A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed (12 m/s), the total force on the driver has a magnitude of 146 N. What is the total vector force (in N) on the driver if the speed is 18 m/s instead?

Answers

Answer 1

Answer:

a1 =  v1^2 / R

a2 =  v2^2 / R

a2 = (v2 / v1)^2 = (3 / 2)^2 = 9/4

F2 = 9/4 * F1 = 9/4 * 146 = N     329 N      since F = m * a


Related Questions

help asap PLEASE I will give u max everything all that

steps if possible

Answers

Explanation:

2. [tex]R_T = R_1 + R_2 + R_3 = 625\:Ω + 330\:Ω + 1500\:Ω[/tex]

[tex]\:\:\:\:\:\:\:= 2455\:Ω = 2.455\:kΩ[/tex]

3. Resistors in series only need to be added together so

[tex]R_T = 8(140\:Ω) = 1120\:Ω = 1.12\:kΩ[/tex]

If you drive first at 40 km/h west and later at 60 km/h west, your average velocity is 50 km/h west.

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and what else? is that all?

c) You wish to put a 1000-kg satellite into a circular orbit 300 km above the earth's surface. (a)
What speed, period, and radial acceleration will it have? (b) How much work must be done to the
satellite to put it in orbit? (c) How much additional work would have to be done to make the

Answers

Answer:

Scalar

Explanation:

No direction

A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is 48.0 kg, air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.50 m/s. Determine the work done (in J) by the child as the tricycle travels down the incline.

Answers

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, [tex]v_i[/tex] = 1.4 m/s

final velocity of the child, [tex]v_f[/tex] = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

total mass, m = 48 kg

frictional force, [tex]f_k[/tex] = 41 N

The work done by the child is calculated as;

[tex]\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech} + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96 \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J[/tex]

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

Are you aware of human rights violation happening in the community and explain​

Answers

Answer:

Individuals who commit serious violations of international human rights or humanitarian law, including crimes against humanity and war crimes, may be prosecuted by their own country or by other countries exercising what is known as “universal jurisdiction.”

Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible?

Answers

Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.

Explanation:

It is known that formula for path difference is as follows.

[tex]\Delta L = (n + \frac{1}{2}) \times \frac{\lambda}{2}[/tex]    ... (1)

where, n = 0, 1, 2, and so on

As John is standing perpendicular to the line joining the speakers. So, the value of [tex]L_{1}[/tex] is calculated as follows.

[tex]L_{1} = \sqrt{(2)^{2} + (5)^{2}}\\= 5.4 m[/tex]

Hence, path difference is as follows.

[tex]\Delta L = (5.4 - 5) m = 0.4 m[/tex]

For lowest frequency, the value of n = 0.

[tex]\Delta L = (0 + \frac{1}{2}) \times \frac{\lambda}{2} = \frac{\lambda}{4}[/tex]

[tex]\lambda = 4 \Delta L[/tex]

where,

[tex]\lambda[/tex] = wavelength

The relation between wavelength, speed and frequency is as follows.

[tex]\lambda = \frac{\nu}{f}\\4 \Delta L = \frac{\nu}{f}\\[/tex]

where,

[tex]\nu[/tex] = speed

f = frequency

Substitute the values into above formula as follows.

[tex]f = \frac{\nu}{4 \Delta L}\\f = \frac{340}{4 \times 0.4 m}\\= 212.5 Hz[/tex]

Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.

b) When the muscles connected to the crystalline lens contract fully, its focal length is 16.5000 cm. With this focal length, how far away must an object be to form sharply focused images on the retina? (Note: this distance is called the far point of vision.)
c) When the muscles connected to the crystalline lens relax, the focal length is 9.0000 cm. With this focal length, how close must an object be to form sharply focused images on the retina? (Note: this distance is called the near point of vision.)
d) As people age, the crystalline lens hardens (a condition called presbyopia or “old-age” eyes) and can only vary in focal length from 12 to 15.60 cm. Calculate range of vision (the new near point and far point) for this older eye.
e) Based on part d) why might an older person hold the newspaper at arm’s length to read it?

Answers

Answer:

I have to go to work and figure it out

The speed of a 2.0-kg object changes from 30 m/s to 40 m/s during a 5.0-second time interval.
During this same time interval, the velocity of the object changes its direction by 90°. What is the
magnitude of the average total force acting on the object during this time interval?
a. 30 N
b. 20 N
c. 15 N
d. 40 N
e. 10 N
Which is the correct answer?

Answers

Answer:

F = 2 * 30 / 5 = 12 N to stop forward motion

F = 2 * 40 / 5 = 16 N to accelerate to 90 degrees

(12^2 + 16^2)^1.2 = 20 N   average force applied

The magnitude of the average total force acting on the object during this time interval is 20 N.

The given parameters:

Mass of the object, m = 2.0 kgInitial velocity, u = 30 m/sFinal velocity, v = 40 m/sTime of motion, t = 5.0 s

The magnitude of the average total force acting on the object during this time interval is calculated as follows;

[tex]F = \frac{mv }{t} \\\\F_1 = \frac{2(40)}{5} \\\\F_1 = 16\ N\\\\F_2= \frac{2(30)}{5} \\\\F_2 = 12 \ N\\\\F = \sqrt{F_1^2 + F_2^2} \\\\F = \sqrt{16^2 + 12^2} \\\\F = 20 \ N[/tex]

Learn more about resultant force here:  https://brainly.com/question/25239010

How many times will the temperature of oxygen with a mass of 1 kg increase if its volume is increased by 4 times, and the pressure is decreased by 2 times?
Round off the answer to the nearest whole number.

Answers

Answer:

9.2 Relating Pressure, Volume,

Figure 1. In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.

Explanation:

hope its help :)

nicsfrom #philippines

A(n) 28.3 g bullet is shot into a(n) 5004 g wooden block standing on a frictionless surface. The block, with the bullet in it, acquires a speed of 1.7 m/s. Calculate the speed of the bullet before striking the block. Answer in units of m/s

Answers

Answer:

the speed of the bullet before striking the block is 302.3 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 28.3 g = 0.0283 kg

mass of the wooden block, m₂ = 5004 g = 5.004 kg

initial velocity of the block, u₂ = 0

final velocity of the bullet-wood system, v = 1.7 m/s

let the initial velocity of the bullet before striking the block = u₁

Apply the principle of conservation of linear momentum to determine the initial velocity of the bullet.

m₁u₁  +  m₂u₂  =  v(m₁  +  m₂)

0.0283u₁  + 5.004 x 0   =  1.7(0.0283  +  5.004)

0.0283u₁   =   8.5549

u₁ = 8.5549 / 0.0283

u₁ = 302.3 m/s

Therefore, the speed of the bullet before striking the block is 302.3 m/s.

5. Tests performed on a 16.0 cm strip of the donated aorta reveal that it stretches 3.37 cm when a 1.80 N pull is exerted on it. (a) What is the force constant of this strip of aortal material

Answers

Answer:

53.41 N/m

Explanation:

From Hooke's law,

Applying,

F = ke............. Equation 1

Where F = Force, e = extension, k = force constant of the aortal material

Make k the subject of the equation

k = F/e............. Equation 2

From the question,

Given: F = 1.8 N, e = 3.37 cm = 0.0337 m

Substitute these values into equation 2

k = 1.8/(0.0337)

k = 53.41 N/m

Hence the force constant of the aortal material is 53.41 N/m

Copy the diagram. add a voltmeter to show how you would measure the voltage of the cell

Answers

Answer: the answer is 23voltage

Explanation: because the voltage and time put together is 23

Pete is investigating the solubility of salt (NaCl) in water. He begins to add 50 grams of salt to 100 grams of
room temperature tap water in a beaker. After adding all of the salt and stirring for several minutes, Pete
notices a solid substance in the bottom of the beaker. Which statement best explains why there is a solid
substance in the bottom of the beaker?
A. The salt he is using is not soluble in water.
B. The salt is changing into a new substance that is not soluble in water,
C. The dissolving salt is causing impurities in the water to precipitate to the bottom
D. The water is saturated and the remaining salt precipitates to the bottom

Answers

Answer:

would the answer be c

Explanation: that what i think in my opian

Answer:

A

Explanation:

A rocket at fired straight up from rest with a net upward acceleration of 20 m/s2 starting from the ground. After 4.0 s, the thrusters fail and the rocket continues to coast upward with insignificant air resistance. (a) What is the maximum height reached by the rocket

Answers

Answer:

The maximum height reached by the rocket is 486.53 m

Explanation:

Given;

initial velocity of the rocket, u = 0

acceleration of the rocket, a= 20 m/s²

duration of the rocket first motion, t = 4 s

The distance traveled by the rocket before its thrust failed

h₁ = ut + ¹/₂at²

h₁ = 0 + ¹/₂ x 20 x 4²

h₁ = 160 m

The second distance moved by the rocket is calculated as follows;

The velocity of the rocket before its thrust failed;

v = u + at

v = 0 +  20 x 4

v = 80 m/s

This becomes the initial velocity for the second stage

At maximum height, the final velocity = 0

[tex]v_f^0 = v_i^2 - 2gh_2\\\\0 = (80)^2 - (2 \times 9.8)h_2\\\\0 = 6400 - 19.6h_2\\\\19.6h_2 = 6400\\\\h_2 = \frac{6400}{19.6} \\\\h_2 = 326.53 \ m[/tex]

The maximum height reached by the rocket = h₁ + h₂

                                                                          = 160 + 326.53

                                                                          = 486.53 m

Transfer of thermal energy between air molecules in closed room is an example of

conduction

convection

radiation


Answer and I will give you brainiliest ​

Answers

Answer: Conduction

Explanation: Conduction is the process by which heat energy is transmitted through collisions between neighboring atoms or molecules. Conduction occurs more readily in solids and liquids, where the particles are closer to together, than in gases, where particles are further apart.






One way families influence healthy technology use is when siblings explain the use of media to each other. Which of these outfits would you expect if this guideline was followed?

Answers

Answer:

The answer would be C.

Explanation:

This is what I would expect when you show someone else how to do something then is also known as teaching.

Please Mark as Brainliest

Hope this Helps

A 2000 kg car experiences a braking force of 10000N and skids to a 14 m stop. What was the speed of the car just before the brakes.

Answers

Answer:

V = 11.83 m/s

Explanation:

Given the following data;

Mass = 2000 kg

Force = 10000N

Distance = 14 m

To find the final velocity of the car;

First of all, we would determine the acceleration of the car;

Acceleration = force/mass

Acceleration = 10000/2000

Acceleration = 5 m/s²

Next, we would use the third equation of motion to find the final velocity;

[tex] V^{2} = U^{2} + 2aS [/tex]

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement measured in meters.

Substituting into the equation, we have;

V² = 0² + 2*5*14

V² = 0 + 140

V = √140

V = 11.83 m/s

A cement block accidentally falls from rest from the ledge of a 53.4-m-high building. When the block is 19.4 m above the ground, a man, 2.00 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way

Answers

Answer:

The time required by the man to get out of the way is 0.6 s.

Explanation:

height of building, H = 53.4 m

height of block, h = 19.4 m

height of man, h' = 2 m

Let the velocity of the block at 19.4 m is v.  

use third equation of motion

[tex]v^2 = u^2 + 2 gh\\\\v^2 = 0 + 2 \times 9.8 \times (53.4 - 19.4)\\\\v = 25.8 m/s[/tex]

Now let the time is t.

Use second equation of motion

[tex]h = u t + 0.5 gt^2\\\\19.4 - 2 = 25.8 t + 4.9 t^2\\\\4.9 t^2 + 25.8 t - 17.4= 0 \\\\t = \frac{-25.8\pm\sqrt{665.64 + 341.04}}{9.8}\\\\t = \frac{-25.8\pm31.7}{9.8}\\\\t = 0.6 s, - 5.9 s[/tex]

Time cannot be negative so time t = 0.6 s.

7. An electric train moving at 20km/hrs
. Accelerates to a speed of 30km/hrs. in
20 sec, find the distance travelled in meters during the period of
acceleration​

Answers

Answer

NB:

- speed, U is measure in m/s

- acceleration, a is measured in m/s²

-time t in seconds , s

Therefore conversation must be made

Speed U = 20km/hrs

=20km÷1hr

But 20km= 20×1000=20000m

1hr= 1×60min×60sec=3600s

U=20000÷3600=5.56m/s

a=30km/hrs

=30km÷1hr

But 30km=30×1000=30000

1hr=3600s

a=30000÷3600=8.33m/s²

From the equation of motion

S=Ut + ½ at².

Where s= distance

S = 5.56m/s × 20s + ½(8.33m/s²)(20s)²

S = 1777.3m

Please help I need this done

Answers

The 3rd option if not sorry I think that is the answer

Calculate the rms speed of helium atoms near the surface of the Sun at a temperature of about 5100 K. Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

[tex]V_{rms}=5.6*10^3m/s[/tex]

Explanation:

From the question we are told that:

Temperature [tex]T=5100K[/tex]

Generally the equation for RMS Speed is mathematically given by

 [tex]V_{rms}=\sqrt{\frac{3kT}{m}}[/tex]

Where

 [tex]K=Boltzman's constant[/tex]

 [tex]K=1.38*10^{-23}[/tex]

And

 [tex]M=molecular mass[/tex]

 [tex]M=4*1.67*10^{-27}[/tex]

 [tex]V_{rms}=\sqrt{\frac{3(1.38*10^{-23})5100}{4*1.67*10^{-27}}}[/tex]

 [tex]V_{rms}=5.6*10^3m/s[/tex]

Its volume is 20 cm3, and its mass is 100 grams. What is the sample’s density?

Answers

Well the density is mass/volume
100/20 = 5
The density is 5

E=kq/r^2 chứng minh điện thế V=kq/r từ mối liên hệ giữa điện trường E và điện thế V

Answers

Answer:

hindi ko maintindihan teh

Equilibrium of forces

Answers

Answer:

If the size and direction of the forces acting on an object are exactly balanced, then there is no net force acting on the object and the object is said to be in equilibrium. Because the net force is equal to zero, the forces in Example 1 are acting in equilibrium.

Equilibrium of forces means that the net force is 0. It can either be when there is no force acting on the object or when the force acting on the object are balanced.

An amusement park ride whisks you vertically upward. You travel at a constant speed of 15 m/s during the entire ascent. You drop your phone 4.0 s after you (and your phone) begin your ascent from ground level.
a. How high above the ground is your phone when you drop it?
b. Find the maximum height above the ground reached by your phone.

Answers

Answer:

a. 60 m

b. 71.48 m

Explanation:

Below are the calculations:

a. The phone's height above the ground = Speed x Time

   The phone's height above the ground = 15 x 4 = 60 m

b. Speed when phone drops, u = 15 m/s

At maximum height, v = 0

Use below formula:

v² = u² -2gh

0 = 15² + 2 × 9.8 × h

h = 11.48 m

Total height = 60 + 11.48 = 71.48 m

Hai điện tích điểm Q1 = 8 C, Q2 = –6
C đặt tại hai điểm A, B cách nhau 0,1
m trong không khí. Tính cường độ điện
trường do hai điện tích này gây ra tại
điểm M, biết MA = 0,2 m

Answers

Answer:

English please

Explanation:

I don't understand the question

The wave functions for states of the hydrogen atom with orbital quantum number l=0 are much simpler than for most other states, because the angular part of the wave.

a. True
b. False

Answers

True. Fun fact. Hope this helps

When two bodies at different temperatures are placed in thermal contact with each other, heat flows from the body at higher temperature to the body at lower temperature until them both acquire the same temperature. Assuming that there is no loss of heat to the surroundings, the heatSingle choice.

(1 Point)

(a) gained by the hotter body will be equal to the heat lost by the colder body

(b) the heat gained by the hotter body will be less than the heat lost by the colder body

(c) the heat gained by the hotter body will be greater than the heat lost by the colder body

(d) the heat lost by the hotter body will be equal to the heat gained by the colder body.​

Answers

Answer:

Part d is correct.

Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate would the car have to accelerate so that a quarter placed on the back wall would remain in place?

Answers

Answer:

25.59 m/s²

Explanation:

Using the formula for  the force of static friction:

[tex]f_s = \mu_s N[/tex] --- (1)

where;

[tex]f_s =[/tex] static friction force

[tex]\mu_s =[/tex] coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

[tex]N = ma_{min}[/tex]

From equation (1)

[tex]f_s = \mu_s ma_{min}[/tex]

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

[tex]F = f_s[/tex]

[tex]mg = \mu_s ma_{min}[/tex]

[tex]a_{min} = \dfrac{mg }{ \mu_s m}[/tex]

[tex]a_{min} = \dfrac{g }{ \mu_s }[/tex]

where;

[tex]\mu_s = 0.383[/tex] and g = 9.8 m/s²

[tex]a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }[/tex]

[tex]\mathbf{a_{min}= 25.59 \ m/s^2}[/tex]

Kaseem is taking his bicycle for a ride. His bicycle is a system, and its main purpose is to provide transportation. What is the main input into this system? What is the desired output of this system?

Answers

His bicycle that operates on a system for transportation
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