Answer:
u=speed, w=wavelenght, f=frequency
It's known that u=w*f => f=u/w
u=20m/s ==> f=20/0,5 => f=40 Hz
w=0,50m
Explanation:
the projectile thrown with the same initial velocity , one at an angle theta and the other at an angle 90-theta can both projectiles strike at the same distance from the projection point? can both projectiles be in air for the same time interval?
Answer:
If theta is equal to 90, then both projectiles strike at the same distance from the projection point and projectiles are in air for the same time interval.
Explanation:
Temperatures expressed in the Kelvin scale are ____ higher than temperatures in the Celsius scale?
A.) 200°
B.) 283°
C.) 460°
D.) 273°
Answer:
Temperatures expressed in the Kelvin scale are D.) 273° higher than temperatures in the Celsius scale.
Explanation:
This answer is 110% correct guaranteed!
⭐⭐⭐⭐⭐ +
a. Recent studies have raised concern about 'heading' in youth soccer (i.e., hitting the ball with the head). A soccer player 'heads' a size 3 ball deflecting it by 40.0°, and keeps its speed of 13.20 m/s constant. A size 3 ball has a mass of approximately 2.000 kg. What is the magnitude of the impulse which the player must impart to the ball?
b. If the player's head has a mass of 2.90 kg, what is the magnitude of the average acceleration of the player's head during the impact. Assume that over the brief time of the impact, 30.40 ms, the player's head can be treated separately from the player's body.
a. The magnitude of the impulse which the player must impart to the ball is 17.25 kg.m/s.
b. The magnitude of average acceleration of the player's head during the impact is 195.66 m/s².
What is impulse?The change in momentum is equal to the product of impact force applied while colliding and time for that impact.
Impulse F. t = m (Vf -Vi)
where, Vf is the final velocity and Vi is the initial velocity.
Given, a soccer player 'heads' a size 3 ball deflecting it by 40.0°, and keeps its speed of 13.20 m/s constant. A size 3 ball has a mass of approximately 2.000 kg.
a. Substitute the values into the expression, we get
Impulse (x) = 2 x (13.20 - 13.20cos40°)
Impulse (x) = 3.088 kg.m/s
Impulse in y direction, is
Impulse (y) = 2 x 13.20sin40°
Impulse (y) = 16.97 kg.m/s
so, the magnitude of impulse is
I = sq rt(Ix² + Iy²)
Put the values, we get
I = 17.25 kg.m/s
Thus, the magnitude of the impulse which the player must impart to the ball is 17.25 kg.m/s
b. The player's head has a mass of 2.90 kg. The brief time of the impact, 30.40 ms,
F.t = I
Put the given values, we have
F = 17.25 / 30.40 x 10⁻³
F = 567.434 N
From the Newtons second law of motion,
F = ma
Plug the values from the question, we have
567.434 N = 2.90 x a
a = 195.66 m/s²
Thus the magnitude of average acceleration of the player's head during the impact is 195.66 m/s².
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question is below............
The total distance traveled by the gardener is 8 meters. the magnitude of total displacement is approximately 7 meters.
Total distance covered by gardener = 2 + 1 +2 + 3= 8 meters
For the magnitude of displacement, we will take the shortest path for the gardener movement.
for displacement-
[tex]=\sqrt{( 0-4^{2}) +( 6-0^{2} )}\\=\sqrt{52} \\=7[/tex](approximately)
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what type of forces that created at shaping machine
Answer:
a magnetic force
Explanation:
When a cold alcohol thermometer is placed in a hot liquid, the column of alcohol goes down slightly before going up. Explain why.
Answer:
See below
Explanation:
When a cold alcohol thermometer is placed in a hot liquid, the following takes place.The glass of thermometer expands because of the heat so the column of alcohol goes down.After some time, the heat travels to alcohol and then alcohol expands.So, Reason why alcohol goes down first:Alcohol first goes slightly down because of the expansion of the glass of thermometer before doing up.
[tex]\rule[225]{225}{2}[/tex]
A box with a mass of 40 kg sits at rest on a frictionless tile floor. with your foot, you apply a 20 N force in a horizontal direction. what is the acceleration of the box?
Answer:
0.5 m/s²
Explanation:
according to Newton's second law, we are goven a relationship between force, mass and acceleration, with the formula:
F = m×a
F for force
m for mass
a for acceleration
we use the given data and get:
20 = 40×a
we find a=20/40=0.5m/s²
acceleration of the box is 20m/s^2
given
mass (m) = 40 kg
force (f) = 20n
acceleration (a) =?
we know,
f = m × a
20 = 40 × a
40 - 20 = a
20m/s^2 = a
Find the magnitude of this
vector:
The magnitude of the vector is 34.092 N
What is resultant?The resultant is the square root of sum of square of the vector component of x and y direction.
R =√ [(Nx)² +(Ny)²]
Given is the vector with horizontal length of the vector = -28.7 m and the vertical length of the vector = 18.4 m
The resultant will be
R =√ [(-28.7)² +(18.4)²]
R = 34.092 N
Thus, the magnitude of the vector is 34.092 N.
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Metals are used in many products because of the characteristic properties that most metals have.
Which product
requires the high luster of metals in order to work?
O mirrors
O pots and pans
O food wrap
O circuit boards
a) Mirrors
Mirror requires the high luster of metals in order to work.
What causes the highest luster of a metal?When light shine on to the surface of a metal, its electrons absorb small amounts of energy and become excited into one of its many empty orbitals.
The electrons immediately fall back down to lower energy levels and emit light.
This process is responsible for the high luster of metals.
Here,
Mirrors have metal coating on its back side, because of high reflective property , high lustrous mirror is preferred . When metal is more lustrous, the less light reflect into it and bounces back.
Coating one side of a piece of glass with shiny metals can turn it into a mirror, reflecting light coming toward it.
Window glass can reflect only eight percent of light hitting it, while mirrors can reflect 95 percent of light hitting them.
The glass in a mirror is usually coated with a layer of silver or aluminum.
Mirrors are shiny because when photons (rays of light) coming from an object , it strike the smooth surface of a mirror, then they bounce back at the same angle. Our eyes see these reflected photons as a mirror image.
Therefore,
Mirror requires the high luster of metals in order to work.
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Answer:A) mirrors
Explanation:
Across:
5. The term used for describing the
bending of an arm or a leg
6.
9.
Down:
1. This is one popular brand of exercise
machine
2.
They are made of two types of fibers
You can achieve your target pulse rate
by varying the number of these
flexes the arms at
10. The arm
the elbows
11. These muscle fibers do not require as
much oxygen
13. He wrote a book on aerobics
14. This type of training works primarily
with the white muscle fibers
15.
extends the legs at the
The leg
knees
16. Aerobic exercises help the body's
ability to use this element
18.
The term used to describe the straight-
ening of your arm or leg
3.
4.
5.
6.
7.
8.
This was the most popular form of aerobic
exercise in the early years of aerobics
This organ is one part of the cardiovascu-
lar system
This exercise conditions the cardiovascu-
lar system
Aerobic exercise became part of this
movement in the 1970s
12.
17.
This accompanies aerobic dance
A specific exercise produces this type of
result
This type of training involves moving
from one exercise machine to another with
little rest
If you just train for strength, you will not
improve this
These muscle fibers contract more slowly
Answer:hight
Explanation:
Helium is the second most abundant element on Jupiter (and in the Universe overall). But it's rare on Earth, being only 0.0005% of the atmosphere. Why is this?
A) Jupiter's stronger gravity results in a faster escape velocity, and almost no helium atoms are moving fast enough to escape its gravity. Earth's escape velocity is lower, so a few helium atoms at a time are moving fast enough to escape, causing Earth to slowly lose its helium.
B) Jupiter's core continually generates helium, while the Earth's core does not.
C) Jupiter has a powerful magnetic field that can attract helium atoms. Earth has a magnetic field as well, but it's not strong enough.
Answer:
Explanation:
Comment
I think the best answer is probably A. It's not the best reason for Earth loosing it's helium, but it is the only one that is close. The Earth derives some of its Helium (most) by the decomposition process of radio active high weight chemical that break down. Helium combines with practically nothing so once it gets moving, very little will stop it. That's the way we loose our Helium.
Still answer A.
An ultraviolet wave traveling through a vacuum has wavelength of 4.0 x 10^-7 m. The waves frequency, written in scientific notation to two significant figures, is ? X10^14Hz.
Answer:
λ = c / f or f = c / λ
f = 3.0E8 / 4.0E-7 = .75E15 / sec = 7.5E14 / sec = 7.5 X 10^14 /sec
Radioactive decay can involve electrons
True or False
Answer:
i believe the answer is true
If a jet is travelling horizontally at 100m/s at a height of 500m above the ground and a jet drops a bomb to the ground. Where does the bomb strike the ground relative to the point atwhich it is released ?A.100m B.1km C.500m D.5km
The bomb strike the ground relative to the point at 1km . B
How to determine the distanceUsing the equation
h = 1/2 gt^2
500 = 1/2 * 10* t^2
500 = 5t^2
t = √500/5
t = √100
t = 10seconds
To find the distance,
Distance = velocity * time
Distance = 100 ÷ 10
Distance = 1000m = 1km
Therefore, the bomb strike the ground relative to the point at 1km . B
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A plant is thrown straight
down from a balcony 7.01 m
high at 8.84 m/s. How long
does it take the plant to hit
the ground?
The time taken for the plant to hit the ground from a distance of 7.01m and at a velocity of 8.84m/s is 1.59s.
How to calculate time?The time taken for a motion to occur can be calculated using the following formula:
v² = u² - 2as
Where;
v = final velocityu = initial velocitys = distancea = acceleration8.84² = 0² + 2 × a × 7.01
78.15 = 14.02a
a = 5.57m/s²
V = u + at
8.84 = 0 + 5.57t
t = 1.59s
Therefore, the time taken for the plant to hit the ground from a distance of 7.01m and at a velocity of 8.84m/s is 1.59s.
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A point charge is placed 3m from a 4uc charge what is the strength of the electric field on the point charge at this distance round to the nearest thousand
The strength of the electric field on the point charge at this distance will be 4000 V/m.
What is the strength of the electric field?
The strength of the electric field is the ratio of electric force per unit charge.
The given data in the problem is;
Qis the unit charge = 4.0 × 10⁻⁶ C
E is the strength of the electric field
R is the distance from point charge = 3 m
The strength of the electric field is;
[tex]\rm E = \frac{KQ}{R^2} \\\\ \rm E = \frac{9 \times 10^9 \times 4 \times 10^{-6} \ C}{3^2} \\\\ E= 4000 V/m[/tex]
Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.
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Jose was out drinking with his friends for nearly the whole night. The next morning he was confused and vomiting, and had a low body temperature.
Answer:
He has a hangover.
Explanation:
Just something I know.
Which graph shows the correct relationship between kinetic energy and speed? A. Graph representing the relationship between speed on the x-axis and kinetic energy on the y-axis. A straight line starts at the y-axis B. Graph representing the relationship between speed on the x-axis and kinetic energy on the y-axis. A straight line starts from the origin C. Graph representing the relationship between speed on the x-axis and kinetic energy on the y-axis. A curve line runs parallel to the x-axis and y-axis D. Graph representing a relationship between speed on the x-axis and kinetic energy on the y-axis. A semi-curve line starts above the origin on the y-axis and curves upwards as it moves forward in speed
The proper connection between kinetic energy and speed is depicted in Graph D.
What is kinetic energy?The kinetic energy (KE) is defined as one-half of the mass times multiplied by the square of velocity.
[tex]\rm KE = \frac{1}{2}mv^2[/tex]
where,
KE is the kinetic energy
m is the mass of each molecule
(V) is the velocity
As the square of the velocity is exactly proportional to kinetic energy. Consequently, the relationship between velocity and kinetic energy must be parabolic.
The proper connection between kinetic energy and speed is depicted in Graph D. Graph showing the link between kinetic energy on the y-axis and speed on the x-axis.
On the y-axis, a semi-curve line begins above the origin and ascends as it accelerates.
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5, the following equation shows the position of a particle in time t, x=at2i + btj where t is in second and x is in meter. A=2m/s2, b=1m/s. Find
A, the average velocity of the particle in the time interval t1=2sec and t2=3sec
B, the velocity and acceleration at any time t.
C, the average acceleration in the time interval given in part (a)
The average velocity is 4i m/sec
The average acceleration is 4i
What is a average velocity ?Average velocity is defined as the change in position or displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs. T
According to the question,
x=at²i + btj ( t is in second and x is in meter).
Average velocity = [tex]\frac{dx}{dt}[/tex]
[tex]\frac{dx}{dt} = 2at i + b j\\\\\frac{dx}{dt} = 2 * 2 * t i +1 j\\\\\frac{dx}{dt} = 4t i +j[/tex]
[tex]\frac{dx}{dt}[/tex] in the time interval of 2 to 3 sec is
at 2 sec,
[tex]\frac{dx}{dt} = 4 * 2 i +j = 8i +j[/tex]
at 3 sec ,
[tex]\frac{dx}{dt}[/tex] = 4 * 3 i + j = 12i+j
Therefore the average velocity is
[tex]\frac{(12i +j - 8i -j)}{(3-2)} = 4i[/tex]
The velocity at any given time is [tex]\frac{dx}{dt} = 4t i +j[/tex]
The acceleration at any given time is [tex]\frac{dx^2}{dt^2}[/tex]
[tex]\frac{dx^2}{dt^2} = 4 i[/tex]
The average acceleration in the time interval t₁=2sec and t₂=3sec
Average Acceleration = 4i
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A hollow cast-iron cylinder 4m long, 300mm outer diameter, and thickness of metal 50mm is subjected to a central load on the top when standing straight. The stress produced is 75000kN/m2. Assume Young's modulus for cast iron as 1.5x 108 kN/m^2 and find (i) magnitude of the load, (ii) longitudinal strain produced, and (iii) total decrease in length.
Here, the calculated Magnitude of the load P is 2945.2 kN, the Longitudinal strain produced is 0.0005 and the decrease in length is 2 mm.
Given,
Length, L = 4 m
Outer diameter, D = 300mm, D= 0.3 m
Thickness, t = 50 mm, t = 0.05 m
Stress produced, σ = 75000 kN/m²
Young's modulus for cast iron, E = 1.5 x 10⁸ kN/m²
Calculating the diameter of the cylinder,
Diameter of cylinder, d = (D) – (2t) = 0.3 –( 2 × 0.05)
d= 0.2 m
(i) Magnitude of the load P:
Using the relation, σ =P/A
P = σ × A = 75000 × π /4 (D² – d² )
P= 75000 × π/4 (0.3² – 0.2²)
P= 75000 × π/4 (0.09 – 0.04)
P = 2945.2 kN
Hence, Magnitude of the load P is 2945.2 kN.
(ii) Longitudinal strain produced, e :
Using the relation, Strain, (e) = stress/E
e= 75000/(1.5 x 10⁸)= 0.0005
Hence, the Longitudinal strain produced is 0.0005.
(iii)Total decrease in length, dL:
The total decrease in length can be calculated using the strain as the ratio of change in length to the original length is known as Strain.
Strain = change in length/original length
e= dL/L
0.0005 = dL/4
dL = 0.0005 × 4m = 0.002m=2mm
Hence,the decrease in length is 2 mm.
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2. Height at point A is 30.0 meters with a cart mass of 200.0 kg initially at rest. Point B is at a height of 25.0 meters. Find the KE and PE at points A and B along with the velocity at point B assuming no friction. Show Your Work
At maximum height, velocity is zero and kinetic energy will be zero.
K.E(A) = 0
P.E(A) = mgh = 200 x 9.8 x 30 = 58,800 J
K.E(B) = P.E(A) - P.E(B)
K.E(B) = 58,800 J - (200 x 9.8 x 25)
K.E(B) = 58,800 J - 49,000 J
K.E(B) = 9,800 J
Velocity at point BK.E = ¹/₂mv²
v² = 2K.E/m
v² = (2 x 9800)/(200)
v² = 98
v = √98
v = 9.9 m/s
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Two students have fitted their scooters with the same engine. Student A and his
scooter have a combined mass of 127.5 kg and a maximum acceleration of
2.40 m/s². Student B has a maximum acceleration of 1.70 m/s on her scooter.
Show that the combined mass of student B and her scooter is 120 kg
You may assume that the frictional forces acting on each scooter are negligible.
The force exerted by student A with his scooter is 306 N and that of student B is 204 N.
Force applied by each studentThe force exerted by each student is calculated from Newton's second law of motion.
F = ma
where;
m is mass a is accelerationF(A) = 127.5 x 2.4
F(A) = 306 N
F(B) = 120 x 1.7
F(B) = 204 N
Thus, the force exerted by student A with his scooter is 306 N and that of student B is 204 N.
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A star near the visible edge of a galaxy travels in a uniform circular orbit. It is 41,200 ly (light-years) from the galactic center and has a speed of 275 km/s. Estimate the total mass of the galaxy based on the motion of the star.
Gravitational constant is 6.674×10−11 m3/(kg·s2) and mass of the Sun Ms=1.99 × 1030 kg.
*Answer in billion solar mass
The total mass of the galaxy is 443.4 Solar mass
Orbital velocity ([tex]v[/tex]) = [tex]\sqrt{\frac{MG}{R} }[/tex]
where M= weight of galaxy
G= gravitational constatnt = [tex]6.674*10^-^1^1[/tex] (given)
R = distance from centre = [tex]41200[/tex] Light years (given)= [tex]4.12*9.5*10^1^6[/tex] km (1 ly= [tex]9.5*10^3[/tex] billion km)
v= orbital velocity = [tex]275[/tex] [tex]km/s[/tex] (given)
∴ According to the formula
[tex](2.75*10^2)^2[/tex] = [tex]\frac{M*6.674*10^-^1^1}{4.12*9.5*10^1^6}[/tex]
⇒ [tex]7.56*10^4*4.12*9.5*10^1^6=M*6.674*10^-^1^1[/tex] (cross multiplying and expanding)
⇒ [tex]29.59*10^2^1=M*6.674*10^-^1^1[/tex]
⇒ [tex]\frac{29.59*10^2^1*10^1^1}{6.674}=M[/tex]
⇒ [tex]4.434*10^3^2=M[/tex]
1 solar mass = [tex]1.989*10^3^0 kg[/tex]
⇒ Mass in solar mass =443.4 Solar mass
⇒ M = 443.4 Solar mass
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A 808 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 632 kg. The two cars lock up and slide together with a speed of 15.0 km/h. What was the speed of the first car just before the collision?
The speed of the first car just before the collision is 26.73 km/h.
What is conservation of momentum principle?When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.
The external force is not acting here, so the initial momentum is equal to the final momentum. For inelastic collision, final velocity is the common velocity for both the bodies.
m₁u₁ +m₂u₂ =(m₁ +m₂) v
Given a 808 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 632 kg. The two cars lock up and slide together with a speed of 15.0 km/h.
Second car is parked, so its velocity will be zero.
808 x u +632 x 0 = (808 +632 ) x 15
u = 26.73 km/h
Thus, the speed of the first car just before the collision is 26.73 km/h
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the 3kg object in figure is released from rest at height of 5m on curved frictionless ramp.at the foot of the ramp is a spring of force constant 400N/m.the object slide down the ramp and into the spring, compressing it a distance x before coming momentarily to rest. a) find x
b)describe the motion of the object (if any) after the block momentarily comes to rest?
The value of x given the data from the question is 0.86 m
How to determine the energyFrom the data given above, we can s determine the energy as follow:
Mass (m) = 3 KgHeight (h) = 5 mAcceleration due to gravity (g) = 9.8 m/s² Energy (E) = ?E = mgh
E = 3 × 9.8 × 5
E = 147 J
How to determine the value of xSpring constant (K) = 400 N/mEnergy (E) = 147 JExtention (e) = x = ?E = ½Ke²
147 = ½ × 400 × x²
147 = 200 × x²
Divide both side by 200
x² = 147 / 200
Take the square root of both sides
x = √( 147 / 200)
x = 0.86 m
Since the block came to rest, there is no motion
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What is the spring constant if 85.0J of energy is stored when a spring is compressed 0.400m?
A.1060N/m
B.212N/m
C.531N/m
D.425N/m
Answer:
212N/m
Explanation:
First of all, what is spring constant
As per Hooke's law, The restoring force applied by the spring is proportional to the change in length, and restoring force is directed towards the equilibrium position. The proportionality constant is called the spring constant and it is a measure of the stiffness of the spring. So as Hooke's law
F = -Kx
Where
F -> Restoring force
K-> Spring constant
x -> change in length
the negative sign indicates it is acting toward the center
Rearranging this for the Spring constant, Spring constant formula is given by
k = Fx
Unit of Spring Constant is Newton/meter
(the problem setup and question are below)
The gauge pressure at bottom of vaccine solution will be 16 kPa
Positive pressure is another name for gauge pressure. When a system's internal pressure exceeds that of its surroundings, it is said to be under positive pressure. Any leak that develops in the positively pressured system will therefore escape into the outside world. In contrast, a negative pressure chamber draws air into it.
Given As seen in the illustration, a syringe is held vertically. The container carries a 3 cm tall column of vaccine solution and has an open inner diameter of 1 cm. The needle contains a 2 cm column of vaccine solution and has an open inner diameter of 0.5 mm. At the needle's open end, the solution is exposed to the air. The vaccination solution has a density of 1200 kg/m3.
We have to find the gauge pressure at bottom of vaccine solution
Since the 5N force is applied to vaccine solution the pressure exerted will be much more
Hence the gauge pressure at bottom of vaccine solution will be 16 kPa
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One support of the behavioral perspective is how B. F. Skinner’s ideas have brought to light how the __________ and __________ influence personality.
One support of the behavioral perspective is how B. F. skinner’s ideas have brought to light how the environment and learning influence personality.
What is the behavioral perspective?The behavioral perspective on personality and individual variations places an emphasis on quantifiable behaviors influenced by the environment and the results that follow.
The theoretical aim of behaviorism is the prediction and control of behavior given its natural science methodology.
Skinner's theories have made it clear how learning and the environment affect personality, which is one argument in favor of the behavioral approach.
One support of the behavioral perspective is how B. F. skinner’s ideas have brought to light how the environment and learning influence personality.
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suppose the velocity of a particle is given by the equation, V=m+nt2, where m=10cms and n=2cm/s3.
A, find the change in velocity of the particle in the initial time interval b/n t₁=2sec and t₂=5sec
B, find the average acceleration in this time interval
C, find the instantaneous acceleration at time t₁=2sec
Hello!
Begin by plugging in the values for m and n. We get the equation for the velocity of the particle to be:
[tex]v(t) = 10 + 2t^2[/tex]
A.
To find the change in velocity over the interval (2s ≤ t ≤ 5s), we can simply find the difference in the velocities at these times.
[tex]\Delta v= v_f - v_i[/tex]
For this situation:
[tex]\Delta v = v(5) - v(2)[/tex]
Substitute these times for 't' into the equation and solve.
[tex]v(5) = 10 + 2(5^2) = 60 \frac{cm}{s}\\\\v(2) = 10 + 2(2^2) = 18 \frac{cm}{s}\\\\\Delta v = 60 - 18 = \boxed{42 \frac{cm}{s}}[/tex]
B.
To find the average acceleration, we must take the SLOPE of the velocity function over this interval using the slope formula:
[tex]a_{avg} = \frac{v_f - v_i}{\Delta t}[/tex]
Plug in the values for the particle's velocity at t = 2 s and 5 s that we solved for above.
[tex]a_{avg} = \frac{60- 18}{5 - 2}\\\\a_{avg} = \frac{42}{3} = \boxed{ 14 \frac{cm}{s^2}}[/tex]
C.
The instantaneous acceleration can be found by taking the derivative of the v(t) function using the power rule. Recall:
[tex]\frac{dy}{dx} x^n = nx^{n-1}[/tex]
Using this rule:
[tex]a(t) = v'(t) = 2(2t) = 4t[/tex]
Substituting in t = 2 s:
[tex]a(2) = 4(2) = \boxed{8 \frac{cm}{s^2}}[/tex]
Velocity (m/s)
50
40
30
20
10
0
A
Velocity vs Time
B
C
1 2 3 4 5 6 7 8 9 10
Time (s)
Use the information presented in the graph to answer
the questions.
Which segments show acceleration?
Which segment indicates that the object is slowing
down?
What is the velocity of segment B?
What is the acceleration of segment B?
Explanation:
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⠀
(a) The segment A shows acceleration as velocity increases with the increase in time.
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(b) The segment C shows the object is slowing down as the time increases in segment C, the velocity decreases and afterwards it comes to rest.
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(c) The velocity is segment B is 40m/s. And in the diagram there is no change in velocity.
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(d) The acceleration of segment B is zero. As there in no change in curve and it is moving with uniform velocity.
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[tex] \: [/tex]
Thank you!Answer:
1 A and C
2 C
3 40m/s
4 zero
Explanation:
