Which of the following would cause electromagnetic induction?
The current in a loop of coils is constant.
A loop of coils is placed near high voltage power lines with alternating current.
A constant current is supplied to a circuit board in your computer.
Your phone is turned on after being fully charged.
Answer:
The loop of coils - Electromagnetic induction is caused a changing magnetic field moving thru a closed loop(s) of coils
In designing subway trains and stations, engineers want to optimize the time and energy use between stations. A train can comfortably accelerate at 5 ft/s2 without causing its passengers to fall over. When coasting, the train will slow down at an acceleration of -2 ft/s2. The train begins at rest and must come to a complete stop at the destination. If two train stations are 2 miles apart, determine the fastest time that the train can arrive at the second station given the provided accelerations. Also, determine the maximum velocity obtained. Draw the a-t, v-t, and s-tgraphs for the train. (122 s, 174 ft/s)
Answer:
Explanation:
Let the time required for acceleration a₁ and deceleration a₂ be t₁ and t₂ .
Since final velocity during acceleration and initial velocity during deceleration are same
a₁ t₁ = a₂ t₂
5t₁ = 2 t₂ ------------------------------------------ ( 1 )
Distance travelled during acceleration = 1/2 a₁t₁²
= 1/2 x 5 x t₁² = 2.5 t₁²
Distance travelled during deceleration = 1/2 a₂t₂²
= 1/2 x 2 x t₂² = t₂²
Total distance travelled = 2 miles = 2 x 1760 x 3 ft = 10560
2.5 t₁² + t₂² = 10560
2.5 ( 2t₂ / 5 )² + t₂² = 10560
.4 t₂² + t₂² = 10560
1.4 t₂² = 10560
t₂ = 86.85 s
t₁ = 2t₂ / 5 = 34.75 s
t₁ + t₂ = 121.6 = 122 s
Total time taken = 122 s .
maximum velocity = a₁t₁
= 5 x 34.75 = 173.75 = 174 m/s .
Tom wanted to buy a hunting dog with a waterproof undercoat and a good sense of smell. Which type of genetic engineering would help Tom find the right dog?
a. GMO
b. biomedical research
C. Selective breeding
D. Mutation research
How much heat is required to raise the temperature of 50 grams of water from 30 °C to 90 °C? C of water 4186 J / kg C.
12558 J
12558000 J
125580 J
1255800 J
Answer:
12558 J
Explanation:
Please do mark as brainliest. Hope this helps! :)
What essential nutrient is so important and makes up approximately 80% of the body.
Protein
it's the main component of muscles, bones, organs ext.
what is the maximum distance we can shoot a dart,from ground level provided our toy dart gun gives a maximum initial velocity of 2.7m/s and air resistance is negligible
Answer:
R = v^2 sin 2 theta / g
The range provides the distance a projectile can travel
R(max) = v^2 / g if theta = 45 deg
R = 2.7^2 / 9.8 = .74 m
oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student observes that a particular spring-mass system has a frequency of oscillation of 10 Hz. the spring constant of the spring is 250 N/m. what is the mass?
Answer:
0.063 Kg
Explanation:
From the question given above, the following data were obtained:
Frequency (f) = 10 Hz
Spring constant (K) = 250 N/m
Mass (m) =?
Next, we shall determine the period of oscillation. This can be obtained as follow:
Frequency (f) = 10 Hz
Period (T) =?
T = 1/f
T = 1/10
T = 0.1 s
Finally, we shall determine the mass of the spring. This can be obtained as follow:
Spring constant (K) = 250 N/m
Period (T) = 0.1 s
Pi (π) = 3.14
Mass (m) =?
T = 2π√(m/K)
0.1 = 2 × 3.14 × √(m/250)
0.1 = 6.28 × √(m/250)
Divide both side by 6.28
0.1 / 6.28 = √(m/250)
Take the square of both side.
(0.1 / 6.28)² = m/250
Cross multiply
m = (0.1 / 6.28)² × 250
m = 0.063 Kg
Therefore, the mass of the spring is 0.063 Kg.
On a distance-time graph, what is shown when the curve is flat going from left to the right?
A. a negative speed
B. no speed
C. a positive speed
D. It does not mean anything.
Please help me !!im on a test
girl is sitting on a tire swing that is attached with a rope that is 2.1 m in length. Her dad pushes
her with a speed of 3.0 m/s. If the centripetal force is 88 N, what is the mass of the girl?
pls help
Answer:
m=29.6kg
Explanation:
your welcome:) have a great day
Se transmiten ondas transversales en una cuerda tensada orientada sobre el eje x. La función de ondas correspondientes es y= 5.00 sen(6.05x + 5.19t + 1.57). Donde y y x están en metros y t en segundos. Cual es la rapidez de las ondas que se transmiten en dicha cuerda?
Answer:
Explanation:
Whats the entire question?
What step occurs first in a scientific investigation and in the development of a new technology?
-Findings are communicated to others.
-A problem or need is identified
-A series of tests are analyzed.
-Information is researched
Answer:
B. A problem or need is identified
Explanation:
hoep this helps
Answer:
B
Explanation:
edge
Please help in one minute will mark brainlist
Answer: Different types of telescopes usually don't take simultaneous readings. Space is a dynamic system, so an image taken at one time is not necessarily the precise equivalent of an image of the same phenomena taken at a later time. And often, there is barely enough time for one kind of telescope to observe extremely short-lived phenomena like gamma-ray bursts. By the time other telescopes point to the object, it has grown too faint to be detected.
Explanation: Trust me
If a Sam goes above and beyond what is expected in her job duties, what might Sam's supervisor note on Sam's next evaluation?
O A. Sam has good attendance.
O B. Sam shows initiative.
O C. Sam stays focused on assigned tasks.
OD. Sam works well with others.
Answer:
2
Exadadadwwwwwwwplanation:
Calculate the potential difference across the 8 ohm resistor
Explanation:
if the current is 1A
V=iR
V= 1 × 8
V = 8volts
Count the equivalent resistance.
[tex]\huge{ \mathfrak{ \underline{question : }}}[/tex]
Find the equivalent resistance .....
[tex]\huge{ \mathfrak{ \underline{ Answer \: \: ✓ }}}[/tex]
Equivalent resistance :
[tex] \dfrac{1}{ R _ t} = \dfrac{1}{3 + 3} + \dfrac{1}{3} [/tex][tex]\dfrac{1}{ R _ t} = \dfrac{1}{6} + \dfrac{1}{3} [/tex][tex]\dfrac{1}{ R _ t} = \dfrac{1 + 2}{6} [/tex][tex]\dfrac{1}{ R _ t} = \dfrac{3}{6} [/tex][tex] R _ t = 2 [/tex]Equivalent resistance = 2 ohms
[tex]\#TeeNForeveR [/tex]
A 4.00-kg block hangs by a light string that passes over a massless, frictionless pulley and is connected to a 6.00-kg block that rests on a frictionless shelf. The 6.00-kg block is pushed against a spring to which it is not attached. The spring has a spring constant of 180 N/m, and it is compressed by 30.0 cm. Find the speed of the block after the spring is released and the 4.00-kg block has fallen a distance of 40.0 cm.
Answer:
v = 2.82 m/s
Explanation:
For this exercise we can use the conservation of energy relations.
We place our reference system at the point where block 1 of m₁ = 4 kg
starting point. With the spring compressed
Em₀ = K_e + U₂ = ½ k x² + m₂ g y₂
final point. When block 1 has descended y = - 0.400 m
Em_f = K₂ + U₂ + U₁ = ½ m₂ v² + m₂ g y₂ + m₁ g y
as there is no friction, the energy is conserved
Em₀ = Em_f
½ k x² + m₂ g y₂ = ½ m₂ v² + m₂ g y₂ + m₁ g y
½ k x² - m₁ g y = ½ m₂ v²
v² = [tex]\frac{k}{m_2} x^2 - 2 \frac{m_1}{m_2} \ g y[/tex]
let's calculate
v² = [tex]\frac{180}{6.00} \ 0.300^2 - 2 \ \frac{4.00}{6.00} \ 9.8 \ (- 0.400)[/tex]
v² = 2.7 + 5.23
v = √7.927
v = 2,815 m / s
using of significant figures
v = 2.82 m/s
A 10kg block is Pulled along a horizontal
Surface by a force
of 50N at an angles
of 37° with the horizontal If the
coefficient of sliding friction b/n the
block and the surface is o.2
(g=10m/s^2 Sin 37=O.6 and cos 37 = 0.8)
A, what frictional forces acting on the block?
B,what is the acceleration of the block?
Answer:
hope u can understand the method
For the circuit in the previous part, what happens to the maximum current if the frequency is doubled and the inductance is halved
Answer:
Following are the responses to these question:
Explanation:
Since the [tex]max^{m}[/tex] is the current of ckt which depend on the reactance which inductor that also enables the ckt and inductor resistance [tex](X_L)[/tex] for capacities
[tex]\to X_{C}=\frac{1}{W L}[/tex]
for
[tex]\to X_{L}=wL[/tex]
When [tex]w \longrightarrow 2w[/tex]
[tex]L\longrightarrow \frac{L}{2}[/tex]
then
[tex]\to X_{L}=2 w \times \frac{L}{2}=wL[/tex]
therefore, [tex]X_{L}[/tex] remains at the same so, the maximum current remains the in same ckt.
Consider two antennas separated by 9.00 m that radiate in phase at 120 MHz, as described in Exercise 35.3. A receiver placed 150 m from both antennas measures an intensity I0 . The receiver is moved so that it is 1.8 m closer to one antenna than to the other. (a) What is the phase difference f between the two radio waves produced by this path difference
Answer:
[tex]\phi=4.52 rad[/tex]
Explanation:
From the question we are told that
Distance b/e antenna's [tex]d=9.00m[/tex]
Frequency of antenna Radiation[tex]F_r=120 MHz \approx 120*10^6Hz[/tex]
Distance from receiver [tex]d_r=150m[/tex]
Intensity of Receiver [tex]i= 10[/tex]
Distance difference of the receiver b/w antenna's [tex](r^2-r^1)=1.8m[/tex]
Generally the equation for Phase difference [tex]\phi[/tex] is mathematically given by
[tex]\phi=\frac{2\pi}{\frac{c}{f_r}} *(r^2-r^1)[/tex]
[tex]\phi=\frac{2*\pi}{\frac{3*10^{8}}{120*10^6}} *1.8[/tex]
[tex]\phi=\frac{4\pi}{5} *1.8[/tex]
[tex]\phi=4.52 rad[/tex]Therefore phase difference f between the two radio waves produced by this path difference is given as
[tex]\phi=4.52 rad[/tex]
The density of table sugar is 1.59g/cm3 what is the volume of 7.85g of sugar?
Answer: 4.94cm³
Explanation:
Data;
ρ = 1.59g/cm³
mass = 7.85g
volume = ?
density = mass / volume
ρ = m / v
v = m / ρ
v = 7.85 / 1.59
v = 4.94cm³
A state trooper is traveling down the interstate at 20 m/s. He sees a speeder traveling at 50 m/s approaching from behind. At the moment the speeder passes the trooper, the trooper hits the gas and gives chase at a constant acceleration of 2.5 m/s^2.
Required:
a. Assuming that the speeder continues at 60 m/s , how long will it take the trooper to catch up to the speeder?
b. How far down the highway will the trooper travel before catching up to the speeder?
Answer:
Explanation:
From the given information;
Let assume that the distance travelled by the speeder prior to the time the trooper catches with it to be = d
the time interval to be = t
Then, the speeder speed = distance/time
Making distance the subject; then:
distance (d) = speed × time
d = (50 m/s)t
d = 50 t --- (1)
Now, for the trooper; Using the equation of motion:
[tex]d = ut + \dfrac{1}{2}at^2[/tex]
[tex]d = (20)t+\dfrac{1}{2}(2.5)t^2[/tex]
d = 20t + 1.25t²
Replace the value of d in (1) to the above equation, we have:
50 t = 20 t + 1.25t²
50t - 20t = 1.25t²
30t = 1.25t²
30 = 1.25t
[tex]t = \dfrac{30}{1.25}[/tex]
t = 24 seconds
From (1), the distance far down the highway the trooper will travel prior to the time it catches up with the speeder is:
= 50t
= 50(24)
= 1200 seconds
g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 kg and is 4.2 m in length. At the other end of the bar sits another 3.6-kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision. Assume that the bar is horizontal when the dropped ball hits it. How high (in meters) will the other ball go after the collision
Answer:
h = 3.5 m
Explanation:
First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:
[tex]2gh = v_f^2 - v_i^2\\[/tex]
where,
g = acceleration due to gravity = 9.81 m/s²
h = height = 3.5 m
vf = final speed = ?
vi = initial speed = 0 m/s
Therefore,
[tex](2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s[/tex]
Now, we will apply the law of conservation of momentum:
[tex]m_1v_1 = m_2v_2[/tex]
where,
m₁ = mass of colliding ball = 3.6 kg
m₂ = mass of ball on the other end = 3.6 kg
v₁ = vf = final velocity of ball while collision = 8.3 m/s
v₂ = vi = initial velocity of other end ball = ?
Therefore,
[tex](3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s[/tex]
Now, we again use the third equation of motion for the upward motion of the ball:
[tex]2gh = v_f^2 - v_i^2\\[/tex]
where,
g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)
h = height = ?
vf = final speed = 0 m/s
vi = initial speed = 8.3 m/s
Therefore,
[tex](2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\[/tex]
h = 3.5 m
Calculate the speed of your cat as it runs towards its food bowl 14.7m away in 4.5 s. Give answer in mph.
Answer:
3.26mph
Explanation:
To calculate speed, use the formula distance/time. In this case, just divide 14.7 by 4.5.
The speed with which the cat runs towards its food bowl is 7.3 Miles per hour.
Given the data in the question
Distance of cat from it's food bowl; [tex]s = 14.7m[/tex]Time taken for the cat to reach the food bowl; [tex]t = 4.5s[/tex]Speed of the cat; [tex]v = \ ?[/tex]
Speed is the time rate at which an object is moving along a path
Speed is the time rate at which an object is displaced.
Speed = Distance / Time
[tex]v = s / t[/tex]
We substitute our values into the equation
[tex]v = \frac{14.7m}{4.5s}\\\\v = 3.267 m/s\\\\v = 7.3mph[/tex]
Therefore, the speed with which the cat runs towards its food bowl is 7.3 Miles per hour.
Learn more: https://brainly.com/question/680492
A soccer ball is released from rest at the top of a grassy incline. After 6.2 seconds, the ball travels 47 meters. One second later, the ball reaches the bottom of the incline.
(a) What was the balls acceleration?(assume that the acceleration was constant).
(b) How long was the incline?
Answer:
(a) a = 2.44 m/s²
(b) s = 63.24 m
Explanation:
(a)
We will use the second equation of motion here:
[tex]s = v_it+\frac{1}{2}at^2[/tex]
where,
s = distance covered = 47 m
vi = initial speed = 0 m/s
t = time taken = 6.2 s
a = acceleration = ?
Therefore,
[tex]47\ m = (0\ m/s)(6.2\ s)+\frac{1}{2}a(6.2\ s)^2\\\\a = \frac{2(47\ m)}{(6.2\ s)^2}[/tex]
a = 2.44 m/s²
(b)
Now, we will again use the second equation of motion for the complete length of the inclined plane:
[tex]s = v_it+\frac{1}{2}at^2[/tex]
where,
s = distance covered = ?
vi = initial speed = 0 m/s
t = time taken = 7.2 s
a = acceleration = 2.44 m/s²
Therefore,
[tex]s = (0\ m/s)(6.2\ s)+\frac{1}{2}(2.44\ m/s^2)(7.2\ s)^2\\\\[/tex]
s = 63.24 m
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it elastically strikes another block that is only one-half its mass. Find the velocity of block M at the bottom of the incline before the collision with the small block m.
Solution :
Given :
M = 0.35 kg
[tex]$m=\frac{M}{2}=0.175 \ kg$[/tex]
Total mechanical energy = constant
or [tex]$K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$[/tex]
But [tex]$K.E._{top} = 0$[/tex] and [tex]$P.E._{bottom} = 0$[/tex]
Therefore, potential energy at the top = kinetic energy at the bottom
[tex]$\Rightarrow mgh = \frac{1}{2}mv^2$[/tex]
[tex]$\Rightarrow v = \sqrt{2gh}$[/tex]
[tex]$=\sqrt{2 \times 9.8 \times 0.35}$[/tex] (h = 35 cm = 0.35 m)
= 2.62 m/s
It is the velocity of M just before collision of 'm' at the bottom.
We know that in elastic collision velocity after collision is given by :
[tex]$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$[/tex]
here, [tex]$m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$[/tex]
∴ [tex]$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}[/tex]
[tex]$=\frac{0.175}{0.525}+0$[/tex]
= 0.33 m/s
Therefore, velocity after the collision of mass M = 0.33 m/s
A solenoid that is 66.2 cm long has a cross-sectional area of 18.0 cm2. There are 1300 turns of wire carrying a current of 8.15 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy in joules stored in the magnetic field there (neglect end effects).
Answer:
(a) Energy Density = 160.94 J/m³
(b) Energy Stored = 0.192 J
Explanation:
(a)
The energy density of the magnetic field inside the solenoid is given by the following formula:
[tex]Energy\ Denisty = \frac{B^2}{2\mu_o}\\[/tex]
where,
B = magnetic field strength of solenoid = [tex]\frac{\mu_oNI}{l}[/tex]
Therefore,
[tex]Energy\ Density = \frac{\mu_oN^2I^2}{2l^2}[/tex]
where,
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
N = No. of turns = 1300
I = current = 8.15 A
L = length = 66.2 cm = 0.662 m
Therefore,
[tex]Energy\ Density = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(1300)^2(8.15\ A)^2}{2(0.662\ m)^2}[/tex]
Energy Density = 160.94 J/m³
(b)
Energy Stored = (Energy Density)(Volume)
Energy Stored = (Energy Density)(Area)(L)
Energy Stored = (160.94 J/m³)(0.0018 m²)(0.662 m)
Energy Stored = 0.192 J
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standard power cycle for which the working fluid is evaporated, passed through a turbine, and subsequently condensed. The system is to be used in very special locations for which the oceanic water temperature near the surface is approximately 300 K, while the temperature at reasonable depths is approximately 280 K. The warmer water is
Answer:
Explanation:
Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.
To objective is to find the:
(i) required heat exchanger area.
(ii) flow rate to be maintained in the evaporator.
Given that:
water temperature = 300 K
At a reasonable depth, the water is cold and its temperature = 280 K
The power output W = 2 MW
Efficiency [tex]\zeta[/tex] = 3%
where;
[tex]\zeta = \dfrac{W_{out}}{Q_{supplied }}[/tex]
[tex]Q_{supplied } = \dfrac{2}{0.03} \ MW[/tex]
[tex]Q_{supplied } = 66.66 \ MW[/tex]
However, from the evaporator, the heat transfer Q can be determined by using the formula:
Q = UA(L MTD)
where;
[tex]LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}[/tex]
Also;
[tex]\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K[/tex]
[tex]\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K[/tex]
[tex]LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}[/tex]
[tex]LMTD = \dfrac{8}{In (5)}[/tex]
LMTD = 4.97
Thus, the required heat exchanger area A is calculated by using the formula:
[tex]Q_H = UA (LMTD)[/tex]
where;
U = overall heat coefficient given as 1200 W/m².K
[tex]66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}[/tex]
The mass flow rate:
[tex]Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}[/tex]
Hypothesis what is it
Explanation:
hope it helps.
stay safe healthy and happy.differences between adhesion and cohesion
Answer:
As for the definitions, the tendency of two or more different molecules to bond with each other is known as Adhesion, whereas the force of attraction between the same molecules is known as Cohesion.
hopefully this helps
Answer:
Adhesion is the force of attraction between molecules of different substances while cohesion is the force of attraction between molecules of same substances.
When a particular hanging mass is suspended from the string, a standing wave with two segments is formed. When the weight is reduced by 2.2 kg, a standing wave with five segments is formed. What is the linear density of the string
Solution :
Mass is varied keeping frequency constant.
Wavelength, λ [tex]$=\frac{2l}{n}$[/tex]
where length of spring = l
number of segments = n
Velocity, v = λ x f
= [tex]$\sqrt{\frac{T}{\mu}}$[/tex]
[tex]$\mu $[/tex] = mass density, T = tension in string
[tex]$T=\frac{4 \mu l^2f^2}{n^2}$[/tex]
[tex]$T=mg = \frac{4 \mu l^2f^2}{n^2}$[/tex] , n = 2
[tex]$T = (m-2.2)g = \frac{4 \mu l^2f^2}{n^2}, n = 5$[/tex]
[tex]$\Rightarrow \frac{m}{m-2.2}=\frac{25}{4}$[/tex]
[tex]$\Rightarrow m = 2.619\ kg$[/tex]
Therefore, μ = 0.002785 kg/ m
Frequency is varied keeping T constant
[tex]$T=\frac{4 \mu l^2f^2}{n^2}, f=60 , \ \ n = 2$[/tex]
[tex]$T=\frac{4 \mu l^2f^2}{n^2}, f=? , \ \ n = 7$[/tex]
[tex]$\Rightarrow \frac{60^2}{4}=\frac{f^2}{49}$[/tex]
f = 210 Hz