Answer:
The new force becomes 4 times the initial force.
Explanation:
The force of attraction or repulsion is given by the relation as follows :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
Where
d is the distance between the interacting charges
F is inversely proportional to the distance between charges.
If the distance is halved, d'=(d/2), new force is given by :
[tex]F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F[/tex]
So, the new force becomes 4 times the initial force.
"45 meters north" is an example of
Answer:
Displacement
Explanation:
The quantity 45m north is a typical example of displacement.
Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.
When we are specifying the displacement of a body, the direction must be indicated accurately. Therefore, the quantity given is displacementA heavy book is launched horizontally out a window from the first floor, a height, h, above the ground, with initial velocity, v0, and it hits the ground a horizontal distance X1 away from the window. Another book is similarly launched (same initial velocity) from the second floor window, a height 2h above the ground. Where does the second book land relative to the first book
Answer:
x₂ / x₁ = √2
Explanation:
To solve this exercise we can use the projectile launch ratios, let's find the time it takes for the second book to reach the ground
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
as the book is thrown horizontally v_{oy} = 0, when it reaches the ground its height is zero y= 0
0 = y₀ - ½ g t²
t = [tex]\sqrt{ \frac{2y_o}{ g} }[/tex]
t = \sqrt{ \frac{2 \ 2h}{ g} }
with this time we calculate the horizontal distance traveled
x = v₀ t
x₂ = v₀ [tex]\sqrt{ \frac{4h}{g} }[/tex]
now let's calculate the time it takes him to get to the floor when he leaves from the first floor
t =\sqrt{ \frac{2y_o}{ g} }
the horizontal distance traveled is
x₁ = v₀ [tex]\sqrt{ \frac{2h}{g} }[/tex]
therefore the difference in distance between the two runs is
Δx = x₂-x₁
Δx = v₀ \sqrt{ \frac{4h}{g} } - v₀ \sqrt{ \frac{2h}{g} }
Δx = v₀ \sqrt{ \frac{2h}{g} } √2
Δx =√2 x₁
the relationship between the two distances is
x₂ / x₁ = √2
A ratio is another name for a decimal true or false
Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer distance
Please show working
Distance = (speed) x (time)
Car A: Distance = (8 m/s) x (43 s) = 344 meters
Car B: Distance = (7 m/s) x (50 s) = 350 meters
350 meters is a longer distance than 344 meters.
Car-B traveled a longer distance than Car-A did.
Answer:
[tex]\boxed {\boxed {\sf Car \ B : 350 \ meters }}[/tex]
Explanation:
Distance is equal to the product of speed and time.
[tex]d=s*t[/tex]
1. Car A
Car A has a speed of 8 meters per second and travels for 43 seconds.
[tex]s= 8 \ m/s \\t= 43 \ s[/tex]
Substitute the values into the formula.
[tex]d= 8 \ m/s *43 \ s[/tex]
Multiply and note that the seconds will cancel out.
[tex]d= 8 \ m*43= 344 \ m[/tex]
2. Car B
Car B has a speed of 7 meters per second and travels for 50 seconds.
[tex]s= 7 \ m/s \\t= 50 \ s[/tex]
Substitute the values in and multiply.
[tex]d= 7 \ m/s * 50 \ s[/tex]
[tex]d= 7 \ m * 50 = 350 \ m[/tex]
350 meters is a longer distance than 344 meters, so Car B traveled the longer distance.
g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal radiation from the filament becomes visible. One bulb filament has a surface area of 30 mm2 and emits 60 W when operating. If the bulb filament has an emissivity of 0.8, what is the operating temperature of the filament
Answer:
2577 K
Explanation:
Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.
So, T = ⁴√(P/σεA)
Since P = 60 W, we substitute the vales of the variables into T. So,
T = ⁴√(P/σεA)
= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)
= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)
= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)
= ⁴√(0.00441 × 10¹⁶K⁴)
= 0.2577 × 10⁴ K
= 2577 K
What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified wavelength encounters the following natural or anthropogenic particles?
Slides 16-31, Lecture 2 ought to help - slides 19, 24, and 31 are key.
Wavelength O2 molecules Smoke particles Cloud droplets Rain droplets
(size 10^-10 m) (size 0.3 (μm) (20 μm) (size 3 mm)
550 nm
11 μm
1600 nm
1 cm
Solution :
1. Rayleigh scattering takes place when the particle size is smaller than the wavelength (λ).
2. Mie scattering takes place when particle size is nearly equal to the wavelength (λ).
3. Non-selective scatter takes place when particle size in greater than the wavelength (λ).
We have the sizes of different particles :
[tex]$O_2 \rightarrow 10^{10} \ m $[/tex]
Smoke particles [tex]$\rightarrow 3 \times 10^{-7} \ m$[/tex]
Cloud droplets [tex]$\rightarrow 2 \times 10^{-5} \ m$[/tex]
Rain droplets [tex]$\rightarrow 3 \times 10^{-3} \ m$[/tex]
Wavelength [tex]$ O_2 $[/tex] Smoke particles Cloud droplets Rain droplets
[tex]$10^{-10} \ m$[/tex] [tex]$ 3 \times 10^{-7} \ m$[/tex] [tex]$ 2 \times 10^{-5} \ m$[/tex] [tex]$ 3 \times 10^{-3} \ m$[/tex]
[tex]$5500 \times 10^{-4} \ m$[/tex] Rayleigh Non-selective Non-selective Non-selective
[tex]$11 \times 10^{-6} \ m $[/tex] Rayleigh Rayleigh Non-selective Non-selective
[tex]$1600 \times 10^{-10} \ m $[/tex] Rayleigh Non-selective Non-selective Non-selective
[tex]$10^{-2} \ m $[/tex] Rayleigh Rayleigh Rayleigh Mie
take a picture of an object in your house, describe the
energy stores and transfers that happen with it. You can be as imaginative as you wish
with the object (choose something unusual), but the stores you identify and transfers
that happen must be real.
pls give me ideas of what to take a photo of for this I'm really stuck :(
Pete applies a 10.9-Newton force to a 1.32-kg mug of root beer in order to accelerate it from rest over a distance of 1.25-m? How much work does Pete do on the mug of root beer?
Answer: 4 J
explanation:
is 0.8 kilograms bigger then 80 grams
Answer:
Yes
Explanation:
0.8 kilograms is equal to 800 grams
Answer:
Yes, 0.8 kilograms is greater than 80 grams
Explanation:
0.8 kilograms is equal to 800 grams and 80 grams is equal to 0.08 kilogrmas.
Sorry if I'm wrong, correct me.
On March 27, 2004, the United States successfully tested the hypersonic X-43A scramjet, which flew at Mach 7.0 (seven times the speed of sound) for 11 seconds. (A scramjet gets its oxygen directly from the air, rather than from fuel.) For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Swim competition. Part A At this rate, how many minutes would it take such a scramjet to carry passengers the approximately 5000 kmkm from San Francisco to New York? (Use 331 m/sm/s for the speed of soun
Answer:
Explanation:
Speed of sound = 331 m /s
speed of jet = 7 .00 Mach = 7 times speed of sound
= 7 x 331 = 2317 m /s
distance to be covered = 5000 x 1000 = 5 x 10⁶ m
Time taken = distance / speed of jet
= 5 x 10⁶ / 2317
= 2.158 x 10³ s
= 35.96 minutes .
Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.
a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?
Answer:
a) ω₁ = ω₂ = 3.7 rad/sec
b) Δθ₁ = Δθ₂ = 18.5 rad
c) d₁ = 14.5 m d₂ = 57.5 m
d) Fc1 = 273.9 N Fc2 = 1069.8 N
e) The boy near the outer edge.
Explanation:
a)
Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.However, linear speeds of points at different distances from the center, are different.Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:[tex]v = \omega*r (1)[/tex]
Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:[tex]\omega = \frac{v_{out} }{r_{out} } = \frac{11.5m/s}{3.14m} = 3.7 rad/sec (2)[/tex]
As we have already said, ωout = ωin = 3.7 rad/secb)
Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:[tex]\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)[/tex]⇒ Δθ₁ = Δθ₂ = 18.5 rad.
c)
The linear distance traveled by each child, will be related with the linear speed of them.Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:[tex]v_{inn} = \omega * r_{inn} = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)[/tex]
vout is a given of the problem ⇒ vout = 11. 5 m/s
Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:[tex]d_{inn} = v_{inn} * t = 2.9m/s* 5.0 s = 14.5 m (5)[/tex]
[tex]d_{out} = v_{out} * t = 11.5 m/s* 5.0 s = 57.5 m (6)[/tex]
d)
The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:[tex]F_{c} = m*\frac{v^{2}}{r} (7)[/tex]
Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:[tex]F_{cin} = m*\frac{v_{in}^{2}}{r_{in}} = 25.4 kg* \frac{(2.9m/s)^{2} }{0.78m} = 273.9 N (8)[/tex]
In the same way, we get Fcout (the force on the boy near the outer edge):[tex]F_{cout} = m*\frac{v_{out}^{2}}{r_{out}} = 25.4 kg* \frac{(11.5m/s)^{2} }{3.14m} = 1069.8 N (9)[/tex]
e)
The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.The maximum friction force is given by the product of the coefficient of static friction times the normal force.Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.As both boys have the same mass, the normal force is also equal.This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:[tex]F_{frs} = \mu_{s} * m* g (10)[/tex]If this force is greater than the centripetal force, the boy will be able to hold on.So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.Two objects travel the same distance. The one that is moving faster will:
Take more time to go the distance
Take less time to go the same distance
Take the same time as the slower object
None of the above
Answer: take less time to go the same distance
Explanation:
Because if it is going faster let’s say mph 60 mph is 60 miles per hour if you are going 40 miles per hour it will take you longer to get to your destination.
what type of waves can only travel through a medium?
Answer:
Mechanical waves
Explanation:
Mechanical waves are the waves that can travel only through a medium. Mechanical waves are disturbance of matter and require medium to transfer the energy. There are three types of mechanical waves that include transverse wave, longitudinal wave and surface wave.
Some of the examples of mechanical waves are sound waves and seismic waves etcetera.
Hence, the correct answer is "Mechanical waves".
Which landform is produced at location E where the Mississippi River enters the Gulf of
Mexico?
a delta a drumlin an out wash an escarpment
Answer:
a delta
Explanation:
The landform produced at the location E where the Mississippi River enters the Gulf of Mexico is a delta.
A delta is a depositional landform where a smaller body of water enters into a larger one.
The Gulf of Mexico contains a larger body of water and as the Mississippi river enters into it, it splits up into many distributaries.
So, this feature is a delta.
In what way did the cloud model resemble the Bohr model
Answer:
While Bohr's atomic model hypothesizes that electrons move in particular energy levels around the nucleus, the electron cloud model suggests that electrons move in an unpredictable pattern but are more likely to be in certain regions than others.
Explanation:
two identical balls are rolling down a hill ball 2 is rolling faster than ball 1 which ball has more kinetic energy
A mass m is gently placed on the end of a freely hanging spring. The mass then falls 33 cm before it stops and begins to rise. What is the frequency of the oscillation
Answer:
Explanation:
The mass falls by .33 m before it begins to rise . At that point loss of potential energy is equal to gain of elastic energy .
1/2 k x² = mgx
.5 x k x .33² = m x 9.8 x .33
k / m = 59.4
frequency of oscillation = [tex]\frac{1}{2\pi} \times\sqrt{\frac{k}{m} }[/tex]
= [tex]\frac{1}{2\pi} \times\sqrt{59.4}[/tex]
= 1.22 per second .
As a person pushes a box across a floor, the energy from the person's moving arm is transferred to the box, and the box and the floor becomes warm. During the process, what happens to energy
Answer:
isnt heat transfer
Explanation:
sorry if im wrong
A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 W. The remainder of the day she is sitting in class, studying or resting; during these activities, she expends energy at a rate of 100 W. If she expends a total of 1.1 x 10^7 J of energy in a 24 hour day, how much of the day did she spend walking
The time of the day she spent walking is equal to 3.70 hrs.
What is power?Power can be explained as the rate of doing work in unit time. The SI unit of measurement of power is J/s or Watt (W). Power can be described as a time based quantity. The mathematical expression for power can be represented as mentioned below.
Power = work/time
P = W/t
Given, the energy spends part of her day walking, Ew = 280 W
The energy is spent by sitting in the class, Es = 100 W
The total energy spends, Et = 1.1 × 10⁷J
[tex]E_w \times t + E_s(24\times 60\times 60-t)= 1.1 \times 10^7J[/tex]
[tex]280 \times t + 100(24\times 60\times 60-t)= 1.1 \times 10^7[/tex]
280 t + 0.86 × 10⁷ - 100 t = 1.1 × 10⁷
180 t = 0.24 × 10⁷
t = 0.24 × 10⁷/180 × 3600
t = 3.70 hr
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An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at rest. After the collision, the 8.00 kg object moves south at 3.90 m/s. (a) What is the velocity of the 10.0 kg object after the collision
Answer:
9.3m/s
Explanation:
Based on the law of conservation of momentum
Sum of momentum before collision = sum of momentum after collision
m1u1 +m2u2 = m1v1+m2v2
m1 = 8kg
u1 = 15.4m/s
m2 = 10kg
u2 = 0m/s(at rest)
v1 = 3.9m/s
Required
v2.
Substitute
8(15.4)+10(0) = 8(3.9)+10v2
123.2=31.2+10v2
123.2-31.2 = 10v2
92 = 10v2
v2 = 92/10
v2 = 9.2m/s
Hence the velocity of the 10.0 kg object after the collision is 9.2m/s
Magnification of lens is 1. What does it mean?
Answer:
It means when you look into the lens your vision magnifies by x1
Explanation:
A student is driving through a mountainous region where the road is at some times flat, at some times inclined upward, and at some time inclined downward. The student maintains a speed of 20 m/s on the roadway, but is required to make an emergency stop on the three sepearte occasions. On levels roadway, it takes 25 m to stop. On a downward-sloping roadway, it takes 40 m to stop. On an upward-sloping roadway, it takes 18 m to stop. Explain why the stopping distances are different. (Focus answer using work and energy, other concepts may be used as well but be sure work and energy are included.)
Answer:
Explanation:
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more .
Hence displacement is more in the downward slopping.
What is Displacement?Displacement is defined as the change in position of an object. It is a vector quantity and has a direction and magnitude.
It is frictional force of the ground that helps in bringing the vehicle to stop . In the process of stopping , negative work is done on the car by friction force to overcome its kinetic energy .
At levelled road , for stoppage
Kinetic energy of vehicle = Work done by frictional force . = friction force x displacement .
At upward slopping road , gravitational force acting downward also helps the vehicle to stop do friction has to do less work .
At upward inclined road , for stoppage
Kinetic energy of vehicle = Work done by frictional force + work done by gravitational force = (friction force + gravitational force ) x displacement .
Hence displacement is less .
At downward slopping road , friction has to do more work because friction has to do work against gravitational force acting downwards wards and kinetic energy of the vehicle also .
At downward inclined road , for stoppage
Kinetic energy of vehicle + work done by gravitational force = Work done by frictional force = friction force x displacement .
Hence displacement is more in the downward slopping.
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if you watch football let me know who you think is going to win super bowl 55 and what do you think the score going to be Kansas city chiefs or tampa bay buccaneers
Answer:
I think the bucs are gonna win because Tom Brady is on their team and it's rigged
but maybe I'm just thinking negatively lol
a wooden block is cut into two pieces, one with three times the mass of the other. a depression is made in both faces of the cut so that a fire cracker can be placed in it and the block is reassembled. the reassembled block is set on rough surface and the fuse is lit. when the fire cracker explodes, the two blocks separate. what is the ratio of distances traveled by blocks?
Answer:
1/9
Explanation:
Let A denote the bigger piece and let B denote the smaller piece.
We are told that one with three times the mass of the other.
Therefore, we have;
M_a = 3M_b
Firecracker is placed in the block and it explodes and thus, momentum is conserved.
Thus;
V_ai = V_bi = 0
Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.
Since initial momentum equals final momentum, we have;
P_i = P_f
Thus;
0 = (M_a × V_af) + (M_b × V_bf)
Since M_a = 3M_b, we have;
(3M_b × V_af) + (M_b × Vbf) = 0
Making V_af the subject, we have;
V_af = -⅓V_bf
The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;
W_f = -½M_b•(v_bf)²
Now, let's express the work is in terms of the force and the distance.
Thus;
W_f = F_f × Δx × cos 180°
Frictional force is also expressed as μmg
Thus;
W_f = -μM_b × g × Δx
Earlier, we saw that;
W_f = -½M_b•(v_bf)²
Thus;
-½M_b•(v_bf)²= -μM_b × g × Δx
Δx = (v_bf)²/2μg
Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b
Thus;
Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)
Δx_a/Δx_b = ((v_af)²/((v_bf)²)
Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²
Δx_a/Δx_b = 1/9
At an airport, two business partners both walk at 1.5 m/sm/s from the gate to the main terminal, one on a moving sidewalk and the other on the floor next to it. The partner on the moving sidewalk gets to the end in 60 ss, and the partner on the floor reaches the end of the sidewalk in 90s.
Required:
What is the speed of the sidewalk in the Earth reference frame?
Answer:
[tex]v=0.8m/s[/tex]
Explanation:
From the question we are told that
Distance [tex]d=1.5m/sm/s[/tex]
Time [tex]t_1=60s[/tex]
Time [tex]t_2=90s[/tex]
Generally the the equation for the distance traveled is mathematically given as
[tex]d=vt[/tex]
[tex]d=1.5*90[/tex]
[tex]d=138m[/tex]
Generally equation for speed of side walk is mathematically given as
[tex]d=(v+u)t[/tex]
[tex]v=\frac{d}{t}-u[/tex]
[tex]v=\frac{138}{60}-1.5[/tex]
[tex]v=0.8m/s[/tex]
A truck travelling down the street suddenly brakes, applying a 14 N force over 3.5 seconds. What was the impulse over the given time.
Answer:
49 Ns
Explanation:
Given data
Force= 14N
time = 3.5seconds
Applying the expression for impulse
P= Ft
substitute
P=14*3.5
P=49 Ns
Hence the impulse is 49 Ns
Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 554 N/C. If the particles are free to move, what are their speeds (in m/s) after 51.6 ns
Answer:
the speed of electron is 5.021 x 10⁶ m/s
the speed of proton is 2733.91 m/s
Explanation:
Given;
magnitude of electric field, E = 554 N/C
charge of the particles, Q = 1.6 x 10⁻¹⁹ C
time of motion, t = 51.6 ns = 51.6 x 10⁻⁹ s
The force experienced by each particle is calculated as;
F = EQ
F = (554)(1.6 x 10⁻¹⁹)
F = 8.864 x 10⁻¹⁷ N
The speed of the particles are calculated as;
[tex]F = ma\\\\F = \frac{mv}{t} \\\\v = \frac{Ft}{m} \\\\v_e = \frac{Ft}{m_e}\\\\v_e = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{9.11 \times \ 10^{-31}}\\\\v_e = 5.021 \times 10^{6} \ m/s[/tex]
[tex]v_p = \frac{Ft}{m_p}\\\\v_p = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{1.673 \times \ 10^{-27}}\\\\v_p = 2733.91 \ m/s[/tex]
please help!!!!!! 26 points
Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a
second identical ball (Ball B). After the collision Ball A continues to move
in the same direction at 2 m/s. What is the magnitude of the velocity for
Ball B after the collision?
Before Collision:
10 m/s
A
After Collision:
2 m/s
O
Answer:
6m/s
Explanation:
Using the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the momentum after collision.
Using the expression
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity after collision
Substitute the given values in the formula
0.35(10)+0.35(2) = (0.35+0.35)v
3.5+0.7 = 0.7v
4.2 = 0.7v
v = 4.2/0.7
v = 6m/s
Hence the magnitude of the velocity for Ball B after the collision is 6m/s
It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball-bearings. Consider a flywheel made of iron, with a density of 7800 kg/m^3 , in the shape of a uniform disk with a thickness of 11.3 cm.
Required:
a. What would the diameter of such a disk need to be if it is to store an amount of kinetic energy of 14.1 MJ when spinning at an angular velocity of 93.0 rpm about an axis perpendicular to the disk at its center?
b. What would be the centripetal acceleration of a point on its rim when spinning at this rate?
Answer:
Explanation:
kinetic energy = 14.1 MJ = 14.1 x 10⁶ J
Let radius of flywheel be r .
volume of flywheel = π r² x t where t is thickness
= 3.14 x r² x .113 m³
= .04 r² m³
mass = volume x density
= .04 r² x 7800 = 312.73 r²kg
moment of inertia I = 1 / 2 mass x radius²
= .5 x 312.73 r² x r²
= 156.37 r⁴ kg m²
angular velocity ω = 2π x 93/60
= 9.734 rad /s
kinetic energy = 1/2 Iω² where ω is angular velocity
= .5 x 156.37 r⁴ x 9.734²
= 7408.08 r⁴
Given
7408.08 r⁴ = 14.1 x 10⁶
r⁴ = .19 x 10⁴
r = .66 x 10
= 6.60 m .
Diameter = 13.2 m
b )
centripetal acceleration of a point on its rim = ω² r
= 9.734² x 6.6
= 625.35 m /s²
