A car starts at 80 m/s but sees a cop and hits the brakes slowing down to 50 m/s in 2 sec.
Answer:
-15m/s/s
Explanation:
Acceleration = change in speed/ change in time
The change in speed is calculated by subtracting the initial speed from the final speed, so the change in speed is: 50 - 80 = -30m/s. The change in time is 2 - 0 = 2.
So the car acceleration is -30/2 = -15m/s/s
It is negative because it is decelerating.
Hope this helped!
A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.394 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s
Answer:
The distance traveled by the second block is 0.197 m
Explanation:
Given;
mass of the small block, m₁ = 0.2 kg
distance traveled by the block, d₁ =0.394 m
time of travel, t = 2 s
mass of the second block, m₂ = 0.4 kg
distance traveled by the second block, d₂ = ?
The work done per unit time on the inclined plane is given by;
[tex]\frac{f_1d_1}{t} = \frac{f_1d_1}{t}\\\\f_1d_1 = f_2d_2\\\\m_1gd_1 = m_2gd_2\\\\m_1d_1 = m_2d_2\\\\d_2 = \frac{m_1d_1}{m_2} \\\\d_2 = \frac{0.2*0.394}{0.4}\\\\d_2 = 0.197 \ m[/tex]
Therefore, the distance traveled by the second block is 0.197 m
Hitting a ball off a tee would be considered which of the following?
A.
A closed serial skill
B.
An open continuous skill
C.
An open discrete skill
D.
A closed discrete skill
What is the total electric charge of 2.5 kg of (a) electrons and (b) protons?
Answer:
a
[tex]Q_e = -4.39 *10^{11} \ C [/tex]
b
[tex]Q_p = + 2.395*10^{-8} \ C [/tex]
Explanation:
Generally the number of electron in the given mass is mathematically evaluated as
[tex]N_e = \frac{2.5}{m_e }[/tex]
Here m_e is the mass of electron with value [tex]m_e = 9.11 * 10^{-31} \ kg[/tex]
=> [tex]N_e = \frac{2.5}{ 9.11 * 10^{-31} }[/tex]
=> [tex]N_e =2.74 *10^{30} \ electrons [/tex]
The total electric charge is mathematically represented as
[tex]Q_e = N_e * e[/tex]
Here e is the charge on a single electron with value [tex]e = 1.60 *10^{-19} \ C[/tex]
So
[tex]Q_e = -2.74 *10^{30} * 1.60 *10^{-19} [/tex]
[tex]Q_e = -4.39 *10^{11} \ C [/tex]
The negative sign is because we are considering electron
Generally the number of protons in the given mass is mathematically evaluated as
[tex]N_p = \frac{2.5}{m_p }[/tex]
Here m_p is the mass of electron with value [tex]m_e = 1.67 * 10^{-27} \ kg[/tex]
=> [tex]N_p = \frac{2.5}{ 1.67 * 10^{-27} }[/tex]
=> [tex]N_p =1.497 *10^{27} \ protons [/tex]
The total electric charge is mathematically represented as
[tex]Q_p = + N_p * e[/tex]
Here p is the charge on a single proton with value [tex]p = 1.60 *10^{-19} \ C[/tex]
So
[tex]Q_p = +1.497 *10^{27} * 1.60 *10^{-19} [/tex]
[tex]Q_p = + 2.395*10^{-8} \ C [/tex]
Using the equation for for Newton's Second law, m=F/a solve the following problem. You have been given an object with a force of 10N and an acceleration of 2 m/s2, what is the mass?
Group of answer choices
1. 8g
2. 3g
3. 20g
4. 5g
Answer:
4. 5g
Explanation:
F=ma so, m=Fa. All you have to do is 10/2. Don't be confused by the units. Mass will normally be in grams.
In electronic circuits it is not unusual to encounter currents in the microampere range. Assume a 36 μA current, due to the flow of electrons. Part A What is the average number of electrons per second that flow past a fixed reference cross section that is perpendicular to the direction of flow?
Answer:
n = 2.25 x 10¹⁴ electrons/s
Explanation:
The amount of electric current is defined as the electric charge passing through an area per unit time. Hence:
I = q/t
where,
I = Current
q = amount of charge
t = time interval
but,
q = ne
therefore,
I = ne/t
where,
n = no. of electrons
e = charge on single electron = 1.6 x 10⁻¹⁹ C
t = 1 s (for electrons passing per second)
I = Current = 36 μA = 3.6 x 10⁻⁵ A
Therefore,
3.6 x 10⁻⁵ A = n(1.6 x 10⁻¹⁹ C)/1 s
n = (3.6 x 10⁻⁵ A)/(1.6 x 10⁻¹⁹ C)
n = 2.25 x 10¹⁴ electrons/s
Question: A) A car is driving too fast on a flat non-banked curve. The car cannot stay on the road. What path will the car take as it leaves the road? See Image. (a) (b) B) The driver wants to go faster around the same curve mentioned above and not run off the road. He adds sandbags to his car to make it weigh more so that the friction force between the tires and the road will be increased. Will this work? Explain your answer.
Explanation:
(a) The car will leave tangentially to the road, so it will take path d.
(b) Sum of forces on the car in the vertical direction:
∑F = ma
N − mg = 0
N = mg
Sum of forces on the car in the centripetal direction:
∑F = ma
Nμ = m v²/r
mgμ = m v²/r
gμ = v²/r
v = √(grμ)
The maximum velocity is independent of mass, so adding sandbags will not work.
What were the physical activities in your childhood that you still do today? Do you spend more time now in doing these activities as compared before?
A column of soldiers, marching at 100 steps per minute, keep in step with the beat of a drummer at the head of the column. It is observed that the soldiers in the rear end of the column are striding forward with the left foot when the drummer is advancing with the right. What is the approximate length of the column? (Take the speed of sound to be 343 m/s.)
Answer:
The value is [tex]D = 205.8 \ m [/tex]
Explanation:
The time taken for the column to take a step mathematically represented as
100 steps => 1 minutes => 60 seconds
1 step => t
=> [tex]t = 0.6 \ s [/tex]
Generally the length of the column is mathematically represented as
[tex]D = v * t[/tex]
substituting 343 m/s for v we have
[tex]D = 343 * 0.6 [/tex]
=> [tex]D = 205.8 \ m [/tex]
An object is released from rest from a top of a building 90 meters high. Neglect air friction.
What is the volume of its acceleration?
Calculate the time it takes to reach the floor?
With what velocity does it reach the floor?
How fast is it moving when it is 50 meter above the floor?
Answer:
1. a = 9.8 m/s²
2. t = 4.28 s
3. Vf = 42 m/s
4. Vf = 28 m/s
Explanation:
1.
Since, the body is under free fall motion. Therefore, the value of its acceleration shall be equal to the acceleration due to gravity.
a = 9.8 m/s²
2.
The time taken by the ball to reach the ground can be calculated by using second equation of motion:
h = Vi t + (1/2)gt²
where,
h = height = 90 m
Vi = initial velocity = 0 m/s
t = time taken = ?
Therefore,
90 m = (0 m/s)t + (1/2)(9.8 m/s²)t²
t = √(18.36 s²)
t = 4.28 s
3.
In order to find final velocity we use first equation of motion:
Vf = Vi + gt
Vf = 0 m/s + (9.8 m/s²)(4.28 s)
Vf = 42 m/s
4.
when the ball is at height of 50 m, it means it has covered:
h = 90 m - 50 m = 40 m
we use third equation of motion at this point:
2gh = Vf² - Vi²
(2)(9.8 m/s²)(40 m) = Vf² - (0 m/s)²
Vf = √(784 m²/s²)
Vf = 28 m/s
Calculate the length of segment RS with midpoint, M , if RM = 5x and MS = x + 12.
A. 3
B. 25
C. 30
D. 36
E. 15
Answer:
the full segment is: 30 units long
which coincides with answer C in your list
Explanation:
If M is the midpoint, then it divides the segment in two equal parts. Then, we can say that:
RM = MS (equality among the two parts of the divided segment)
replacing RM with "5 x" and MS with "x + 12", we get:
5 x = x + 12
solving for x:
5 x - x = 12
4 x = 12
then x = 3
With this info, we can calculate the length of each half of the segment and consequently its full length:
RM = 5 x = 5 (3) = 15
then the full segment is: 30 units long
why is evidence good for practice
Answer:
HEYO BRO! (my bad for da caps)
Explanation:
Evidence-based practice (EBP) is the idea that occupational practices ought to be based on scientific evidence. That at first sight may seem to be obviously desirable, but the proposal has been controversial.
Happy to Help From, Adam
What does this picture show?
A. Good accuracy, poor precision
B. Poor accuracy, good precision
C. Good accuracy, good precision
D. Poor accuracy, poor precision
A magnetic field is created by ____.
A moving electric charges
b elctromagnetic pulses
c a strong current
d a change in the current of a wire
A pendulum built from a steel sphere with radius r cm 5 and density stl kg m S 3 7800 is attached to an aluminum bar with length l m 1 thickness t cm 0 8. and width w cm 4 and density . al kg m S 3 2820 a. Calculate the mass moment of inertia of the pendulum about its center of mass, . cm I b. Calculate the mass moment of inertia of the pendulum about its pivot point, o I .
Answer:
a) I = 0.0198 kg m² , b) I = 21.85 kg m²
Explanation:
For this exercise we will use the definition of moment of inertia
I = ∫ r² dm
For body with high symmetry they are tabulated
sphere I = 2/5 m r²
bar with respect to center of mass I = 1/12 m L²
let's calculate the mass of each body
bar
ρ = m / V
m = ρ V
m = ρ l w h
where we are given the density of the bar rho = 32840 kg / m³ and its dimensions 1 m, 0.8 cm and 4 cm
m = 32820 1 0.008 0.04
m = 10.5 kg
Sphere
M = ρ V
V = 4/3 pi r³
M = rgo 4/3 π r³
give us the density 37800 kg / m³ and the radius of 5 cm
M = 37800 4/3 π 0.05³
M = 19.8 kg
a) asks us for the moment of inertia of the sphere with respect to its center of mass
I = 2/5 M r²
I = 2/5 19.8 0.05²
I = 0.0198 kg m²
b) the moment of inertia with respect to the turning point, for this we will use the theorem of parallel axes
I = I_cm + M d2
where d is the distance from the body to the point of interest
I_cm = 0.0198 kg m²
the distance to the pivot point is
l = length of the bar + radius of the sphere
l = 1 + 0.05 = 1.005 m
I = 0.0198 + 19.8 1.05²
I = 21.85 kg m²
For any object in projectile motion, the vertical velocity is independent of gravity?
Answer:
the y axis, every moment is subjected to a vertical acceleration directed towards the center of the Earth, which is called the acceleration of gravity
Explanation:
In projectile launching, the movement is separated into two movements, one on the x-axis and the other on the y-axis, related through time.
In horizontal movement, the speed is constant, because there is no acceleration in this axis, the effect of air friction is almost always eliminated.
In the other movement on the y axis, every moment is subjected to a vertical acceleration directed towards the center of the Earth, which is called the acceleration of gravity.
This value of the eceleration of gravity is constant for small distances
compared to the radius of the Earth, for higher altitudes an expansion in beings of the distance is used, giving a linear dependence.
3) An explorer walks 13 km due east, then 18 km north, and finally 3 km west.
a) What is the total distance walked?
b) What is the resulting displacement of the explorer from the starting point?
Answer: 34 km, 21 km 61 degrees north of east
Explanation: distance = 13 + 3 + 18 = 34
displacement = 13 - 3 = 10
10^2 + 18^2 = 424
find the square root of 424 ( 20.5 rounded to 21 )
The total distance walked is 34 km and the resulting displacement is 20.6 km.
a) The total distance is gotten by summing up all the distance.
Total distance = Distance moved east + distance moved north + distance moved west
Total distance = 13 km + 18 km + 3 km = 34 km
b) The displacement is the distance from the beginning point to end point.
Displacement² = 18² + (13 - 3)² = 18² + 10²
Displacement² = 424
Displacement = 20.6 km
Therefore the total distance walked is 34 km and the resulting displacement is 20.6 km.
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sam exerts a force of 65 N on a lawn mower with a mass of 25 kg. which formula can be used to calculate the acceleration of a lawn mower
Answer:
F = ma - with this you will get 2.6 m/s^2
Explanation:
Use this formula to get the acceleration...
where F is force, M is mass and A is acceleration
by using this we get...
65 = 25 * a
so, a = 65/25
Therefore, the acceleration is 2.6 m/s^2
Hope that helped :)
The acceleration of a lawn mower will be 2.6 m/s². It is the ratio of force and the mass.
What is acceleration?The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.
The given data in the problem is;
The force act on a lawn mower,[tex]\rm F = 65 \ N[/tex]
Mass of lawn mower,[tex]\rm m = 25 kg[/tex]
The acceleration of a lawn mower is,[tex]\rm a = ?[/tex]
Acceleration, is found as the ratio of force and the mass.
[tex]\rm F= ma \\\\ a = \frac{F}{m} \\\\ a= \frac{65}{25} \\\\ a = 2.6 \ m/s^2[/tex]
Hence, the acceleration of a lawn mower will be 2.6 m/s².
To learn more about acceleration, refer to the link;
https://brainly.com/question/2437624
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The rain gauge has 2 inches of precipitation.
Qualitative or Quantitative
Answer:
Quantitative is the correct term
how many legs a cow has
what would the answer be ?
Answer:
im going between 2 of them b and c but i would have choose b
what is machinery
............
what does Aristotle say about the good life?Does it still stand contemporary world?
Answer:
Explanation:
Aristotle was a philosopher and a scientist who believed good life for humans is ultimately about happiness and this happiness can be achieved by virtue or high moral standards.
In the contemporary world, generally, everyone's ultimate goal is still to be happy but achieving this feat by being virtuous or having a high moral standard is fading away in our contemporary world as people just want to be happy by any means possible. Some amass wealth illegally just to be happy while others do things that are considered immoral (by the society) to be happy hence people still want a good life but not a way described by Aristotle.
Aristotle stated that a good life is about happiness and this can only be achieved through high moral standards.
The statement that a good life is about happiness and this can only be achieved through high moral standards doesn't stand anymore in the contemporary world. People now engage in immoral things as long as it brings them happiness. Even though the ultimate goal of people is to be happy, there are different ways that people go about in achieving this. There are some people who get their wealth illegally just to ensure that they are happy. This source of happiness is gotten through vices.In conclusion, Aristotle's statement about the good life is not applicable in the contemporary world anymore.
Read related link on:
https://brainly.com/question/21792802
An unstrained horizontal spring has a length of 0.34 m and a spring constant of 180 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.024 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges. g
Answer:
a
[tex]q_1 = -1.389 *10^{-5} \ C [/tex] , [tex]q_2 = -1.389 *10^{-5} \ C [/tex]
OR
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] , [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
b
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] and [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
Explanation:
Generally the force exerted on the string is mathematically represented as
[tex]F = k * e[/tex]
substituting values 180 N/m for k and 0.024 m for e
[tex]F = 180 * 0.024[/tex]
[tex]F = 4.32 \ N[/tex]
This force can also equivalent to the electrostatic force between the charges i.e
[tex]F = k * \frac{q^2}{ r^2}[/tex]
substituting [tex]9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex] for k and ( 0.34 + 0.024 = 0.364 m) for r we have
[tex] 4.32= 9*10^{9} * \frac{q^2}{ (0.364)^2}[/tex]
[tex]q = \sqrt{1.929 *10^{-10}}[/tex]
[tex]q = 1.389 *10^{-5} \ C [/tex]
Given the spring was stretched it means that the force between the charges is a repulsive for which tell us that both charge are of the same sign thus the possible algebraic signs of the charges are
[tex]q_1 = -1.389 *10^{-5} \ C [/tex] , [tex]q_2 = -1.389 *10^{-5} \ C [/tex]
OR
[tex]q_1 = 1.389 *10^{-5} \ C [/tex] , [tex]q_2 = 1.389 *10^{-5} \ C [/tex]
2. The components of vector A are given as follows:
Ax = 5.6 Ay = -4.7
What is the angle between vector A and positive direction of x-axis?
Answer:
50 degree.
Explanation:
Given that the components of vector A are given as follows: Ax = 5.6 Ay = -4.7
The angle between vector A and B in the positive direction of x-axis will be achieved by using the formula:
Tan Ø = Ay/Ax
Substitute Ay and Ax into the formula above.
Tan Ø = -4.7 / 5.6
Tan Ø = -0.839
Ø = tan^-1(-0. 839)
Ø = - 40 degree
Therefore, the angle between vector A and B positive direction of x-axis will be
90 - 40 = 50 degree.
An oil pump is drawing 44kW while pumping oil with a density of 860 kg/m3 at a rate of 0.1 m3/s. The inlet and outlet diameters of the pipe are 8cm and 12cm, respectively. If the pressure increases by 500kPa going through the pump and the motoreffi ciency is 90%, determine the mechanical efficiency of the pump
Answer:
The mechanical efficiency of the pump is 91.8 %
Explanation:
Given;
input power, p = 44 kw
density of oil, ρ = 860 kg/m³
motor efficiency, η = 90 %
inlet diameter, d₁ = 8 cm
outlet diameter, d₂ = 12 cm
volume flow rate, V = 0.1 m³/s
pressure rise, P = 500kPa
output power = motor efficiency x input power
output power = 0.9 x 44 = 39.6 kW
Thus, the mechanical input power = 39.6 kW
The mechanical output power is given by change in mechanical energy;
[tex]E = mgh + \frac{m}{2} (v_2^2 - v_1^2) \\\\E = \rho V g h + \frac{\rho V}{2} [(\frac{V_2}{\pi r_2^2} )^2 - (\frac{V_1}{\pi r_1^2})^2]\\\\E = PV + \frac{\rho V^3}{2\pi^2} [\frac{1}{ r_2^4} - \frac{1}{ r_1^4}]\\\\E = (500 *10^3)(0.1) + \frac{(860)(0.1)^3}{2\pi^2} [\frac{1}{ 0.06^4} - \frac{1}{ 0.04^4}]\\\\E = 50000 -13653.51\\\\E = 36346.48 \ W\\\\E = 36.347 \ kW[/tex]
The mechanical efficiency is given by
η = mechanical output power / mechanical input power
η = 36.347 / 39.6
η = 0.918
η = 91.8 %
Therefore, the mechanical efficiency of the pump is 91.8 %
- Over the weekend you go for an urban hike, traveling 3 miles north then 4 miles west before stopping to take a break. If this hike takes 1 hour, what is your average velocity during that time? - You leave your house to go for a drive to clear your head, driving 10 miles over the course of 30 minutes before arriving back home. What is your average velocity?
Answer:
A.) 5 mph
B.) 20 mph
Explanation:
Given that you are traveling 3 miles north then 4 miles west before stopping to take a break. If this hike takes 1 hour,
Let's first calculate the displacement be using pythagorean theorem
D = sqrt ( 3^2 + 4^2 )
D = sqrt ( 9 + 16 )
D = sqrt ( 25 )
D = 5
Average velocity = displacement/time
Average velocity = 5/1
Average velocity = 5 mph
Therefore, your average velocity during that time is 5 mph
- You leave your house to go for a drive to clear your head, driving 10 miles over the course of 30 minutes before arriving back home.
Average velocity = 10/ 0.5
Average velocity = 20 mph
What the differences between static and kinetic friction?
Answer:
Static friction prevents a stationary object from moving while kinetic or dynamic friction slows down a moving object.
Explanation:
Static Friction is the maximum force that must be overcome before a stationary object begins to move, while kinetic or dynamic friction is the maximum force that must be overcome for an object in motion to continue moving at a uniform velocity.
Static friction keeps a stationary object at rest, once the Force of Static friction is overcome, the Force of Kinetic friction is what slows down the moving object.
Which of the following is the best thermal conductor?
A. Fiberglass.
B. Stainless.
C. Steel.
D. Wood.
E. Silver
Answer:
steel because being alloy in metal it has free electrons in metals
A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 41 km/h at the 61-m mark. He then maintains this speed for the next 83 meters before uniformly slowing to a final speed of 34 km/h at the finish line. Determine the maximum horizontal acceleration which the sprinter experiences during the run. Where does this maximum acceleration value occur
Answer:
2.13 m/s^2.
Find the remaining answer in the explanation
Explanation:
Given that a sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 41 km/h at the 61-m mark. He then maintains this speed for the next 83 meters before uniformly slowing to a final speed of 34 km/h at the finish line.
1.) The maximum horizontal acceleration will be when she attained her maximum speed. That is,
Speed = 41 km/h
Convert km/h to m/s
(41×1000) / 3600
Speed = 11.39 m/s
Using speed formula to calculate time
Speed = distance/time
11.39 = 61/t
t = 61/11.39
t = 5.36s
Where the distance = 61 m
Maximum acceleration = velocity /time
Maximum acceleration = 11.39/5.36
Maximum acceleration = 2.13 m/s^2
The maximum acceleration value occurs when the sprinter is starting from rest and attaining the maximum speed.