Answer:
Vb = 0.334 m/s
Va = -1.265 m/s
Vc = 1.424 m/s
Explanation:
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Initial momentum = 255(2) – 242.5(1.5) = 146.25
Final momentum = 255Va + 290Vb + 242.5 Vc = 146.25
Vb - Va = 0.8(2) = 1.6
Vc - Vb = 0.8(1.5) = 1.2
Va = Vb -1.6
Vc = Vb + 1.2
255(Vb -1.6) + 290Vb + 242.5(Vb + 1.2) = 146.25
255 Vb – 408 + 290 Vb + 242.5 Vb + 291 = 146.25
787.5 Vb = 263.25
Vb = 0.334 m/s
Va = Vb -1.6 = 0.334 – 1.6 = -1.265 m/s
Vc = Vb + 1.2 = 0.224 + 1.2 = 1.424 m/s
if you ride a bike at 2 km/h and travel a total distance of 20km, how long willnit takr ( in second) you to teach your destination
Time = (20 km) x (1 hr/2 km) x (3,600 sec/hr)
Time = (20 x 1 x 3,600 km-hr-sec) / (2 km-hr)
Time = (20 x 1 x 3,600 / 2) (km-hr-sec / km-hr)
Time = 36,000 seconds
(That's 10 hours.)
Why is fluorine more chemically reactive than nitrogen
Answer:
Me encanta el pollo, de todos modos, un día que es hoy, por cierto, fui a la tierra de los pollos y caminé en México, ¿sabías que es mi sueño? ¿Olvidé que mi nombre es jemma y me gusta caminar? párrafo así que nos vemos
Explanation:
Answer:
Fluorine needs to gain only 1 electron to have a full octet, while nitrogen needs to gain 3.
Waves are ?
that can travel through matter.
Answer:
A wave can be thought of as a disturbance or oscillation that travels through space-time, accompanied by a transfer of energy. The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.
Gravity does not actually "pull" objects at
all, rather gravity is a result of the warping.
or curving, of
A Stars and planets
B. The solar system
C Spacetime
D. Any object
What is the gravitational potential energy of an object that has a mass of 8 kg and is 11.2 meters above Earth? Round your answer the nearest whole number.
A. 878 J
B. 30 J
C. 680 J
A ball of mass m=10g, carrying a charge q =-20μe is suspended from a string of length L= 0.8m above a horizontal uniformly charged infinite plane sheet of charge density σ = 4μe/m^2. The ball is displaced from the vertical by an angle and allowed to swing from rest.
Required:
a. Obtain the equations of motion of the charged ball based on Newtonian laws of motion.
b. Assume the displaced angle θ is small and simplify the results obtained in part (a) to obtain the frequency of oscillations of the charged ball.
Answer:
a) [tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] = [tex]\frac{d^2 \theta}{d t^2}[/tex]
b) f = 2π [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]
Explanation:
a) To have the equations of motion, let's use Newton's second law.
Let's set a reference system where the x-axis is parallel to the path and the y-axis is in the direction of tension of the rope.
For this reference system the tension is in the direction of the y axis, we must decompose the weight and the electrical force.
Let's use trigonometry for the weight that is in the vertical direction down
sin θ = Wₓ / W
cos θ = W_y / w
Wₓ = W sin θ
W_y = W cos θ
we repeat for the electric force that is vertical upwards
F_{ex} = F_e sin θ
F_{ey} = F_e cos θ
the electric force is
F_e = q E
where the field created by an infinite plate is
E = [tex]\frac{ \sigma}{2 \epsilon_o}[/tex]
let's write Newton's second law
Y axis
T - W_y = 0
T = W cos θ
X axis
F_{ex} - Wₓ = m a (1)
we use that the acceleration is related to the position
a = dv / dt
v = dx / dt
where x is the displacement in the arc of the curve
substituting
a = d² x /dt²
we substitute in 1
q E sin θ - mg sin θ = m [tex]\frac{d^2 x}{dt^2}[/tex]
we have angular (tea) and linear (x) variables, if we remember that angles must be measured in radians
θ = x / R
x = R θ
we substitute
sin θ (q E - mg) = m \frac{d^2 R \ theta}{dt^2}
[tex]- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{sin \theta}{R }[/tex] = [tex]\frac{d^2 \theta}{d t^2}[/tex]
this is the equation of motion of the system
b) for small oscillations
sin θ = θ
therefore the solution is simple harmonic
θ = θ₀ cos (wt + Ф)
if derived twice, we substitute
- ( g - \frac{q}{m} \frac{\sigma }{ 2 \epsilon_o} ) \frac{\theta}{R } θ₀ cos (wt + Ф) = -w² θ₀ cos (wt + Ф)
w² = [tex]\frac{g}{R}[/tex] - [tex]\frac{q}{m} \frac{ \sigma }{2 \epsilon_o} \frac{1}{R}[/tex]
angular velocity is related to frequency
w = 2π f
f = 2π / w
f = 2π/w
f = 2π [tex]\sqrt{ \frac{R}{ g - \frac{q}{m} \frac{\sigma }{2 \epsilon_o} } }[/tex]
PLEASE HELP!!!
What are some compound machines that are used today?
Answer:
Examples of compound machines include bicycles, cars, scissors, and fishing rods with reels. Compound machines generally have lower efficiency but greater mechanical advantage than simple machines
Brilianst
Answer:
screws, inclined planes , etc
Explanation:
no explanation needed
Explain which energies work together to bake the bread
and cook the eggs.
Answer:
Thermal Energy
Explanation:
The energies that work together to bake the bread and cook the eggs is the thermal energy. Thermal energy is basically heat energy, which makes the food warmer.
Thermal energy and chemical energy work together to bake the bread and cook the eggs.
What is thermal energy?The energy present in a system that determines its temperature is referred to as thermal energy. Thermal energy flows as heat. Thermodynamics is a whole field of physics that studies how heat is transmitted across various systems and how work is done in the process.
What is chemical energy?Chemical energy is described as: the power that is kept in chemical compound bonds (molecules and atoms). It is released during the chemical process, which is referred to as an exothermic reaction, which mostly generates heat as a byproduct.
When the bread is baked or the eggs are cooked, thermal energy is provided from outside and chemical energy inside the molecules of bread or eggs come to play. So, these two energies work together to bake the bread and cook the eggs.
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Define speed and what is it’s SI unit.
A screw-jack used to lift a bus is a
A) first order lever
B) second order lever
C) pulley
D) screw
Answer:
c
Explanation:
The local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bumper car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.
Calculate the time required for bumper car A to travel the 10 m to collide with bumper car B.
Calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s.
Is the direction of motion for bumper car A following the collision with bumper car B to the right, to the left, or is bumper car A at rest?
Is the collision elastic? Justify your answer.
Answer:
a. 20 s
b. 0 m/s
c. right
d.no its inelastic because when the car b was at rest and a was coming in at it, since b had no force what so ever car a swept it away with it moving to the right
Explanation:
im not sure though
By applying conservation of linear momentum, the answers are:
1. Time = 2 s
2. 3 m/s
3. same direction
4. Inelastic collision
COLLISIONThere are for types of collision. They are;
Elastic CollisionPerfectly elastic collisionInelastic collisionPerfectly Inelastic collisionGiven that a local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bumper car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.
1. The time required for bumper car A to travel the 10 m to collide with bumper car B can be calculated by using first equation of linear motion.
V = U + at
Where
V = 4 m/s
U = 2 m/s
a = 1 m/[tex]s^{2}[/tex]
Substitute all the parameters into the formula
4 = 2 + t
t = 4 - 2
t = 2s
2. To calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s, we will apply conservation of linear momentum
[tex]m_{1}u_{1}[/tex] = [tex]m_{1}v_{1}[/tex] + [tex]m_{2}v_{2}[/tex]
300 x 4 = 300V + 200 x 3
1200 = 300V + 300
300V = 1200 - 300
300V = 900
V = 900/300
V = 3 m/s
3. Since the final velocity of car A is positive, the direction of motion for bumper car A follows the collision with bumper car B to the right.
4. Since the both move at the same velocity, the collision inelastic.
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Problem 05.086 - Water pumped from a lower reservoir to a higher reservoir Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 21 kW of useful mechanical power to the water. The free surface of the upper reservoir is 45 m higher than the surface of the lower reservoir. If the flow rate of water is measured to be 0.03 m3/s, determine the irreversible head loss of the system and the lost mechanical power during this process. Take the density of water to be 1000 kg/m3. The irreversible head loss of the system is 26.35 m. The lost mechanical power in this process is kW.
Answer:
Lost Mechanical Power = 7.7565 KW
Head Loss = 26.35 m
Explanation:
First, we will find the useful mechanical power used to transport water to the higher reservoir:
[tex]P_{useful} = \rho ghV[/tex]
where,
P_useful = Useful mechanical Power = ?
ρ = density of water = 1000 kg/m³
g = acceleration due to gravity = 9.81 m/s²
h = height = 45 m
V = Volume flow rate = 0.03 m³/s
Therefore,
[tex]P_{useful} = (1000\ kg/m^3)(9.81\ m/s^2)(45\ m)(0.03\ m^3/s)\\P_{useful} = 13243.5\ W = 13.2435\ KW[/tex]
Now, the lost mechanical power will be:
[tex]Lost\ Mechanical\ Power = Total\ Mechanical\ Power - Useful\ power\\Lost\ Mechanical\ Power = 21\ KW - 13.2435\ KW\\[/tex]
Lost Mechanical Power = 7.7565 KW
Now, for the head loss:
[tex]Lost\ Mechanical\ Power = \rho g(Head\ Loss)V\\Head\ Loss = \frac{Lost\ Mechanical\ Power}{\rho gV} \\\\Head\ Loss = \frac{7756.5\ W}{(1000\ kg/m^3)(9.81\ m/s^2)(0.03\ m^3/s)} \\[/tex]
Head Loss = 26.35 m
A city near the ocean experiences thunderstorms on a frequent basis. Which of these is the BEST explanation for this?
A.
The interactions of the waves with the landmass change the air pressure to cause thunderstorms.
B.
The cold fronts that form only over coastal areas cause thunderstorms.
C.
The amount of sunshine received by coastal areas causes thunderstorms.
D.
The uneven heating of land and water in coastal areas creates fronts that cause thunderstorms.
Answer:
d
Explanation:
The uneven heating of land and water in coastal areas creates fronts that cause thunderstorms. So, option D is correct.
What is meant by heat energy ?Heat energy of a system is defined as the energy transferred between the molecules due to the temperature difference between them. Heat energy is the energy responsible for the temperature of the molecules.
Here,
A city near the ocean experiences thunderstorms on a frequent basis. This is because,
The atmosphere near to the land is highly humid and warm and thus creates a relatively low pressure in the land. In order to balance the pressure variation the air from the oceans make movements towards the land, and as a result, sea breeze are formed. During the evenings when the water in the ocean is more warm and when land begins to cool, the air from land rushes towards the sea, in order to balance the pressure variation occurred here. This makes the air above the ocean highly humid and thus would lead to the formation of thunderstorms.
Hence,
The uneven heating of land and water in coastal areas creates fronts that cause thunderstorms.
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When you are doing either a chest or a bounce pass where are your thumbs at the end of the
motion?
Answer:
yes
Explanation:
When an object is raised vertically, work is done against __________. What one word completes the sentence?
please help quickly!!
When an object is raised vertically, work is done against _gravitational force _________
What is gravitational force ?The force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface is called gravitational force .
On every body there is a force acting named gravitational force which attracts the body downward .
If a object / body is thrown up it always comes downward if only gravitational force is acting on it which act due to earth's gravity .
When an object is raised vertically, work is done against _gravitational force _________
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HELP
HELP
SORRY TO BEG :/
HELP ASAP :) PLZ
Answer:
It's b because he discovered galaxies .
What variables affect density weight,conductivity,color,volume, or mass
Answer:
Density is affected by volume and mass.
Explanation:
Density is defined as the quantity of mass per unit of volume, or expressed mathematically, d = m/v.
During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke's law with a spring constant of 112 N/m. If the hose is stretched by 4.70 m and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length
Answer:
1237 J
Explanation:
The work done by the hose on the balloon is the work done by a spring which is
W = 1/2k(x₀² - x₁²) where k = spring constant = 112 N/m, x₀ = 4.70 m and x₁ = 0 m.
Substituting the values of the variables into the equation, we have
W = 1/2k(x₀² - x₁²)
W = 1/2 × 112 N/m((4.70 m)² - (0 m)²)
W = 56 N/m(22.09 m² - 0 m²)
W = 56 N/m(22.09 m²)
W = 1237.04 J
W ≅ 1237 J
PLEASE HELP ME WITH THIS ONE QUESTION
Which pendulum would have the highest frequency?
1) 1 kg mass and 0.5-meter length
2) 0.5 kg mass and 0.75-meter length
3) 0.5 kg mass and 1-meter length
4) 1 kg mass and 0.75-meter length
Answer:
B is the Answer
Follow me please
Mark brainliest
Answer:
The answer is A
Explanation:
This is because f = 1/T,
or frequency equals 1 over period, T
Since the highest T would be the answer with the greatest length and mass does not matter the answer would be c IF they were asking for the highest period.
F is found by 1 over T so the greatest F would be the smallest number, Hence answer A
Please Mark me brainiest
3. What is the power of a derby race car that uses a 50 N force to travel 20 meters in 10 seconds?
A. 1000 W
B. 100 W
C. 80 W
A dump truck, whose bed is made of steel, holds an old steel watering trough. The bed of the truck is slowly raised until the trough begins to slide. For dry steel to steel μs= 0.80, μk= 0.60. What is the acceleration of the trough as it slides down the truck bed? Express your answer with the appropriate units.
Answer:
a = 1,538 m / s²
Explanation:
Let's use Newton's second law, let's set a reference system where the x-axis is parallel to the sloping floor of the truck and the positive direction is in the direction of movement of the trough, for this case the weight is the only force to decompose
sin θ = Wₓx / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
Y axis
N -W_y = 0
N = mg cos θ
X axis
Wₓ - fr = m a
the friction force has the expression
fr = μ N
There are values of the friction coefficient (μ_s) one for when the movement has not started and it takes a smaller value for when the bodies are moving.
In this case we first find the angle for which the movement begins, in this part we use the static coefficient and the acceleration is zero
Wₓ - μ_s N = 0
m g sin θ = μ_s mg cos θ
tan θ = μ_s
θ = tan⁻¹ μ_s
we calculate
θ = tan⁻¹ 0.8
θ = 38.7º
For this angle, how the trough begins to move, the coefficient is reduced to the dynamics coefficient (μ_k) and the acceleration is different from zero.
we substitute
mg sin θ - μ_k mg cos θ = m a
a = g (sin θ - μ_k cos θ)
let's calculate
a = 9.8 (sin 38.7 - 0.6 cos 38.7)
a = 1,538 m / s²
The main difference between a chest and a bounce pass is what?
Answer: The main difference between the three is the mode of transmission. The chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate. Finally, the overhead pass is projected high in the air to avoid defenders.
Explanation:
Answer:
The main difference is: the chest pass is straight through the air towards your teammate. While the bounce pass is directed toward the ground and then at your teammate.
The secondary main difference is the amount of power from the ball recived from the person reciving
Please answer :>
40 POINTS
Answer:
rotates
Explanation:
I'm so bored
yrfgggghhgghhyuj
TAKING 100 PONTS!!! PLEASE SOMEONE HELP.JUST LOOK THE PICTURE
Explanation:
oh no i hope i dont fall and the i agree with the other dude
Answer: Okay, thanks I really appreciate the generosity.
Object 1 has a momentum of 10 kg m/s and Object 2 has a momentum of 25 kg m/s. Will it be easier to change the direction of movement of Object 1 or of Object 2?
Answer:
I think its object 1
Explanation:
Because the object that has more weight has a greater momentum and the lightest object that has a less momentum will be easier to change because its lighter.
Missy Diwater, the former platform diver for the Ringling Brother's Circus had a kinetic energy of 15,000 J just prior to hitting the bucket of water. If Missy's mass is 50 kg, then what is her speed?
Answer:
24.5 m/s
Explanation:
KE=1/2mv^2
15000=1/2(50)v^2
30000=(50)v^2
600=v^2
sqrt600=v
v=24.5 m/s!!
Planets don't collide into
the sun because they
A. Are moving
B. Have too much mass
C. Have their own gravity
D. Are more attracted to each other
What is the wavelength of this wave
[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Waves.
Wavelength is the distance between two consecutive crest or trough.
hence, here the distance is 10cm
So the wavelength is 10cm
===> 10 cm
student c precise or accurate darts
Answer:
i don't get it
Explanation:
A ball is thrown straight up with an initial velocity of 128 ft/sec, so that its height (in feet) after t sec is given by s(t) = 128t-16t2. (a) What is the average velocity of the ball over the following time intervals? [3,4] ft/sec [3,3.5] ft/sec [3,3.1] ft/sec (b) What is the instantaneous velocity at time t = 3? ft/sec (c) What is the instantaneous velocity at time t = 6? ft/sec Is the ball rising or falling at this time? rising falling (d) When will the ball will hit the ground? t = sec
Answer:
one sec let me think
Explanation:
(a)The average velocity of the ball over the following time intervals will be [3,4] ft/sec.
(b)The instantaneous velocity at time t = 3 will be32 ft/sec.
(c)The instantaneous velocity at time t = 6 will be -64 ft/sec.
(d)The ball will hit the ground at 13.4 sec.
What is velocity?The change of displacement with respect to time is defined as the velocity. velocity is a vector quantity. it is a time-based component.
The given data in the question will be ,
u is the initial velocity by which ball thrown=128 ft/sec.
V₃ is the instantaneous velocity at time t=3 sec.
V₆ is the instantaneous velocity at time t=6 sec.
t is the time when ball hits the ground=?
(a) Given equation for the displacement
s(t) =128t-16t² (on differenting got the velocity )
v(t) = 128-32t
Time when velocity is zero will be
[tex]\rm{ t=\frac{128}{32}[/tex]
[tex]\rm{ t=4 sec[/tex]
If the velocity got in the equation is 128 and 32 ft /sec. it can be only when the average velocity is [3,4] ft/sec .
Hence the average velocity obtained from the problem will be [3,4] ft/sec
(b)
s(t) =128t-16t² (on differenting got the velocity )
v(t) = 128-32t
At time( t=3 sec)
v(3) = 128-32×3
v(3) =32 m/sec.
Hence the instantaneous velocity at time t = 3 will be32 ft/sec.
(c)
s(t) =128t-16t² (on differenting got the velocity )
v(t) = 128-32t
At time( t=6 sec)
v(6) = 128-32×6
v(6) = -64 m/sec.
Hence the instantaneous velocity at time t = 6 will be -64 ft/sec.
(d)
According to Newtons third equation of motion we got
v=u+gt
If the body returens from a certain height at max height its velocity must be zero; ( u=0)
[tex]\rm t=\frac{(v-u)}{g} \\\\\ \rm t=\frac{(128-0)}{9.81}\\\\\rm t=13.04 sec.[/tex]
Hence the ball will hit the ground at 13.4 sec.
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