Answer:
1.) 4m
2.) 37 m
3.) 62m
4.) 2.5 s
Explanation:
1.) Given that the
Thinking distance = 1m
Breaking distance = 3m
Stopping distance = breaking distance + thinking distance
Stopping distance = 1 + 3 = 4m
2.) Given that the
Stopping distance = 52 m
Thinking distance = 15m
Breaking distance = 52 - 15 = 37m
3.) The stopping distance = 76m
Thinking distance = 14m
Breaking distance = 76 - 14 = 62m
It take the brakes 62m to slow the car down to a stop.
4.) Given that a lorry travels 28m when stopping from a speed of 4m/s. If its braking distance was 18m, what was the driver’s reaction time?
Thinking = stopping distance - braking distance
Thinking distance = 28 - 18 = 10m
Speed = distance/time
4 = 10/reaction time
Reaction time = 10/4
Reaction time = 2.5 s
5.) Question incomplete
Three remote control cars are identical size and weight. The motors installed in the three cars are 100 watts, 150 watts and 200 watts respectively. Describe what will be true about the maximum speed of each car? Explain.
Answer:
The true statement about the maximum speed of each car will be that the power of each car directly proportional to the speed of each car.
Explanation:
Given that,
Power of first car = 100 watt
Power of second car = 150 watt
Power of third car = 200 watt
We know that,
Power :
The power is directly proportional to the cube of speed.
In mathematically,
[tex]P\propto v^3[/tex]
Where, P = power
v = speed
If the power is increased then the speed will be increase.
We need to find the maximum speed of car
According to question,
The power of car is maximum so the speed of car will be maximum.
For first car,
The power of first car is 100 watt.
So, the speed of first car will be 1000000 m/s.
For second car,
The power of second car is 150 watt.
So, the speed of second car will be 3375000 m/s.
For third car,
The power of third car is 200 watt.
So, the speed of first car will be 8000000 m/s.
Hence, The true statement about the maximum speed of each car will be that the power of each car directly proportional to the cube of speed of each car.
How can accuracy be limited?
Answer:
accuracy refers to the deviation of a measurement from a standard or true value of the quantity being measured
examples of electrostatic force
Answer:
When we run a piece of paper with the oil in our head with the help of a comb produces electrostatic force.
Balloons get attracted to another balloon when one of them are rubbed with hair.
Explanation:
Those are 2 examples of electrostatic force.
Calculate the workdone by a bricklayer who lifted cement block of mass 10kg from the floor to the height of 3cm g=9.8m
Answer:
The work done is 2.94 Nm.
Explanation:
First, you have to find the force of a brick layer using the formula, Force = mass×gravity :
[tex]f = m \times g[/tex]
[tex]let \: m = 10,g = 9.8[/tex]
[tex]f = 10 \times 9.8[/tex]
[tex]f = 98N[/tex]
Given that the force of a brick layer is 98N. Next, you have to apply work done formula, Work done = force×distance :
[tex]w.d = f \times d[/tex]
[tex]let \: f = 98,d = 0.03m[/tex]
[tex]w.d = 98 \times 0.03[/tex]
[tex]w.d = 2.94Nm[/tex]
A doctor ran 100 meters in 20 seconds a duck slowly waddled 20 feet in 2 minutes some dude flew 100 miles in 2 hours Dave rolled down a hill he rolled 300 meters 30 seconds
Answer:
1. 50 m/s 2. 10 ft/min 3. 50 miles/hour 4. 10 m/s
Explanation:
We are required to calculate the speed in each instant
Since speed = distance/time,
1. For the doctor, speed = distance/time = 100 m/20 s = 50 m/s
2. For the duck, speed = distance/time = 20 feet/ 2 min = 10 ft/min
3. For the dude who flew, speed = distance/time = 100 miles/ 2 hours = 50 miles/hour = 50 mi/h
4. For Dave who rolled down the hill, speed = distance/time = 300 metres/ 30 s = 10 metres/ second = 10 m/s
5.3 two 30 kg children in a 20 kg cart are stationary at the top of a hill. They start rolling down the 80 m tall hill and they are travelling at 30 km h-1 when they reach the bottom. how muvh work was done on the cart by friction during it's travel down the hill
Answer:
60008.4 J
Explanation:
The mass of each kid = 30 kg
mass of the cart = 20 kg
The speed of the cart down the hill = 30 km/hr = 30 x 1000/3600 = 8.33 m/s
The height of the hill = 80 m
The potential energy of the boys at the top of the hill = mgh
where
m is the total mass of the kids and the cart = (30 x 2) + 20 = 80 kg
g is the acceleration due to gravity = 9.81 m/s^2
h is their height above the ground = 80 m (on the top of the hill)
substituting, we have
potential energy PE = 80 x 9.81 x 80 = 62784 J
At an instance at the bottom of the hill
their kinetic energy = [tex]\frac{1}{2} mv^2[/tex]
where
v is their velocity = 8.33 m/s
m is their total mass = 80 kg
substituting, we have
kinetic energy KE = [tex]\frac{1}{2}*80*8.33^2[/tex] = 2775.6 J
Total work done on the cart is equal to the energy lost by the cart when it reached the bottom of the hill
work done by friction = PE - KE = 62784 - 2775.6 = 60008.4 J
a dictatorship can also be a democracy
Which is not a piece of evidence that most scientists use to prove the origin of our universe?
A. Movement of the galaxies (expansion)
B. Cosmic microwave background
C. Presence of hydrogen and helium in the
correct percentages throughout the
universe
D. The contraction of objects in space
Answer:
D) The contraction of objects in space
Explanation:
A) Expansion of the universe as originated from the Big Bang is an important evidence, as B) the Cosmic microwave background which is thought to be left over from the Big Bang. The right proportion of hydrogen and helium as unprocessed material that has been detected as the basic elements from where the universe started.
The only out of context option is option D
how fast does the light travel in a glass refractive index of 1.5?
Answer:
the light travels at a velocity of [tex]2\,\,10^8 \frac{m}{s}[/tex] in the glass
Explanation:
Recall that the refractive index of a material (n) is defined as the quotient between the speed of light in vacuum (c), and the speed of light in the medium (v). Therefore we can solve for the speed of light in the medium if we know the index of refraction it has:
[tex]n=\frac{c}{v} \\v=\frac{c}{n} \\v=\frac{3\,\,10^8}{1.5} \frac{m}{s} \\v=2\,\,10^8 \frac{m}{s}[/tex]
A car drives 30 meters north, then turns around and drives 50 meters south. What was its distance?
Answer:
The total distance moved by the car is 80 metres
Explanation:
Given
North = 30 meters
South = 50 meters
Required
Determine the distance
Distance is a scalar quantity and hence, the total distance is the summation of distance moved in both directions
i.e.
[tex]Total\ Distance = North + South[/tex]
[tex]Total\ Distance = 30\ metres + 50\ metres[/tex]
[tex]Total\ Distance = 80\ metres[/tex]
Hence;
The total distance moved by the car is 80 metres
Complete the statements using data from Table A of your Student Guide. The speed of the cart after 8 seconds of Low fan speed is cm/s. The speed of the cart after 3 seconds of Medium fan speed is cm/s. The speed of the cart after 6 seconds of High fan speed is cm/s.
Answer:
8 seconds = 72cm
3 seconds = 36cm
6 seconds = 96cm
Explanation:
lab assignment i just took.
Speed is the rate of distance over time. The speed of the cart is as follows:
[tex]72cms^{-1}[/tex] at 8 seconds.[tex]36cms^{-1}[/tex] at 3 seconds.[tex]96cms^{-1}[/tex] at 6 seconds.Let:
[tex]s \to[/tex] speed
[tex]d \to[/tex] distance
[tex]t \to[/tex] time
After 8 seconds
For low fan speed; the distance (d) at 8 seconds is:
[tex]d =576cm[/tex]
Speed (s) is calculated using:
[tex]s = \frac dt[/tex]
[tex]s = \frac{576cm}{8s}[/tex]
[tex]s = 72cms^{-1}[/tex]
Hence, the speed after 8 seconds is [tex]72cms^{-1}[/tex]
After 3 seconds
For medium fan speed; the distance (d) at 3 seconds is:
[tex]d = 108cm[/tex]
The speed (s) is :
[tex]s = \frac dt[/tex]
[tex]s = \frac{108cm}{3s}[/tex]
[tex]s = 36cms^{-1}[/tex]
Hence, the speed after 3 seconds is [tex]36cms^{-1}[/tex]
After 6 seconds
For high fan speed; the distance (d) at 6 seconds is:
[tex]d = 108cm[/tex]
The speed (s) is :
[tex]s = \frac dt[/tex]
[tex]s = \frac{576cm}{6s}[/tex]
[tex]s = 96cms^{-1}[/tex]
Hence, the speed at 6 seconds is [tex]96cms^{-1}[/tex]
Read more about speed at:
https://brainly.com/question/7359669
To minimize damage during a broadside collision __________ speed to put the point of impact behind the passenger section.
Answer:
Positive
Ghhhhhhbn
Ggvbbn
To minimize damage during a broadside collision decrease the speed to put the point of impact behind the passenger section.
What is broadside collision?A broadside collision happens when the front of one vehicle bangs into the side of another, usually at a high rate of speed. Although catastrophic injuries can occur in any automobile accident on the road, broadside crashes are especially dangerous.
This is due to the fact that there is not much room or protection between the person inside the hit automobile and the moment of contact. The injury happens to the vehicles depends on their speed.
As the speed is reduced, many such collisions can stopped in time and we can prevent broadside collision as well and giving enough time to pass other vehicles through the intersection.
Find more on broadside collision:
https://brainly.com/question/28271581
#SPJ2
Question 5 of 10
Which option is an example of a conceptual model?
A. A clay ball with a slice cut out, showing the layers of Earth
B. A mental image of gas molecules as tiny balls
C. A diagram of a flagpole and its shadow with measurements
provided to calculate the length of the shadow
D. A weather map created by Doppler radar
Answer:
B. A metal image of gas molecules as tiny balls
Answer:
B. A mental image of gas molecules as tiny balls
You are driving at 25 m/s with your cruise control on when you see a fallen tree in the road. It takes you 0.30 s to put on the brakes, your car decelerates at a rate of 5.0 m/s2, and the tree is 75 m ahead. Will you hit the tree? Choose the correct answer and reason below.
Explanation:
Given data
velocity v= 25m/s
The time it takes to put on brake t= 0.3s
the distance covered when the brake was put on is
v=s/t
s= v*t
s= 25*0.3s
s= 7.5m
hence the distance covered is 7.5m
Also the rate of decrease in aceleration is 5m/s^2
we can also calculate the distance covered at this rate
v^2=u^2+2as
25^2= 0+2*5*s
625=10s
divide both sides by 10
s=625/10
s= 62.5m
The total distance covered between putting on the brakes and decelareation is 7.5+62.5= 70m
Given that the tree is 75m ahead, the car would not hit the tree
aroma of perfume is solid or liquid or gas?
Tell me about your favorite science experiment. Give me a detailed account of the experiment and what you like about it.
Answer:
nngh have
bjruh hjrhhj be rnrnnrnrnnnrnjrjnnnnnnnnrnrn n n and I was nrn
Explanation:
jbbbbbhhhjjnnnnnnnnnnjhvcc
Explanation:
A hot–air balloon is moving at a speed of 10.0 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–direction at a velocity of +4 meters/second with respect to himself. If the ball lands after 30 seconds, about how far does it land horizontally from the point at which it was released?
A. 120 meters
B. 125 meters
C. 300 meters
D. 420 meters
Answer:
Option (D) is correct.
Explanation:
The balloon lands horizontally at a distance of 420 m from a point where it as released.
Velocity of air balloon along +X axis =10 m/s
velocity of ball=4 m/s along + X axis
the velocity of balloon gets added to the velocity of ball. So the resultant velocity of the balloon=10+4 = 14 m/s
time taken= 30 s
The distance traveled is given by d= v t
d= 14 (30)
d= 420 m
Thus the balloon lands horizontally at a distance of 420 m from a point where it as released.
Define the problem. Look around your house or outside. Is there anything that you're wondering about or have questions about?
Answer:
Water
Explanation:
How do we have water at our disposal, like that's crazy.
What is groups of organs that work together to complete a process in the body
Answer:
A System
Explanation:
There are many systems including the circulatory and nervous system where multiple organs work together to perform a job.
When you stand outside your are standing in the, Stratosphere. Troposphere. Mesosphere no
Muscle fibers move back and forth during _____.
Answer:
it must move back and forth during sex
Calculate the speed for each segment on the graph include the units (m/s). Show your calculations, (such as 4/2= 2 m/s)
a. (t=0-10s)
b. (t=10-15s)
c. (t=15-30s)
d. (t=30-40s)
e. (t=40-55s)
Answer:
v_A = 6 m/s , v_B = 0 m/s , v_C = - 4 m/s , v_D = -4 m/s , v_E = 2.66 m / s
Explanation:
For this exercise we use the definition of average velocity in each segment
v_average = (x₂-x₁) / Δt
segment A of the graph you take the value of distance and time
x₁ = 0 m t₁ = 0 s
x₂ = 60 m t₂ = 10 s
using the average velocity equation and calculate
v_A = (60-0) / (10-0)
v_A = 6 m / s
Segment B
x₁ = 60m
x₂ = 60 m
since the displacement variation is zero (particle stopped) the velocity is zero
v_B = 0 m / s
Segment C and D
x₁ = 60 m t1 = 15 s
x₂ = -40 m t2 = 40 s
v_C = (-40 -60) / (40 -15)
v_C = - 4 m / s
Segment D
how is the same line
v_D = -4 m / s
Segment E
x₁ = -40 m t₁ = 40 s
x₂ = 0 m t₂ = 55 s
v_E = (0- (-40)) / (55 -40)
v_E = 2.66 m / s
A toy car is moving across a table with a velocity of 7.0 m/s and drives off the end. The table is 1.8 m tall. How long will the car be in the air before it crashes on the floor? a 0.6 seconds b 0.8 seconds c 0.4 seconds d 0.2 seconds
Answer:
Option D - 0.2 s
Explanation:
We are given;
Initial velocity; u = 7 m/s
Height of table; h = 1.8m
Now,since we want to find the time the car spent in the air, we will simply use one of Newton's equation of motion.
Thus;
h = ut + ½gt²
Plugging in the relevant values, we have;
1.8 = 7t + ½(9.8)t²
4.9t² + 7t - 1.8 = 0
Using quadratic formula to find the roots of the equation gives us;
t = -1.65 or 0.22
We can't have negative t value, thus we will pick the positive one.
So, t = 0.22 s
This is approximately 0.2 s
why do the kites and eagles fly without flapping their wings?
Answer:
because they both have large area to capture air
Answer:
Because they have a large surface area to capture moving air when going at a certain speed.
Explanation:
When your trying to fly a kite and the wind isn't that much, you have to run so that the kite can come in contact with the fast moving air which pushes the kite to go up. Just like a eagle because when a eagle is trying to fly, it has to first flap its wings to push air downwards so that it can create lift and at a certain altitude, the eagle doesn't have to flap it's wings though it tilts its body downwards a little so that gravity can do the work, so the eagle is basically gliding downward.
16. Which of the following is equivalent to 36 km/hr?
a. 10 m/s
b. 10 cm/s
c. 36 m/s
d. 36 mph
Lee skated 36 miles in 3h. What was the speed in which lee stated.
Answer: 12 mph
Explanation: 36 miles divided by 3 hours would be 12 mph
what is the mass of 1 litter of liquid in grams, if the density of liquids is 13.6g/ cm^3
Explanation:
1 L = 1000 cm³
m = ρV
m = (13.6 g/cm³) (1000 cm³)
m = 13,600 g
GIVING BRAINLIST PLZ HELP The steps below identify the part of the process to become a U.S. citizen, the naturalization process. Use the steps below to answer the following question. 1. Be a legal resident in the U.S. for at least 5 years. 2. Complete an application 3. ? 4. Take the Oath of Allegiance What is missing in the process above? A: Register for Selective Service System B: Interview and pass civics test C: Complete a background check D: Receive an appointment letter is da correct answer B
Answer:
Yes the correct answer is B
Explanation:
A car travels north along a straight highway at an average speed of 85 km/h. After driving 2.0 km, the car passes a gas station and continues along the highway. What is the car’s position relative to the start of its trip 0.25 h after it passes the gas station?
Answer:
x_total = 23250 m
Explanation:
This is a uniform motion exercise, the equation that describes the motion is
v = x / t
the position is
x = v t
let's reduce the magnitude to the SI system
v = 85 km / h (1000 m / 1km) (1h / 3600 s) = 23.61 m / s
t = 0.25 h (3600 s / 1h) = 900 s
let's calculate
x₁ = v t₁
x₁ = 23.61 900
x₁ = 21250 m
The distance from the origin is
x_total = x_station + x₁
x_total = 2000 + 21250
x_total = 23250 m
A motorbike is traveling to the left with a speed of 27.0\,\dfrac{\text m}{\text s}27.0 s m 27, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction when the rider slams on the brakes. The bike skids 41.5\,\text m41.5m41, point, 5, start text, m, end text with constant acceleration before it comes to a stop. What was the acceleration of the motorbike as it came to a stop?
Answer:
[tex]\frac{729\frac{m^{2} }{s^{2} } }{83m}[/tex]≈8.8[tex]\frac{m}{s^{2} }[/tex]
Explanation: