PHYSICS COLLEGE

A typical automobile under hard braking loses speed at a rate of about 7.2 m/s2; the typical reaction time to engage the brakes is 0.55 s. A local school board sets the speed limit in a school zone such that all cars should be able to stop in 3.6 m. What maximum speed does this imply for a car in the school zone?

Answers

Answer 1

Answer:

Answer:

4.3 m/s

Explanation:

a = rate at which the automobile loses speed = - 7.2 m/s²

v₀ = initial maximum speed of automobile

t' = reaction time for applying the brakes = 0.55 s

d = distance available for stopping the vehicle = 3.6 m

d' = distance traveled while applying the brakes = v₀ t' = (0.55) v₀

v = final speed after the vehicle comes to stop = 0 m/s

Using the equation

v² = v₀² + 2 a (d - d' )

0² = v₀² + 2 (- 7.2) (3.6 - (0.55) v₀)

v₀ = 4.3 m/s

Related Questions

How do you perceive the importance of online safety security ethics and etiquette
Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.
Point charge 3.5 μC is located at x = 0, y = 0.30 m, point charge -3.5 μC is located at x = 0 y = -0.30 m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.0 μC at x = 0.40 m, y = 0?
A 5.49 kg ball is attached to the top of a vertical pole with a 2.15 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.654.65 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take ????=9.81g=9.81 m/s2. angle: °
Neon signs require about 12,000 V for their operation. Consider a neon-sign transformer that operates off 120- V lines. How many more turns should be on the secondary compared with the primary?

A basketball has a mass of 575 g. Moving to the right and heading downward at an angle of 31° to the vertical, it hits the floor with a speed of 4 m/s and bounces up with nearly the same speed, again moving to the right at an angle of 31° to the vertical. What was the momentum change ? (Take the axis to be to the right and the axis to be up. Express your answer in vector form.)

Answers

Answer:

Taking the x axis to the right and the y axis to be up, the total change of momentum is $\Delta \vec{ p} = 3.9429 (kg \ m)/(s) \hat{j}$

Explanation:

The momentum $\vec{p}$ is given by:

$\vec{p} = m \ \vec{v}$

where m is the mass and $\vec{v}$ is the velocity. Now, taking the suffix i for the initial condition, and the suffix f for the final condition, the change in momentum will be:

$\Delta \vec{ p} = \vec{p}_f - \vec{p}_i$

$\Delta \vec{ p} = m \ \vec{v}_f - m \ \vec{v}_i$

$\Delta \vec{ p} = m (\ \vec{v}_f - \ \vec{v}_i)$

As we know the mass of the ball, we just need to find the initial and final velocity.

Knowing the magnitude and direction of a vector, we can obtain the Cartesian components with the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ is the magnitude of the vector and θ is the angle measured from the x axis.

Taking the x axis to the right and the y axis to be up, the initial velocity will be:

$\vec{v}_i = 4 (m)/(s) ( \ cos ( - (90 \ °- 31 \°)) , sin( - (90 \ ° - 31\°) ) ) =$

where minus sign appears cause the ball is going downward, and we subtracted the 31 ° as it was measured from the y axis

So, the initial velocity is

$\vec{v}_i = 4 (m)/(s) ( \ cos ( - 59 \°) , sin( - 59 \°)) =$

$\vec{v}_i = ( \ 2.0601 \ (m)/(s) , - 3.4286 (m)/(s)) =$

The final velocity is

$\vec{v}_i = 4 (m)/(s) ( \ cos ( 90 \ °- 31 \°) , sin( 90 \ ° - 31\°)) =$

$\vec{v}_i = 4 (m)/(s) ( \ cos ( 59 \°) , sin( 59 \°)) =$

$\vec{v}_i = ( \ 2.0601 \ (m)/(s) , 3.4286 (m)/(s)) =$

So, the change in momentum will be

$\Delta \vec{ p} = m (\ \vec{v}_f - \ \vec{v}_i)$

$\Delta \vec{ p} = 0.575 \ kg (\ ( \ 2.0601 \ (m)/(s) , 3.4286 (m)/(s) - ( \ 2.0601 \ (m)/(s) , - 3.4286 (m)/(s)))$

$\Delta \vec{ p} = 0.575 \ kg (\ ( \ 2.0601 \ (m)/(s) - \ 2.0601 \ (m)/(s), 3.4286 (m)/(s) + 3.4286 (m)/(s))$

$\Delta \vec{ p} = 0.575 \ kg (\ 0 , 2 * 3.4286 (m)/(s) )$

$\Delta \vec{ p} = 0.575 \ kg * 2 * 3.4286 (m)/(s) \hat{j}$

$\Delta \vec{ p} = 3.9429 (kg \ m)/(s) \hat{j}$

Why are continental rocks much older than oceanic crust?A. Oceanic crust is continually recycled through convection in the earth's mantle
B. Oceanic crust is made out of much less dense material than continental crust
C. Continental crust is continually renewed through convection in the earth's mantle
D. Continental crust eats oceanic crust for breakfast

Answers

Answer:

A. Oceanic crust is continually recycled through convection in the earth's mantle

Explanation:

The oceanic plate is constantly being recycled through the forces of convection within the earth's mantle.

New oceanic plate are formed mid-oceanic ridge for example. As the magma cools and solidifies, they are moved away continually.

This is not the case for the continental curst.

Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s. How much time does the driver of the car measure for his trip between the poles?

Answers

Answer:

Observed time, t = 5.58 s

Explanation:

Given that,

Speed of light in a vacuum has the hypothetical value of, c = 18 m/s

Speed of car, v = 14 m/s along a straight road.

A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.

We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :

$T=\frac{t}{\sqrt{1-(v^2)/(c^2)} }$

t is observed time.

$t=T* \sqrt{1-(v^2)/(c^2)} \n\nt=8.89* \sqrt{1-(14^2)/(18^2)} \n\nt=5.58\ s$

So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generated by seven-time-winner Lance Armstrong (m = 75.0 kg) is 6.50 W per kilogram of his body mass. (a) How much work does he do during a 85-km race in which his average speed is 10.5 m/s? J (b) Often, the work done is expressed in nutritional Calories rather than in joules. Express the work done in part (a) in terms of nutritional Calories, noting that 1 joule = 2.389 10-4 nutritional Calories. nutritional Calories

Answers

Answer: a) work done = 3946429.5 J

b) work done = 943.22 nutritional calories

Explanation:

If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.(a) What is the heat capacity of this object?
(b) What is the object's specific heat?

Answers

When an object gets heated by a temperature ΔT energy needed, E = mcΔT

Here energy is given E = 2050 J

Mass of object = 150 g

Change in temperature ΔT = 15 $^0C$ = 15 K

a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.

So heat capacity = E/ΔT = 2050/15 = 136.67 J/K

b) We have E = mcΔT

c = $(2050)/(150*10^(-3)*15) = 911.11 J/kgK$

So object's specific heat = 911.11 J/kgK

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