Answer:
The value is [tex]F_t = 76024 \ N[/tex]
Explanation:
From the question we are told that
The flow rate is [tex]v_1 = 5 \ m/s[/tex]
The entrance diameter is [tex]d = 0.7 \ m[/tex]
The exit diameter is evaluated as [tex]d_e = 0.6 * 0.7 = 0.42 \ m[/tex]
The entrance gauge pressure is [tex]P_g = 350 \ kPa = 350*10^{3} \ Pa[/tex]
The droped gauge pressure is [tex]P_d = 50 \ kPa = 50*10^{3} \ Pa[/tex]
The presure at the exist is evaluated as [tex]P_e =(350 - 50 ) \ kPa = 300*10^{3} \ Pa[/tex]
Generally the entrance cross-sectional area is mathematically represented as
[tex]A = \pi \frac{d^2}{4}[/tex]
[tex]A = 3.142 \frac{0.7^2}{4}[/tex]
[tex]A =0.385 \ m^2[/tex]
Generally the exit cross-sectional area is mathematically represented as
[tex]A_e = \pi \frac{d_e^2}{4}[/tex]
[tex]A_e = 3.142 \frac{0.42^2}{4}[/tex]
[tex]A_e =0.139\ m^2[/tex]
Generally from the continuity equation
[tex]v_1 * A = v_2 * A_e[/tex]
=> [tex]v_2 = \frac{v_1 * A}{A_e}[/tex]
=> [tex]v_2 = \frac{5 * 0.385}{0.139}[/tex]
=> [tex]v_2 = 13.8 m/s [/tex]
Generally the net force in horizontal axis is equivalent to the net momentum change in the horizontal direction
So
[tex]P_g * A + F_t - P_e * A_e = P_d [ A_e * v_2^2 - A* v_1^2][/tex]
Here [tex]F_t[/tex] is the horizontal thrust on the fitting
So
[tex]350*10^{3} * 0.385 + F_t -0.385 * 0.139 = 50*10^{3} [ 0.385* 13.8^2 - 0.385* 5^2][/tex]
What forces are acting on a book sitting on a table? Are action - reaction force involved in this situation? Newton's third law
Answer:Newton’s law also states that larger bodies with heavier masses exert more gravitational pull, which is why those who walked on the much smaller moon experienced a sense of weightlessness, as it had a smaller gravitational pull. To help explain his theories of gravity and motion, Newton helped create a new, specialized form of mathematics.
Explanation: I think I'm very stupid
An observer in frame S sees lightning simultaneously strike two points 100 m apart. The first strike occurs at xx1 = yy1 = zz1 = tt1 = 0 and the second at xx2 = 100 mm, yy2 = zz2 = tt2 = 0. (a) What are the coordinates of these two events in a frame S’ moving in the standard configuration (motion along the common xx − xx’ axis) at 0.70c relative to S? (b) How far apart are the events in S’? (c) Are the events simultaneous in S’? If not, what is the difference in time between the events, and which event occurs first?
Answer:
a) 0, = -0.33 us
b) 140m
c) No, The event are not simultaneous i.e they did not occur at the same time, the second even (-0.33 usec) occurs 0.33 usec earlier than the first event.
Explanation:
a)
the lorentz factor expression is written as;
y = 1₀ / √(1 - (v²/c²))
where v is the relative speed of an observer and c is the speed of light
so we were given that relative speed to be o.7c
therefore
y = 1 / √(1 - ((0.7c)² / c²))
y = 1 / √(1 - (0.49c² / c²))
y = 1 / √(1 - 0.49)
y = 1 / 0.7141
y = 1.4
1 - the coordinates of the first event, the s' frame of reference is,
x1 ' = y(x1 - vt1) = 0
y1 ' = y1, z1' = z1 and
t1 ' = y [t1 - v/c²x1]
= 0
2 - the coordinates of the second event, the s ' frame of reference is'
x2 ' = y(x2-vt2)
= 1.4(100m - 0)
= 140m
y2 ' = y2, z2 ' = z2
t2 ' = y [ t2 - v/c²x2 ]
= 1.4 [ 0 - 0.7c/c²(100) ]
using speed of light c as 3*10^8
1.4 [ 0 - (0.7*3*10^8) / (3*10^8)²(100) ]
= -0.33 us
b)
distance between
delltaX' = X2' - X1'
= 140m - 0
= 140m
c)
No, The event are not simultaneous i.e they did not occur at the same time.
the second even (-0.33 us) occurs 0.33 us earlier than the first event.
a light bulb is a kind of lever is the statement true or false
Answer:
False
Explanation:
It would be a screw
Specialized visual receptors that assist mostly in nighttime vision are __________. A. cones B. rods C. fovea D. optic disks Please select the best answer from the choices provided A B C D
Answer:
B
Explanation:
EDGE 2020
The rod photoreceptor cell type, which is found in the retina, transmits low-light vision and is primarily in charge of the neural transmission of night vision. Thus, option B is correct.
What visual receptors that assist mostly in nighttime vision?We can see at night thanks to rods, and we can see color during the day thanks to cones. Although daylight vision is far more important for us, the topic of why there are roughly 20 times more rods than cones in a human retina has typically been met with a shrug.
Rods function in extremely dim lighting. Because just some light particles (photons) may trigger a rod, we employ these for night vision. Because rods hinder color vision, we experience a grayscale world at night. More than 100 million rod cells make up the human eye.
Therefore, The retina's rod photoreceptor cell type conveys vision in low light and is principally in charge of the neuronal activity of night vision.
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Suppose you wish to whirl a pail full of water in a vertical circle without spilling any of its contents. If your arm is 0.82 m long (from shoulder to fist) and the distance from the handle to the surface of the water is 18.5 cm, what minimum speed is required?
Answer:
Approximately [tex]3.1\; \rm m \cdot s^{-1}[/tex].
Explanation:
The content of this pail is in a centripetal motion because its path forms part of a vertical circle. Let [tex]m[/tex] denote the mass of the contents of this pail, let [tex]v[/tex] denote the (linear) velocity of the content, and let [tex]r[/tex] denote the radius of this circle. The net force on the contents of this pail will thus be:
[tex]\displaystyle F(\text{net}) = \frac{m\, v^2}{r}[/tex] towards the center of the circle.
Assume that there is no friction between the content and walls of the pail. The only two possible forces on the contents pail towards the center would be:
The downwards gravitational pull from the earth, Normal force between walls of the pail and the contents (except at the top and bottom of the circle,) andIf the rotation is fast enough, the normal force from the bottom of the pail, which also points downwards.Note that at the top of the circle, both the gravitational pull and the normal force from the bottom point towards the center of the circle. On the other hand, the normal force from the walls of the pail would be perpendicular to the line towards the center of the circle. At that point in the circle, there's no upward force to support the content of the pail. The uniform rotation will be sufficiently fast if it could allow the content to stay in the pail at the top of the circle.
Let [tex]g[/tex] denote the gravitational field strength at the top of this circle. The size of the gravitational pull on the content would be [tex]m\cdot g[/tex]. Let [tex]F(\text{normal})[/tex] denote the normal force from the bottom of the pail on the contents. The sum of these two forces should be equal to the vertical net force on the contents of this pail. That is:
[tex]F(\text{net}) = m\cdot g + F(\text{normal})[/tex].
From the centripetal motion of the content:
[tex]\displaystyle \frac{m\, v^2}{r} = m\cdot g + F(\text{normal})[/tex].
Rearrange to obtain an expression for the normal force:
[tex]\displaystyle F(\text{normal}) = \frac{m\, v^2}{r} - m\cdot g[/tex].
Note, that the normal force the bottom of the pail exerts on the contents should be greater than or equal to zero. While the pail is at the top of the circle, the normal force from the bottom of the pail cannot pull the contents upwards. Hence:
[tex]\displaystyle \frac{m\, v^2}{r} - m\cdot g = F(\text{normal}) \ge 0[/tex].
[tex]\displaystyle \frac{m\, v^2}{r} - m\cdot g \ge 0[/tex].
Rearrange and simplify to obtain:
[tex]\displaystyle \frac{v^2}{r} - g \ge 0[/tex].
[tex]v^2 \ge g\cdot r[/tex].
[tex]v \ge \sqrt{g \cdot r}[/tex].
In other words, if the gravitational field strength is [tex]g[/tex] and the radius of the circle is [tex]r[/tex], the minimum linear velocity required to keep the content in the pail at the top of the circle is [tex]\sqrt{g \cdot r}[/tex].
If [tex]g = 9.81\; \rm N \cdot kg^{-1} = 9.81\; \rm m \cdot s^{-2}[/tex] and [tex]r = 0.82 \; \rm m + 0.185\; \rm m \approx 1.005\; \rm m[/tex], then the minimum value of [tex]v[/tex] would be approximately:
[tex]\sqrt{9.81 \; \rm m \cdot s^{-1} \times 1.005\; \rm m} \approx 3.1\; \rm m \cdot s^{-1}[/tex].
Galileo's telescopes were not of high quality by modern standards. He was able to see the moons of Jupiter, but he never reported seeing features on Mars. Use the small-angle formula to find the angular diameter of Mars when it is closest to Earth. How does that compare with the maximum angular diameter of Jupiter? (Assume circular orbits with radii equal to the average distance from the sun.)
Answer:
Angular diameter of Mars = 15.80 * 10^5 arc seconds
The Angular diameter of Mars is 3 times the angular diameter of Jupiter
Explanation:
Average distance of the earth from sun = 150.67 * 10^6 km
assuming the radius of Mars ( average distance from sun) = 209.33 * 10^6 km
assuming the radius of Jupiter(average distance from sun) = 768.71 * 10^6 km
The small-angle formula for mars
angular diameter = ( linear diameter / distance ) * (2.06 * 10^5 )
distance between earth and mars = 54.6 * 10^6 km
linear diameter = 2 * radius = 418.66 * 10^6 km
angular diameter = ( 418.66 / 54.6 ) * 2.06 * 10^5
= 15.80 * 10^5 arc seconds
small angel formula for Jupiter
Angular diameter = ( linear diameter / distance ) * (2.06 * 10^5)
distance between Jupiter and earth = 588 * 10^6 km
linear diameter = 2 * radius = 1537.42 * 10^6 km
Angular diameter = ( 1537.42 / 588) * 2.06*10^5
= 5.39 * 10^5 arc seconds
comparing the angular diameter of the Mars and that of Jupiter :
The angular diameter of mars / angular diameter of Jupiter
= 15.80 / 5.39 = 2.931 ≈ 3
A 28-g rifle bullet traveling 190 m/s embeds itself in a 3.1-kg pendulum hanging swing upward l an arc. Determine the vertical and horizontal components of the pendulum's maximum displacement.
Answer:
V = 0.15m
H = 0.85m
Explanation:
We will be using the conservation of momentum to solve the problem.
m(i).v(i) = m(f).v(f) where
0.028 * 190 = 3.1 * v(f)
v(f) = 5.32/3.1
v(f) = 1.72 m/s
next, we use the conservation of energy to find the vertical displacement.
½mv² = mgh
½v² = gh, making h the subject of formula, we have
h = v²/2g
h = 1.72²/ 2 * 9.81
h = 2.9584 / 19.62
h = 0.15 m
The height, h is the vertical displacement.
Next, we find the angle of the pendulum at the top of the swing.
Φ = cos^-1 [(2.5 - 0.15)/2.5] *the height at which the pendulum is hanging is not given, so I assumed it to be 2.5m, yo can substitute that for whatever it is
Φ = cos^-1(2.35/2.5)
Φ = cos^-1 (0.94)
Φ = 19.95°
This gotten angle is used to find the horizontal displacement
x = 2.5 sinΦ
x = 2.5 sin 19.95
x = 2.5 * 0.34
x = 0.85 m
Lucy took 3 hours to cover 2/3 of a journey. She covered the remaining 60 miles in 2 hours. What was the average speed for the whole journey?
The correct answer is 36 miles per hour
Explanation:
To find the average speed, the first step is to find the total time and the total distance Lucy covered because the formula for average speed is the total distance divided into the total time.
Total Distance
It is known Lucy covered 2/3 of the total distance first, and then she completed the remaining distance (60 miles). According to this, the last distance or 60 miles represents 1/3 of the total distance.
[tex]\frac{2}{3} + \frac{1}{3} = \frac{3}{3}[/tex] (Total distance)
Also, if 60 miles is equivalent to 1/3 this number can be used to find the total distance. In this case, just multiply 60 by 3 = 180 miles as 60 is one-third.
This means the total distance is 180 miles
Total Time
It is known the first part of this journey took 3 hours and the second took 2 hours. This means the total time was 5 hours (3 + 2 = 5)
Find the average speed
[tex]Average speed =\frac{Distance}{time}[/tex]
[tex]Average speed = \frac{180 m}{5 h}[/tex]
[tex]Average speed = 36m/h[/tex]
A train accelerated from 10km/hr to 40km/hr in 2 minutes. How much distance does it cover in this period? Assume that the tracks are straight?
Lesley doesn't like having to wait for other people to
A piston-cylinder device containing a fluid is fitted with a paddle wheel stirring device operated by the fall of an external weight of mass 51kg. As the mass drops by a height of 5.6m, the paddle wheel makes 10100 revolutions. Meanwhile the free moving piston (frictionless and weightless) of 0.51m diameter moves out by a distance of 0.71m. Determine the net work for the system if atmospheric pressure is 101 kPa.
Answer:
The value is [tex]W_N = 11849 \ J [/tex]
Explanation:
From the question we are told that
The mass of the external weight is [tex]m = 51 \ kg[/tex]
The height through which the mass drops is [tex]h = 5.6 \ m[/tex]
The number of revolution made is [tex]N = 10100 \ kg[/tex]
The diameter of the free moving piston is [tex]d = 0.51 \ m \ kg[/tex]
The distance moved by the free moving piston is [tex]s = 0.71 \ m \ kg[/tex]
The atmospheric pressure is [tex]P = 101 \ kPa = 101*10^{3}\ Pa [/tex]
Generally the workdone by the external weight is mathematically represented as
[tex]W_w =m * g * h[/tex]
= [tex] 51 * 9.8 * 5.6 [/tex]
= [tex]2799 N [/tex]
Generally the workdone by the free moving piston is mathematically represented as
[tex]W_p =P * A * s[/tex]
Here A is the cross-sectional area with value
[tex] A = \pi * \frac{d^2}{4}[/tex]
[tex] A = 3.142 * \frac{0.51^2}{4}[/tex]
So
[tex]W_p =101*10^{3} * 3.142 * \frac{0.51^2}{4} * 0.71[/tex]
=> [tex]W_p = 14651 [/tex]
So
The net workdone is mathematically evaluated as
[tex]W_N = -W_w+W_p [/tex]
The negative sign shows that it is acting in opposite direction to [tex]W_N[/tex]
So
[tex]W_N = -2799+14651 [/tex]
[tex]W_N = 11849 \ J [/tex]
What is the difference between mass and weight?
A) Mass measures the force of gravity and weight measures volume.
B) Mass is measured in metric units and weight is measured in US Standard System.
C) Mass measures amount of matter in an object and weight measures the force of gravity.
D) Mass measures volume and weight measures the force of gravity.
Answer:
C. Mass is independent of the force acting on the mass.
Weight depends on the gravitational force acting on the mass.
Example: A mass of 1 kg on the earth still has a mass of 1 kg on the moon;
whereas the weight of that mass on earth is 9.8 N and only 1/6 of that amount on the moon.
The amount of space between two points is measure in unit?
The measurement of distance is the length between two places. To measure is to establish the distance between two geometric objects. With a ruler, you can measure distance the most frequently. Ordinarily, eighth-inch (or 0.125 in) segments make up inch rulers.
What space between two points is the measure unit?When calculating the distance to stars, many astronomers choose to use parsecs (abbreviated pc). This is due to the definition's tight connection to a technique for determining the separation between stars. The distance at which 1 AU subtends an angle of 1 arc sec is measured in parsecs.
A unit of measurement is a specific magnitude of a quantity that is established and used as a standard for measuring other quantities of the same kind.
Therefore, Any additional amount of that type can be stated as a multiple of the measurement.
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A force vector points due east and has a magnitude of 140 newtons. A second force is added to . The resultant of the two vectors has a magnitude of 420 newtons and points along the (a) east/ (b) west line. Find the magnitude and direction of . Note that there are two answers. (a) Number newtons (b) Number newtons
Answer:
(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East
(b) When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West
Explanation:
Given;
a force vector points due east, [tex]F_1[/tex] = 140 N
let the second force = [tex]F_2[/tex]
let the resultant of the two vectors = F
(a) When the resultant force is pointing along east line
the second force must be pointing due east
[tex]F = F_1 + F_2\\\\F_2 = F - F_1\\\\F_2 = 420 \ N - 140 \ N\\\\F_2 = 280 \ N[/tex]
[tex]F_2 = 280 \ East[/tex]
(b) When the resultant force is pointing along west line
the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.
[tex]F = F_2 - F_1\\\\F_2 = F + F_1\\\\F_2 = 420 \ N + 140 \ N\\\\F_2 = 560 \ N[/tex]
[tex]F_2 = 560 \ West[/tex]
Which of the following is NOT an important part of exercise clothing? A belts B. shoes C. shirts D. socks
Answer:
BELTS
Explanation:
why would you need belts for running and stuff
Belts are not an important part of exercise clothing.
What is exercise?An exercise is a physical activity performed to increase strength and agility of the muscles as well as to enhance fitness.
During an exercise, it is important to wear clothes that will not hinder movement or cause injury.
Some items of clothing needed in exercise include:
shoes shirts socksTherefore, belts are not important during exercise.
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A cruise ship sails due south at 2.50 m/s while a coast guard patrol boat heads 19.0° north of west at 4.80 m/s. What are the x-component and y-component of the velocity of the cruise ship relative to the patrol boat? (Assume that the +x-axis is east and the +y-axis is north. Enter your answers in m/s.)
Answer:
Explanation:
We shall represent the velocity of cruise ship and coast guard petrol boat in vector form .
velocity of cruise ship
Vcs = - 2.5 j
Vpb = - 4.8 cos 19 i + 4.8 sin 19 j = - 4.54 i + 1.56 j
velocity of the cruise ship relative to the patrol boat
= Vcs - Vpb
= - 2.5 j - ( - 4.54 i + 1.56 j )
= - 2.5 j + 4.54 i - 1.56 j
= 2.04 i - 1.56 j .
x-component of the velocity of the cruise ship relative to the patrol boat
= 2.04 m /s
y-component of the velocity of the cruise ship relative to the patrol boat
= - 1.56 m /s .
A certain ultrasound device can measure a fetal heart rate as low as 50 beats per minute. This corresponds to the surface of the heart moving at about 4.0 x 10-4 m/s. If the probe generates ultrasound that has a frequency of 2.0 MHz (1 MHz = 1 megahertz = 106 Hz), what frequency shift must the machine be able to detect
Answer:
The. Machine must detect a shift of
1 Hz
Explanation:
Frequency shift is given as
={ ( Vsound +V/ V sound -V) -1}f emitted
So by substitution we have
= { 1540+4E-4/1540-4E)-1) 2*10^6
= 1Hz
50 points =) The morning after a massive snowstorm, Michaela gets into her car to drive to work. The storm caused her windows to freeze, so she first needs to defrost the car. While the engine is running, she checks the thermometer. It shows the air inside of her car has a temperature of 0 °C. Does this mean the air inside of her car has no kinetic energy? Explain your answer.
Answer:
YES!
THE AIR INSIDE A CAR THE AIR RESISTANCE WILL CONVERT SOME OF THE KINETIC ENERGY FROM THE CAR INTO THE KINETIC ENERGY IN THE AIR AS IT'S PUSHED AWAY FROM THE FRONT OF THE CAR AND SUCKED BACK INTO THE GAP BEHIND IT.this TOO DEGRADES INTO HEAT.PLEASE I NEED A BRAINLIEST
If the distance d (in meters) traveled by an object in time t (in seconds) is given by the formula d = A + Bt^2, the SI units of A and B must be:_______
Answer:
The SI units of the “A” is m (meters)
The SI units of the “B” is m/s^2
Explanation:
Given the distance = d meters.
Time taken to travel = t (seconds)
Function of the distance, d = A + Bt^2
Now we have given the above information and from the given distance function, we have to find the SI units of the A and B. Here, below are the SI units.
Thus, the SI units of the “A” is = m (meters)
The SI units of the “B” is = m/s^2
Give an example of each type of electromagnetic radiation.
Answer:
Nearly all frequencies and wavelengths of electromagnetic radiation can be used for spectroscopy. Radio waves, infrared rays, visible light, ultraviolet rays, X-rays, and gamma rays are all types of electromagnetic radiation. Radio waves have the longest wavelength, and gamma rays have the shortest wavelength.
Explanation:
You walk to the corner store at 2 m/s for a total time of 76 seconds. What is the distance to the store?
Answer:
152.
Explanation:
2 x 76 = 152
bruh its not that hard
Answer:
152 m
Explanation:
Distance = Speed x Time
[tex]2 \times 76 \\ 152[/tex]
Thus the distance is equal to 152 m.
How long would it take for Sofia to walk 300 meters if she is walking at a velocity of 2.5 m/s?
Answer:
Time=120seconds
Explanation:
S=300m
V=2.5m/s
t=?
V=S/t
t=S/V
t=300/2.5
t=120 second
Answer:
120 seconds
Explanation:
.....
...
.....
Why are chemical processes unable to produce the same amount of energy flowing out of the sun as nuclear fusion?
Answer:
Explanation:
One of the major differences between nuclear reactions and chemical reactions is that nuclear reactions involve larger amount of energy than chemical energy. This is because the force between the protons and neutrons in the nucleus of an atom is much higher than the force of attraction between electrons and the positively charged nucleus, hence nuclear reactions involves/requires a larger amount of energy (because it's reactions involve the nucleus) than chemical reactions (because it's reactions involve the electrons).
Thus, during nuclear fusion, two light nuclei are bombarded against one another to produce a larger/heavier nuclei with the release of large amount of energy (because the forces between the protons and neutrons are much higher) unlike when two atoms/molecules are chemically combined together to form a new molecule with the rearrangement of electrons in the valence shells of the participating molecules.
What is the velocity of a plane that traveled 1500 miles from New York City to Orlando in 10.0 hours?
Answer:
v = (1500/10) = 150 [miles/h]
Explanation:
The velocity can be easily calculated using the following kinematics expression:
v = x/t
where:
v = velocity [miles/h]
t = time = 10 [hr]
x = distance = 1500 [miles]
v = (1500/10) = 150 [miles/h]
sovle it please help me
Explanation:
(a)
F = (GM1M2)/r²
F = (6.7 X 10^-11 x 30.3 x 40.17)/0.5²
F = 326.197 x 10^-⁹N
(b)
F =750N
G = 6.7 x 10^-11
M1 = 2050kg
r = 70.3m
M2 = ?
F = (GM1M2)/r²
F x r² = GM1M2
(F x r²)/GM1 = M2
M2 = (750 x 70.3²)/(6.7 x 10^-11 x 2050)
M2 = 26.99kg
Convert 47 meters to gigameters
Answer: 4.7× 10-8 gigameter.
Exgiga meter. Hope this helps :)
A truck moving at 13.3 m/s hits a concrete wall. As a result of the collision, a 6-kg wrench moves forwards and strikes the wall of the tool compartment. If the wrench stops after being in contact with the wall for 0.07 s, what is the average force exerted on the wrench by the wall
Answer:
Explanation:
The velocity of the wrench must be equal to the velocity of the truck . So momentum of the wrench before it hits the wall
= mv = 6 x 13.3 = 79.8 kg m /s
If resisting force of wall be F , impulse on the wrench = F x time
= F x .07
Impulse = change in momentum of the wrench = mv - 0 = mv = 79.8 kgm/s
So F x .07 = 79.8
F = 1140 N .
Consider a thin circular disk that has been heated up to 400 °C and then left inside a chamber to cool down. The chamber surface is kept at 40 °C and the air inside is maintained at 25 °C. Assume that the disk is held at the center of the chamber, and we can ignore heat conduction effects in this problem. The disk surface property is known as ε=0.65. For the diameter of D=200 mm, what is the total rate of heat transferfrom the disk?
Answer:
hello your question lacks the required diagram attached below is the complete question with the required diagram
answer : Qtotal = 807.4 Mw
Explanation:
Given Data :
disk properties :
∈ = 0.65
D = 200 mm
Ts = 400⁰c
attached below is the detailed solution
The total rate of Heat transferred from the disk
Qtotal = 807.4 Mw
What is the weight in newtons of lily’s convertible, which has a mass of 1800 kg?
Answer:
Explanation:
W=?
Mass=1800kg
g=(9.8m/s^2 constant)
W=mg
W=1800*9.8
W=17649N
a car speeds up from 8.50 m/s to 22.2 m/s in 3.84 s. what is the acceleration of the car
Answer:
3.57
Explanation:
22.2-8.50=13.7
13.7/3.84=3.57
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt.
