Can someone help with number 4 please

Answers

Answer 1

Answer: The way I look at these things is like this:

-- The runner covered 12 meters in 4 seconds.
Average speed = 3 meters per second.

-- Speed at the beginning = zero.
In order to make the average 3, Speed at the end = 6 meters per second.

-- Speed increased from zero to 6 meters per second in 4 seconds.
It must have increased 1.5 meters per second each second.

That's choice-#2.

Answer 2

Answer: Here we have -
Acceleration (a) = ?
Initial velocity (u) = 0 m/s (runner starts from rest)
Distance (s) = 12 m
Time (t) = 4 seconds

So, using the second equation of motion, we have-

$s = ut + (1)/(2)at^(2)$

$12 = 0*4 + (1)/(2)*a*4 ^(2)$

$12 = 0+8a$

$(12)/(8) = a$

$1.5m/ s^(2)$

So, option (2) is correct.

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The reflection of light waves from a rough surface is called _____ reflection.a. diffuse
c. regular
b. scattered
d. fragmented

Answers

The correct answer is (a.) diffuse. The reflection of light waves from a rough surface is called DIFFUSE reflection. This type of reflection reflects light diffusely mostly from a surface that is a non-absorbing material such as plaster and polycrystalline.

An object traveling in a straight line accelerates. What will definitely happen due to the acceleration?

Answers

"Acceleration" means any change in speed or direction of motion.
If the object is known to be accelerating even though it's traveling
in a straight line, then we know that its speed must be changing.

A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after being released?

Answers

(M G H)=(0.5 x 9.8 x 10) = 49 joules.

Newton's Third Law of Motion is often called the law of intertia.

True
False

Answers

This is False. Newton's First Law of Motion is often called the law of inertia.

A) A block of mass, m, is pushed up against a spring, compressing it a distance x, and is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed v. The same spring projects a second block of mass 4m, giving it a speed 3v. What distance was the spring compressed in the second case?B) Initially a body moves in one direction and has kinetic energy K. Then it moves in the opposite direction with three times its initial speed. What is the kinetic energy now?

Answers

KE=1/2mv^2

F= -kx

KE= -kx

-kx= 1/2mv^2

when u quadruple the m, the x also quadruples. when the speed is multiplied by 3, since it is v^2, the x is multiplied by 3^2 , which is 9

so distance the spring compressed is x×4×9= 36x

b) using the KE formular too,

when the v is tripled, since formula is v^2, the resulting KE will be multiplied by 3^2, so KE now= 9K

A motorist traveling at 17 m/s encounters a deer in the road 39 m ahead. If the maximum acceleration the vehicle’s brakes are capable of is −7 m/s 2 , what is the maximum reaction time of the motorist that will allow her or him to avoid hitting the deer? Answer in units of s. 019 (part 2 of 2) 10.0 points If his or her reaction time is 1.21983 s, how fast will (s) he be traveling when (s)he reaches the deer? Answer in units of m/s.

Answers

Answer:

1. t_reaction = 1.08 s

2. v₀₁ = 16.365 m/s

Explanation:

1. This is a kinematics exercise, let's analyze the situation a bit, we can calculate the braking distance and the rest of the distance we can use to calculate the reaction time.

Braking distance

v² = v₀² + 2 a x

when he finishes braking the speed is v = 0

0 = v₀² + 2 a x

x = -v₀² / 2a

x = - 17²/2 (-7)

x = 20.64 m

the distance for the reaction is

d = x_reaction + x

x_reaction = d - x

x_reaction = 39 - 20.64

x_reaction = 18.36 m

as long as it has not reacted the vehicle speed is constant

v = x_reaction / t_reaction

t_reaction = x_reaction / v

t_reaction = 18.36 / 17

t_reaction = 1.08 s

2. Let's find the distance traveled in the reaction time of t1 = 1.21983 s

as the speed is constant

v = x / t

x₁ = v t₁

the distance traveled during braking is

v² = v₀² + 2a x₂

0 = v₀² + 2 a x₂

x₂ = -v₀² / 2a

v = v₀

the total distance is

x_total = x₁ + x₂

x_total = v₀ t₁ + v₀² / 2a

39 = v₀ 1.21983 + v₀²/14

v₀² + 17.08 vo - 546 =0

we solve the second degree equation

v₀ = [ -17.08 ±√(17.08² + 4 546) ]/2

v₀ = [-17.08 ± 49.81 ]/2

v₀₁ = 16.365 m/s

v₀₂ = - 33.445 m/s

as the acceleration is negative the correct result is v₀₁ = 16.365 m/s

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