to the water below.a. If we ignore air resistance, how much time does it take for the water to go from the top of the falls to the bottom? b. Express your answer to two significant figures and include the appropriate units. c. How far does the water move horizontally during this time? d. Express your answer to two significant figures and include the appropriate units.
An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where there is a ceiling fan overhead. He relieves his boredom by throwing the spitwad up at the ceiling fan where it collides with, and sticks to, the end of one of the blades of the stationary ceiling fan. Its horizontal velocity vector is perpendicular to the long axis of the blade. If the fan is free to rotate (no friction at all) and has moment of inertia I=1.4kgm2 , if the spitwad has horizontal velocity 4 m/s, and if the spitwad sticks to the fan blade at a distance of 0.6 m from the rotation axis of the fan, how much time will it take the fan to move through one complete revolution after the spitwad hits it (closest answer)?a. 1minb. 2minc. 3mind. 4mine. 5minf. 6min
An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.

Enter your answer in the provided box. In water conservation, chemists spread a thin film of certain inert materials over the surface of water to cut down on the rate of evaporation of water in reservoirs. This technique was pioneered by Benjamin Franklin three centuries ago. Franklin found that 0.10 mL of oil could spread over the surface of water of about 32.0 m2 in area. Assuming that oil forms a monolayer (that is, a layer that is only one molecule thick) estimate the length of each oil molecule in nanometers. Assume that oil molecules are roughly cubic. (1 nm = 1 × 10−9 m)

Answers

Answer:

≅3.2 nm

Explanation:

Using the converter units as know for this case that:

1 ml is 1 cubic centimeter ⇒ 0.1 ml is 0.1 cubic centimeters

32.0 m² so :

32.0 m² *100 *100 cm² ⇒ 0.1 / ( 32.0 * 100 *100 ) = 100,000,000 * 0.1 / (32.0 * 100 * 100 ) nm

v = 100/32.0 nm = 3.125 nm thick.

v ≅3.2 nm

As oil is one molecule thick and the molecules are cubic, length of each oil is 3.2 nm

wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength

Answers

Complete question:

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:

a) the midpoint between the two rings?

b) the center of the left ring?

Answer:

a) the electric field strength at the midpoint between the two rings is 0

b) the electric field strength at the center of the left ring is 2712.44 N/C

Explanation:

Given;

distance between the two rings, d = 25 cm = 0.25 m

diameter of each ring, d = 10 cm = 0.1 m

radius of each ring, r = $(0.1)/(2) = 0.05 \ m$

the charge on each ring, q = 20 nC

Electric field strength for a ring with radius r and distance x from the center of the ring is given as;

$E = (kxQ)/((x^2 +r^2)^(3/2))$

The electric field strength at the midpoint;

the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m

$E = (kxQ)/((x^2 +r^2)^(3/2)) \n\nE = (8.99*10^(9)*0.125*20*10^(-9))/((0.125^2 + 0.05^2)^(3/2)) \n\nE = 9210.5 \ N/C$

$E_(left) = 9210.5 \ N/C$

The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;

$E_(right) = -9210.5 \ N/C$

The electric field strength at the midpoint;

$E_(mid) = E_(left) + E_(right)\n\nE_(mid) = 9210.5 \ N/C - 9210.5 \ N/C\n\nE_(mid) = 0$

(b)

The distance from the right ring to center of the left ring, x = 0.25 m.

$E = (KxQ)/((x^2 +r^2)^(3/2)) \n\nE = (8.99*10^(9) *0.25*20*10^(-9))/((0.25^2 + 0.05^2)^(3/2)) \n\nE = 2712.44 \ N/C$

Which of the following statements are true?A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping. B. The increase in amplitude of an oscillation by a driving force is called forced oscillation. C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced. D. An oscillation that is maintained by a driving force is called forced oscillation.

Answers

Statements that are right as regards oscillation are:

A. The decrease in the amplitude of an oscillation caused by dissipative forces is called damping.

B. The increase in amplitude of an oscillation by a driving force is called forced oscillation.

C. In a mechanical system, the amplitude of an oscillation diminishes with time unless the lost mechanical energy is replaced.

D. An oscillation that is maintained by a driving force is called forced oscillation.

Amplitude can be regarded as magnitude of change that is been experienced by oscillating variable with each oscillation.

When there is a decrease in the amplitude of an oscillation as a result dissipative forces, then it is regarded as damping.

When there is increase in amplitude of an oscillation as a result of driving force then it is termed forced oscillation.

Therefore, the options are correct.

Learn more at:

brainly.com/question/15272453?referrer=searchResults

Answer:

right A, B, C, D

Explanation:

They ask which statements are true

A) Right. The decrease in amplitude is due to the dissipation of energy by friction and is called damping

B) Right. In resonant processes the amplitude of the oscillation increases, being a forced oscillation

C) Right. In a system with energy loss, the amplitude must decrease, therefore energy must be supplied to compensate for the loss.

D) Right. It is a resonant process the driving force keeps the oscillation of the system

The graphs display velocity data Velocity is on the y-axis (m/s), while time is on the x-axis (s). Based on the graphs, which data set represents constant acceleration?

Answers

Answer:

The first graph is showing the constant acceleration (1 m/s)

Explanation:

The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4

The last graph is showing constant velocity therefore there is no acceleration (a = 0)

Answer:

Explanation:

on edge

A small child weighs 60 N. If mommy left him sitting on top of the stairs, which are 12 m high, how much energy does the child have!Please help ASAP

Answers

Answer:

6000 joules

Explanation:

I jus learned dis

Answer:6000j

Explanation:

Hope that helps

Testosterone is an example of what kind of biomolecule?

Answers

Answer:

Among the four biomolecules: carbohydrates, lipids, nucleic acids, and proteins it falls on the category of protein.

Explanation:

Testosterone, also known as 17-beta-hydroxy-4-androstene-3-one, is an androgen steroid hormone. It is largely released by the testes in males and the ovaries in females, although it is also secreted in minor amounts by the adrenal glands.

Other Questions

Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.380 m. The engine of the car applies a torque of 456 N·m to this wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?

What does a planet need in order to retain an atmosphere? How does an atmosphere affect the surface of a planet and the ability of life to exist?

an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and the push lasts 0.8 seconds. The skater's mass is 55 kg. what is the skater's speed after she stops pushing on the wall

A piece of soft iron is placed in a solenoid increasing the magnetic field in an arrangement that can be switched on and off. Such a device is calleda. a hysteresis loop.b. an electrosolenoid.c. a permanent magnet.d. a ferromagnet.