(a). The equivalent resistance of a series circuit is the SUM of the individual resistances in the circuit.
(4.75Ω + 15Ω + 21Ω ) = 39.75Ω
(b). Current = (voltage) / (resistance)
Current = (12 v) / (39.75Ω)
Current = 0.302 Amperes (rounded to the nearest MilliAmpere))
What x rays travel at the speed of
Which word best completes the sentence?
Select the word from the drop-down menu
He is quite
Choose...
despite never having left his smalL TOWEN
Answer:
it’s cosmopolitan
Explanation:
k12
6xy from -12xy
please give me a answer this question
6 floors down from 12 floors underground = 18 floors underground.
6 degrees colder than 12 degrees below zero = 18 degrees below zero
6 brown cows taken away from -12 brown cows = -18 brown cows
6 cars sold from a dealer that 12 cars were stolen from = 18 cars gone
6xy taken away from -12xy = -18xy
Please help me with this 29 points
Answer:
)Give the definition of poverty line as defined by the World Bank.
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
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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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
3. If a spring extends by 3 cm when a 4 N weight is suspended from it, find the extension
when the weight is changed to
(a) 8 N
(b) 10 N
(c) 14 N
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
A screw-jack used to lift a bus is a
A) first order lever
B) second order lever
C) pulley
D) screw
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
when is an object considered to be in motion
Answer:
An object is considered to be in motion only when its position changes over time with reference to a point which will be known as orgin
Please answer :>
40 POINTS
Answer:
rotates
Explanation:
I'm so bored
yrfgggghhgghhyuj
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
What are characteristics of wide-angle lenses?
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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You are driving your car on a very cold late Fall day. You clear a turn and see a couple of pedestrians standing at the cross walk. They are eager to cross the road and to get into the warmth of their apartment as soon as possible. You have two options: continue driving your car as you were without lowering your speed and drive right by the pedestrians OR slow down, stop right at the crosswalk, and yield to the pedestrians. Although by Virginia law the choice is clear, what about Physics laws? Which scenario (passing by or slowing down and stopping at the crosswalk to yield) will minimize the time the pedestrians are out in the cold freezing before they can cross the road?
Make the following assumptions in your argument. Before you noticed the pedestrians, you are moving with a constant velocity v=22 miles/hour. The distance at which you noticed the pedestrians is D=23 meters. Write down a symbolic expression for the amount of time, tpass , the pedestrians will have to wait till they cross the road if you simply drive by without slowing down or speeding up.
Write down a symbolic expression for the amount of time, tstop, the pedestrians will have to wait till they cross the road if you slow down, come to a complete stop at the crosswalk and yield to the pedestrians.
Answer:
t_pass = 2.34 m
t_stop = 4.68 s
Thus, for the car passing at constant speed the pedestrian will have to wait less.
Explanation:
If the car is moving with constant speed, then the time taken by it will be given as:
[tex]t_{pass} = \frac{D}{v}[/tex]
where,
t_pass = time taken = ?
D = Distance covered = 23 m
v = constant speed = (22 mi/h)(1609.34 m/1 mi)(1 h/3600 s) = 9.84 m/s
Therefore,
[tex]t_{pass} = \frac{23\ m}{9.84\ m/s} \\[/tex]
t_pass = 2.34 m
Now, for the time to stop the car, we will use third equation of motion to get the acceleration first:
[tex]2as = v_{f}^{2} - v_{i}^2\\a = \frac{v_{f}^{2} - v_{i}^2}{2D}\\\\a = \frac{(0\ m/s)^{2}-(9.84\ m/s)^2}{(2)(23\ m)}\\\\a = -2.1\ m/s^2[/tex]
Now, for the passing time we use first equation of motion:
[tex]v_{f} = v_{i} + at_{stop}\\t_{stop} = \frac{v_{f}-v_{i}}{a}\\\\t_{stop} = \frac{0\ m/s - 9.84\ m/s}{-2.1\ m/s^2}[/tex]
t_stop = 4.68 s
Constant velocity is the velocity which covers the same distance for each interval of the time.
The time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.
What is constant velocity?Constant velocity is the velocity which covers the same distance for each interval of the time.
It can be given as,
[tex]v=\dfrac{x}{t}[/tex]
As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.
Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.
Thus amount of time, [tex]t_{pass}[/tex] is,
[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]
As the distance covered by the car is 23 meters and the constant velocity of the car is 22 miles per second.
Convert the unit of velocity in m/s the value obtained will be 9.84 m/s.
Thus amount of time, [tex]t_{pass}[/tex] is,
[tex]9.84=\dfrac{23}{t_{pass} } \\t_{pass} =\dfrac{23}{9.84} \\t_{pass} =2.34[/tex]
According to the third equation of the motion acceleration can be given as,
[tex]v^2-u^2=2ax\\a=\dfrac{v^2-u^2}{2x}\\a=\dfrac{0^2-9.84^2}{2\times 23}\\a=-2.1 \rm \; m/s^2[/tex]
Now, use the first equation of motion, to get the required time,
[tex]v=u+at\\0=9.84+(-2.1)t\\t=4.68\rm \; s[/tex]
Therefore, the time required to pass is 2.34 seconds and the time to stop is 4.68 seconds.
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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!!
An official major league baseball has a mass of 0.14 kg. A pitcher throws a 40 m/s fastball which is hit by the batter straight back up the middle at a speed of 46 m/s.
a) What is the change in momentum of the ball during the collision with the bat?
b) If this collision occurs during a time of 0.012 seconds, what is the average force exerted by the bat on the ball?
Answer:
(a) The change in momentum is 12.04 kg-m/s
(b) The force exerted by the bat is 1003.33 N
Explanation:
Given that,
The mass of a ball, m = 0.14 kg
Initial speed of the ball, u = 40 m/s
Final speed of the ball, v = -46 m/s
(a) The change in momentum of the ball during the collision with the bat is given by :
[tex]\Delta p=m(v-u)\\\\=0.14(-46-40)\\\\=-12.04\ kg-m/s[/tex]
(b) Time for collision, t = 0.012 s
Now the force can be calculated as follows :
[tex]F=\dfrac{\Delta p}{t}\\\\F=\dfrac{12.04}{0.012}\\\\=1003.33\ N[/tex]
Hence, this is the required solution.
Answer:
a. = 12.04 kg*m/s
b. = 1,003.3N
Explanation:
The answer above is correct.
What is the weight of a 25 kg object on Earth with an acceleration due to gravity of 9.8m/s/s?
2.45 n
24.5 n
245 n
2450 n
Which statement best describes this situation
Answer:
what situation?
Explanation:
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.
Help please :)
Infrared waves are transmitted by ....... ?
weight is measured in units called
The weight is measured in a unit called Newton. The newton is a standard unit of weight measurements.
The International System of Units (SI) uses Newtons (N) as the unit of weight measurement. Weight is frequently expressed in pounds (lb) or ounces (oz) in some nations that employ the Imperial system. However, Newton is the accepted weight measurement unit in scientific and international contexts.
The International System of Units (SI) uses Newton as the primary unit of weight measurement. Sir Isaac Newton, a great physicist who significantly influenced our understanding of classical mechanics, is honored by having his name attached to it. Newton is the force needed to accelerate a mass of one kilogram by one meter per second squared in the SI system.
Hence, the weight is measured in a unit called Newton. The newton is a standard unit of weight measurements.
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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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it takes 840s to walk completely around a circular track, moving at a speed of 1.20m/s? what is the radius of the track?
Answer:
160.43 meters
Explanation:
T=(2*pi*r)/v
840=(2*pi*r)/1.2
1008=2*pi*r
504=pi*r
160.43=radius
The distance around the circular track is equal to its circumference. If it takes840 s to complete the path with a velocity of 1.20 m/s then the radius of the path will be 160 m.
What is velocity?Velocity is a physical quantity measuring the distance traveled by unit time. It is a vector quantity thus, characterized by its magnitude and direction. Velocity is usually expressed in m/s.
Velocity is the ratio of distance to the time taken to complete the travel. Thus, greater distance within short time indicates greater velocity.
Given that the time take to complete the circular path = 840 s
velocity = 1.20 m/s
distance = circumference of the circular path = 2πr.
distance = time × velocity
2πr = 840 s × 1.20 m/s
= 1008 m.
Therefore, the radius r of the circular path = 1008 m/ 2 π = 160 m.
Hence, the radius of the circular path 160 m.
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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]
One of the smallest planes ever flown was the Bumble Bee II, which had a mass of 180 kg. If the pilot’s mass was 70 kg, what was the velocity of both plane and pilot if their momentum was 20,800 kg∙m/s to the west?
Answer:
83.2 m/s to the West
Explanation:
From the question given above, the following data were obtained:
Mass of plane = 180 Kg
Mass of pilot = 70 Kg
Momentum = 20800 Kg∙m/s West
Velocity =?
Next, we shall determine the total mass. This can be obtained as follow:
Mass of plane = 180 Kg
Mass of pilot = 70 Kg
Total mass =?
Total mass = Mass of plane + Mass of pilot
Total mass = 180 + 70
Total mass = 250 Kg
Finally, we shall determine the velocity. This can be obtained as follow:
Total mass = 250 Kg
Momentum = 20800 Kg∙m/s West
Velocity =?
Momentum = mass × Velocity
20800 = 250 × Velocity
Divide both side by 250
Velocity = 20800 / 250
Velocity = 83.2 m/s West
Thus, the velocity of both plane and pilot is 83.2 m/s to the West
Suppose that a uniform rope of length L resting on a frictionless horizontal surface, is accelerated along the direction of its length by means of a force F, pulling it at one end. A mass M is accelerated by the rope. Assuming the mass of the rope to be m and the acceleration is a. Stated in terms of the product ma, what is the tension in the rope at the position 0.3 L from the end where the force F is applied if the mass M is 1.5 times the mass of the rope m?
Answer:
2.2 ma
Explanation:
Given :
Length of the rope = L
Mass of the rope = m
Mass of the object pulled by the rope = M
M = 1.5 m
So, L [tex]$\rightarrow$[/tex] m
For unit length [tex]$\rightarrow \frac{m}{L}$[/tex]
∴ 0.3 L = [tex]$0.3 \ L \left(\frac{m}{L}\right)$[/tex]
= 0.3 m
And for remaining 0.7 L = [tex]$0.7 \ L \left(\frac{m}{L}\right)$[/tex]
= 0.7 m
By Newtons law of motion,
F - T = ( 0.3 m) a .........(1)
T = ( M + 0.7 m) a
T = ( 1.5 m + 0.7 m) a
T = ( 2.2 m ) a ..............(2)
So from equation (1) and (2), we have
Tension on the rope
T = 2.2 ma