Answer:
D. Friction
Explanation:
Answer:
d. friction
I took it on study island and got it correct thanks to the other guy that has answered
Explanation:
One reason why it’s often easy to miss an action-reaction pair is because of the ________ of one of the objects.
Answer:
an action-reaction pair is because one of the objects is often much more massive and appears to remain motionless when a force acts on it. It has so much inertia, or tendency to remain at rest, that it hardly
Points A, B, and C lie along a line from left to right, respectively. Point B is at a lower electric potential than point A. Point C is at a lower electric potential than point B. What would best describes the subsequent motion, if any, of a positively-charged particle released from rest at point B?
Answer:
Please see below as the answer is self-explanatory.
Explanation:
If the potential at B is lower than A, and the potential at C is lower than B, this means that there is an electric field, directed from A to C.If a positively-charged particle is released at rest at point B, it will be accelerated by the electric field (which is a force per unit charge, so it produces an acceleration) in the same direction than the field (because it is a positive charge) towards point C.A car travels 100 km due East in 2 hours. It then travels 50 km South in 1 hour. What is its average velocity?
The average velocity of the car is 37.27 km/h.
The given parameters;
Initial displacement of the car, x = 100 kmTime of motion, t = 2 hoursFinal displacement of the car, y = 50 kmtime of motion, t = 1 hourThe average velocity of the car is calculated as follows;
[tex]average \ velocity = \frac{Total \ displacement}{Total \ time}[/tex]
The total displacement of the car is calculated as follows;
[tex]d = \sqrt{x^2 + y^2} \\\\d = \sqrt{100^2 \ + \ 50^2} \\\\d = 111.803 \ km[/tex]
The average velocity of the car is calculated as follows;
[tex]v = \frac{111.803}{3} \\\\v = 37.27 \ km/h[/tex]
Learn more about average velocity here: https://brainly.com/question/6504879
A car traveling 85 km/h is 250 m behind a truck
traveling 73 km/h.
Time needed = t = 20.83 s
Further explanationGiven
car speed = 85 km/h
truck speed = 73 km/h
Required
the time it takes for the car to reach the truck
Solution
When the car reaches the truck, the distance between them will be the same
x car - 250 m = x truck
General formula for distance (d) :
d = v.t
So the equation becomes :
85t-250 = 73t
12t=250
t = 20.83 s
Elizabeth has always believed that people's thoughts can help heal them. She wants to help people use positive thinking to positively affect their
illnesses. What type of psychology would be MOST appropriate for Elizabeth to study?
Answer: Family
Explanation:
Carousel conveyors are used for storage and order picking for small parts. The conveyorsrotate clockwise or counterclockwise, as necessary, to position storage bins at the storageand retrieval point. The conveyors are closely spaced, such that the operators travel timebetween conveyors is negligible. The conveyor rotation time for each item equals 1 minute;the time required for the operator to retrieve an item after the conveyor stops rotatingequals 0.25 minute. How many carousel conveyors can one operator tend without creatingidle time on the part of the conveyors
Answer:
the number of carousel conveyors that an operator can operate without any idle time is 5
Explanation:
Given the data in the question;
first we express the equation for number of carousel conveyors that can be operated by an operator;
n' = [tex]\frac{(a + t)}{( a + b)}[/tex]
where a is the concurrent activity time ( 0.25 minute )
b is the independent operator activity time
t is the independent machine activity time( 1 )
Now independent activity time is zero as the operator is not performing any inspection or packaging tasks.
So time taken for the operator to retrieve the finished item at the end of the process is the concurrent activity and independent machine activity time, the conveyor rotation time of each item
so
we substitute
0.25min for a, 1 for t and 0min for b
n' = [tex]\frac{(0.25min + 1min)}{( 0.25min+ 0 min)}[/tex]
n' = 1.25 min / 0.25
n' - 5
Therefore, the number of carousel conveyors that an operator can operate without any idle time is 5
A plastic rod 1.6 m long is rubbed all over with wool, and acquires a charge of -9e-08 coulombs. We choose the center of the rod to be the origin of our coordinate system, with the x-axis extending to the right, the y-axis extending up, and the z-axis out of the page. In order to calculate the electric field at location A = < 0.7, 0, 0 > m, we divide the rod into 8 pieces, and approximate each piece as a point charge located at the center of the piece.
Solution :
Length of the plastic rod , L = 1.6 m
Total charge on the plastic rod , Q = [tex]$-9 \times 10^{-8}$[/tex] C
The rod is divided into 8 pieces.
a). The length of the 8 pieces is , [tex]$l=\frac{L}{8}$[/tex]
[tex]$=\frac{1.6}{8}$[/tex]
= 0.2 m
b). Location of the center of the piece number 5 is given as : 0 m, -0.09375 m, 0 m.
c). The charge q on the piece number 5 is given as
[tex]$q=\frac{Q}{L}\times l$[/tex]
[tex]$q=\frac{-9 \times 10^{-8}}{1.6}\times0.2$[/tex]
= [tex]$-1.125 \times 10^{-8}$[/tex] C
d). WE approximate that piece 5 as a point charge and we need to find out the field at point A(0.7 m, 0, 0) only due to the charge.
We know, the Coulombs force constant, k = [tex]$8.99 \times 10^9 \ N.m^2/C^2$[/tex]
So the X component of the electric field at the point A is given as
[tex]$E_x = 8.99 \times 10^9 \times 1 \times 10^{-8} \ \cos \frac{187.628}{0.70625}$[/tex]
= -126.15 N/C
The Y component of the electric field at the point A is
[tex]$E_y = 8.99 \times 10^9 \times 1 \times 10^{-8} \ \sin \frac{187.628}{0.70625}$[/tex]
= -16.93 N/C
Now since the rod and the point A is in the x - y plane, the z component of the field at point A due to the piece 5 will be zero.
∴ [tex]$E_z=0$[/tex]
Thus, [tex]$E= <-126.15,-16.93,0>$[/tex]
A laser is placed at the point $(3,5)$. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off the $y$-axis, then hit and bounce off the $x$-axis, then hit the point $(7,5)$. What is the total distance the beam will travel along this path
Answer:
4units
Explanation:
To calculate the total distance the beam will travel along this path, you will use the formula for calculating the distance between two coordinates expressed as;
D = √(x2-x1)²+(y2-y1)²
Given the coordinate points
(3,5) and (7,5)
Substitute
D = √(7-3)²+(5-5)²
D = √(7-3)²+0²
D = √4²
D = √16
D = 4
Hence the total distance the beam will travel along this path is 4units
The surface area of a postage
stamp is 0.00600 m^2, and the air
exerts 1.00 atm of pressure on it.
How much force does it exert on
the stamp?
(Hint: The standard unit for
pressure is Pa.)
(Unit = N)
Answer:
Force = 607.95 Newton
Explanation:
Given the following data;
Area = 0.00600 m^2
Pressure = 1 atm to Pascal = 101325 Pa
To find the force;
Pressure = Force/area
Force = pressure * area
Substituting into the equation, we have;
Force = 101325 * 0.00600
Force = 607.95 Newton.
Therefore, the amount of force exerted by the air on the stamp is 607.95 Newton.
define stress engineering science
Answer:
Stress, in physical sciences and engineering, force per unit area within materials that arises from externally applied forces, uneven heating, or permanent deformation and that permits an accurate description and prediction of elastic, plastic, and fluid behaviour.
I hope it's helpful!
Fred's lightbulb is 45% efficient, and Fran's is 75% efficient. If they both use the same amount of electric energy, which produces more light energy?
Answer:
Frank's 75% efficient light bulb will shine brighter.
Explanation:
The brightness of a bulb is gotten from the power equation;
P = I²R
The more the power rating in watts, the more the brightness.
Now, if they both use the same amount of energy but yet have different efficiency, it means we will just multiply the efficiency by the power.
Thus, 75% efficiency will yield more power than a 45% efficient one.
Therefore, Frank's light bulb will shine brighter.
A truck travels on a straight road at a velocity of 17 meters per second. Over 20
seconds, it accelerates uniformly to 27 meters per second. What distance did the truck
travel during this acceleration?
Answer:
Distance, S = 440 meters.
Explanation:
Given the following data;
Initial velocity, u = 17m/s
Time, t = 20 seconds
Final velocity, v = 27m/s
To find the distance;
First of all, we would determine the acceleration of the truck.
Acceleration = (v-u)/t
Substituting the given values into the equation, we have;
Acceleration = (27 - 17)/20
Acceleration = 10/20
Acceleration = 0.5m/s²
Now, we would use the second equation of motion to find the distance traveled.
S = ut + ½at²
S = 17*20 + ½*0.5*20²
S = 340 + 0.25*400
S = 340 + 100
S = 440m
The equations of motion can be used to obtain the distance covered as 440 m.
We have to use of the equations that are used for uniformly accelerated motion in solving the problem. The chosen equation must be;
v^2 = u^2 + 2as and v = u + at
v = final velocity
u = initial velocity
a = acceleration
s = distance
To obtain the acceleration;
27 = 17 + 20(a)
27 - 17 = 20a
a = 0.5 ms-2
Now, to obtain the distance;
v^2 = u^2 + 2as
v^2 - u^2/as = s
s = (27)^2 - (17)^2/2(0.5)
s = 440 m
Learn more about the acceleration: https://brainly.com/question/12134554
Through what potential difference should electrons be accelerated so that their speed is 1.0 % of the speed of light when they hit the target
Answer:
Explanation:
Considering non - relativistic approach : ----
Speed of electron = 1 % of speed of light
= .01 x 3 x 10⁸ m /s
= 3 x 10⁶ m /s
Kinetic energy of electron = 1/2 m v²
= .5 x 9.1 x 10⁻³¹ x ( 3 x 10⁶ )²
= 40.95 x 10⁻¹⁹ J
Kinetic energy in electron comes from lose of electrical energy equal to
Ve where V is potential difference under which electron is accelerated and e is electronic charge .
V x e = kinetic energy of electron
V x 1.6 x 10⁻¹⁹ = 40.95 x 10⁻¹⁹
V = 25.6 Volt .
The "problem of perception" is best characterized as?
Answer:
making sense of a 3-d world from 2-d data
Explanation:
A friend comments to you that there was a beautiful, thin sliver of a Moon visible in the early morning just before sunrise. Which phase of the Moon would this be, and in what direction would you look to see the Moon (in the southern sky, on the eastern horizon, on the western horizon, high in the sky, etc.)?
Answer: Waning Crescent
Explanation:
According to Coulomb's Law, if the distance between two charged particles is doubled, the electric force will be _________. *
Answer: reduced by 1/4
Explanation:
The force will be reduced by 1/4. Try plugging in 2r, then squaring it. You will get 4r^2, which is essentially dividing the force by 4
A car traveling 85 km/h is 250 m behind a truck
traveling 73 km/h.
Time needed = t = 20.83 s
Further explanationGiven
car speed = 85 km/h
truck speed = 73 km/h
Required
the time it takes for the car to reach the truck
Solution
When the car reaches the truck, the distance between them will be the same
x car - 250 m = x truck
General formula for distance (d) :
d = v.t
So the equation becomes :
85t-250 = 73t
12t=250
t = 20.83 s
What is the mass of an object if it is moving at a speed of 10 m/s and has 400 J of kinetic energy?
Answers:
8 kg
Explanation:
Kinetic Energy = (mass × velocity × velocity) ÷ 2
We know that Kinetic Energy = 400 J and velocity = 10 m/s.
KE = (m × v × v) ÷ 2
400 J = (m × 10 m/s × 10 m/s) ÷ 2
400 J = m × 50 m^2/s^2
To find the mass you will divide 400 J and 50 m^2/s^2.
m = 8 kg
You can also check it if it gives you 400 J.
KE = (m × v × v) ÷ 2
KE = (8 kg × 10 m/s × 10 m/s) ÷ 2
KE = 400 J
So this means that the mass is 8 kg. I know that it is a bit confusing, but when you do J (joules) ÷ m^2/s^2 = kg (kilograms). Hope this helps, thank you !!
A bird lands on a bird feeder which is connected to a spring. The mass of the bird is exactly the same as the mass of the bird feeder. How does the added mass affect the period of oscillation of the bird feeder?
Answer:
The added mass will mean a longer period of oscillation.
Explanation:
The period of oscillation here is given by the formula;
T = 2π√(m/k)
Where m is mass and k is spring constant
From the equation of oscillation period above, it's obvious that when we increase the mass, the oscillation period will also increase.
Thus, the added mass will mean a longer period of oscillation.
A train 350 m long is moving on a straight track with a speed of 84.1 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 15.8 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.
Answer:
t = 25.0 s
Explanation:
Assuming that the engineer applies the brakes just over the crossing, the train moves exactly 350 m at a constant acceleration, with a final speed (when the last car of the train leaves the crossing) of 15.8km/h.Since we know the initial and final speeds, and the horizontal distance traveled (the length of the train) we can use the following kinematic equation to get the acceleration:[tex]v_{f}^{2} - v_{o}^{2} = 2*a* \Delta x (1)[/tex]
Since we need to find the time in seconds, it is advisable to convert vf and vo to m/s first, as follows:[tex]v_{o} = 84.1 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 23.4 m/s (2)[/tex]
[tex]v_{f} = 15.8 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 4.4 m/s (3)[/tex]
Replacing (2) and (3) in (1), since Δx =350m, we can solving for a:[tex]a = \frac{(4.4m/s)^{2} - (23.4m/s)^{2}}{2*350m} = -0.76 m/s2 (4)[/tex]
In order to get the time, we can simply use the definition of acceleration, and rearrange terms:[tex]t =\frac{v_{f}-v_{o}}{a} = \frac{(4.4m/s)-(23.4m/s)}{-0.76m/s2} = 25.0 s (5)[/tex]
Choose the best explanation from among the following:_________.
1. Charge is conserved, and therefore the mass of the object will remain the same.
2. A positive charge increases an object's mass; a negative charge decreases its mass.
3. To give the object a negative charge we must give it more electrons, and this will increase its mass.
Answer: 3. To give the object a negative charge we must give it more electrons, and this will increase its mass.
Explanation:
Suppose we have an object and we negatively charge it.
Then we are "adding" N electrons to the object.
Remember that the mass of an electron is:
m = 9.11*10^(-31) kg
Then if we add N electrons to an object of mass M, the new mass of the object will be:
Mass = M + N*9.11*10^(-31) kg
So we will have an (almost negligible) increase of the mass of the object.
(Something similar can happen if the object is positively charged, where we remove electrons, then the mass of the object decreases)
Then the correct option is:
3. To give the object a negative charge we must give it more electrons, and this will increase its mass.
A simple pendulum of length 5.5 m makes 10.0 complete swings in 25 s what is the acceleration due to gravity at the location of the pendulum ?
Answer:
The acceleration due to gravity at the location of the pendulum is 34.74 m/s².
Explanation:
Given that,
The length of a simple pendulum, l = 5.5 m
It makes 10.0 complete swings in 25 s.
Frequency of pendulum,
[tex]f=\dfrac{10}{25}\\\\f=0.4\ Hz[/tex]
The time period of a simple pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
Frequency,
[tex]f=\dfrac{1}{T}\\\\f=\dfrac{1}{2\pi \sqrt{\dfrac{l}{g}} }\\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}[/tex]
g is the acceleration due to gravity at the location where the pendulum is placed. So,
[tex]f^2=\dfrac{g}{4\pi^2l}\\\\g=f^2\times 4\pi^2l\\\\g=0.4^2\times 4\pi^2\times 5.5\\\\g=34.74\ m/s^2[/tex]
So, the acceleration due to gravity at the location of the pendulum is 34.74 m/s².
a ball is thrown upward with a beginning speed of 40m/s. The graph below shows how the speed of the ball changes until it reaches its maximum height.
use the graph to find
a) the time when the ball reaches its maximum height
b) the acceleration of the ball
c) the maximum height the ball went
Answer:
a) 4.0816s
b) -9.8 ms^-1
c) 81.63265m
At what speed, in m/s, would a moving clock lose 1.3ns in 1.0 day according to experimenters on the ground?
Answer:
v=0.14c
Explanation:
A runner completes the 200-meter dash with a time of 19.80 seconds. What was the runner's average speed in miles per hour?
Answer:
v = 22.54 mph.
Explanation:
Given that,
Distance moved, d = 200 m
Time, t = 19.8 s
We need to find the runner's average speed.
We know that,
1 mile = 1609.34 m
200 m = 0.124 miles
19.8 seconds = 0.0055 h
So,
Speed = distance/time
[tex]v=\dfrac{0.124}{0.0055}\\\\v=22.54\ mph[/tex]
So, the runner's average speed is 22.54 mph.
What is the maximum height achieved if a 0.400 kg mass is thrown straight upward with an initial speed of 40.0 m⋅s−1? Ignore the effect of air resistance
The maximum height : 81.63 m
Further explanationGiven
0.4 kg mass
vo = initial speed = 40 m/s
Required
the maximum height
Solution
We can use the law of conservation energy(ME=PE+KE) or use parabolic motion
For parabolic motion :
h max = (vo²sin²θ)/2g
θ = 90°(straight upward)
Input the value :
h max = (40²sin²90°)/2 x 9.8
h max = 81.63 m
What is the difference between elastic PE and gravitational PE?
A baseball player hits a 0.15 kg 0.15kg0, point, 15, start text, k, g, end text baseball that is initially at rest, changing its momentum by 11 kg ⋅ m s 11 s kg⋅m 11, start fraction, start text, k, g, end text, dot, start text, m, end text, divided by, start text, s, end text, end fraction.
Answer:
73.3m/s
Explanation:
We can find the velocity of the player.
Momentum = mass * velocity
Given
Mass = 0.15kg
Momentum = 11kgm/s
Get the velocity
Velocity = Momentum/Mass
Velocity = 11/0.15
Velocity = 73.3m/s
Hence the velocity of the player is 73.3m/s
Connective Tissue in a tendon is
A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log floating by. Always with the same engine throttle setting, the boater continues to travel upstream for another 1.45 km, which takes her 69.1 min. She then turns the boat around and returns downstream to her starting point, which she reaches at the same time as the same log does. How much time does the boater spend traveling back downstream
Answer:
t ’= [tex]\frac{1450}{0.6499 + 2 v_r}[/tex], v_r = 1 m/s t ’= 547.19 s
Explanation:
This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.
By the time the boat goes up the river
v_b - v_r = d / t
By the time the boat goes down the river
v_b + v_r = d '/ t'
let's subtract the equations
2 v_r = d ’/ t’ - d / t
d ’/ t’ = 2v_r + d / t
[tex]t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }[/tex]
In the exercise they tell us
d = 1.22 +1.45 = 2.67 km= 2.67 10³ m
d ’= 1.45 km= 1.45 1.³ m
at time t = 69.1 min (60 s / 1min) = 4146 s
the speed of river is v_r
t ’= [tex]\frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \ + 2 \ v_r}[/tex]
t ’= [tex]\frac{1450}{0.6499 + 2 v_r}[/tex]
In order to complete the calculation, we must assume a river speed
v_r = 1 m / s
let's calculate
t ’= [tex]\frac{ 1450}{ 0.6499 + 2 \ 1}[/tex]
t ’= 547.19 s