Answer:
the magnitude of the magnetic field is 1.23 × 10⁻⁸ T { in the +y direction }
Explanation:
Given the data in the question;
the relation between the electric field and the magnetic field in an electromagnetic wave can be written as follows;
c = E₀/B₀
where c is speed of light, E₀ is the amplitude of the electric field and B₀ is the amplitude of the magnetic field.
given that; E₀ = 4.70 V/m, we know that; speed of light c = 3 × 10⁸ m/s
so from our relation; c = E₀/B₀
B₀ = E₀/c
we substitute
B₀ = 4.70 / 3 × 10⁸
B₀ = 1.23 × 10⁻⁸ T { in the +y direction }
Therefore, the magnitude of the magnetic field is 1.23 × 10⁻⁸ T { in the +y direction }
Which function represents g(x), a reflection of f(x) = 6(one-third) Superscript x across the y-axis?
Answer:
g(x)=6(3)°x
[tex]g(x) = 6(3){x} [/tex]
A honey bee's wings beat at 230 beats per second. If the speed of sound in air is 340 m/s, what is the wavelength of
the sound waves?
Answer:
[tex]from \: the \: wave \: equation \\ velocity = frequency \times wavelength \\ 340 = 230 \times \lambda \\ \lambda = \frac{340}{230} \\ \lambda = 1.5 \: m[/tex]
solve this with figure.help me ......
Answer:
[tex] \huge\mathfrak\pink{Hope \: it \: helps}\ [/tex]
A classic demonstration illustrating eddy currents is performed by dropping a permanent magnet inside a conducting cylinder. The magnet does not go into free fall. Instead it reaches terminal velocity and can take a few seconds to drop a length of about a meter. Suppose the mass of the magnet is 70 g and width of 1.0 cm. It falls with a terminal velocity of 10 cm/s and the length of the pipe is 80 cm. The magnitude of the Joule heating from the eddy currents is approximately:________
Answer:
The correct solution is "0.69 N".
Explanation:
The given values are:
Mass of magnet,
m = 70 g
or,
= 0.07 kg
Width,
= 1.0 cm
Velocity,
= 10 cm/s
Length of the pipe,
= 80 cm
Whenever the velocity is constant, then the net force which is acting on the magnet will be "0".
On the magnet,
The up-ward force will be:
⇒ [tex]F=mg[/tex]
On substituting the values, we get
⇒ [tex]=0.07\times 9.8[/tex]
⇒ [tex]=0.69 \ N[/tex]
A 1220-N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. A 1960-N crate hangs from the far end of the beam. FTis the Tension in the cable. FVand FHare the vertical and horizontal components of the force on from the wall on the beam. Why is the direction of FVup and not down
Answer:
Suppose the cable was vertical.
Vertical forces have to cancel to zero.
If you added weight to the beam the vertical force (supposed down)
the cable has to support the added weight as well as the weight added to the cable.
One could create a perpetual motion machine by using the additional downward force on the wall from weight applied to the beam. That is because
2 forces must be offset by the force Fv. That is, both the cable and the additional weight would contribute to the downward force at Fv.
The equation used to predict the theoretical period Ty of a simple pendulum assumes a small amplitude of oscillation. A student builds a pendulum by attaching one end of a string of known length to the ceiling and the other end to a small solid sphere. The sphere is pulled back so the string makes a 5 angle with the vertical and released from rest with a theoretical amplitude of Oscillation At The student measures the resulting experimental period Tg and after the sphere makes several oscillations, measures the experiment oscillation. In the absence of air resistance, T T and A. Which of the following best explains how air resistance affects the results of the experiment?
A) with resistance because the sphere will speed up
B) With air resistance As , because although the sphere slows down, it also will travel for a greater time
C) With resistance because the sphere will travel a greater distance
D) With air resistance, 7>7y, because the sphere will slow down
E) with resistance because although the sphere slows down, also will travel shorter distance
Answer:
The answer is "Choice E".
Explanation:
In this situation the option e is right because its resistance decreases through time, however, the time is the same for the same reason, whereas the sphere deteriorates, somehow it travels shorter distances however if the air resistance becomes are using the amplitude of movement declines, that's why other choices were wrong.
why do we preserve food?
Answer:
The primary objective of food preservation is to prevent food spoilage until it can be consumed. Gardens often produce too much food at one time—more than can be eaten before spoilage sets in. Preserving food also offers the opportunity to have a wide variety of foods year-round. It's economic.
Explanation:
solve this with figure.help me ......
[tex] \huge\mathfrak\purple{Hope \: it \: helps}\ [/tex]
What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
Answer:
Wavelength = 1.36 * 10^{-34} meters
Explanation:
Given the following data;
Mass = 0.113 kg
Velocity = 43 m/s
To find the wavelength, we would use the De Broglie's wave equation.
Mathematically, it is given by the formula;
[tex] Wavelength = \frac {h}{mv} [/tex]
Where;
h represents Planck’s constant.
m represents the mass of the particle.
v represents the velocity of the particle.
We know that Planck’s constant = 6.6262 * 10^{-34} Js
Substituting into the formula, we have;
[tex] Wavelength = \frac {6.6262 * 10^{-34}}{0.113*43} [/tex]
[tex] Wavelength = \frac {6.6262 * 10^{-34}}{4.859} [/tex]
Wavelength = 1.36 * 10^{-34} meters
A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is trans- formed into elastic potential energy of the bending board, if the board bends far enough, it breaks. Applying a force to the center of a particular pine board deflects the center of the board by a distance that increases in proportion to the fore. Ultimately the board breaks at an applied force of 800 N and a deflection of 1.2 cm.
a. To break the board with a blow from the hand, how fast must the hand be moving? Use 0.50 kg for the mass of the hand.
b. If the hand is moving this fast and comes to rest in a dis- tance of 1.2 cm, what is the average force on the hand?
(a) The velocity is "6.2 m/s".
(b) The average force is "800.83 N".
According to the question,
Force,
F = 800 NDeflection,
x = 1.2 cm= [tex]1.2\times 10^{-2} \ m[/tex]
As we know,
The work done,
→ [tex]W = F\times d[/tex]
[tex]= 800\times 1.2\times 10^{-2}[/tex]
[tex]= 9.6 \ J[/tex]
(a)
Given:
Mass of hand,
m = 0.50 kgNow,
→ [tex]\frac{1}{2} mv^2 = 9.6 \ J[/tex]
[tex]v = \sqrt{\frac{2\times 9.6}{0.50} }[/tex]
[tex]= 6.2 \ m/s[/tex]
(b)
→ [tex]v^2 = u^2 +2ax[/tex]
→ [tex]a= \frac{v^2}{2x}[/tex]
[tex]= \frac{(6.2)^2}{2\times 1.2\times 10^{-2}}[/tex]
[tex]= 1601.67 \ m/s^2[/tex]
hence,
The average force will be:
→ [tex]F_{avg} = m\times a[/tex]
[tex]= 0.50\times 1601.67[/tex]
[tex]= 800.83 \ N[/tex]
Thus the above answers are correct.
Learn more:
https://brainly.com/question/23115729
Which two mixtures are homogeneous?
Answer:
sand and air
Explanation:
air is a mix of carbon dioxide and oxygen and sand is a mix of rocks and stuff
move the compass around the edge of the black screen far away from the bar magnet does the needle still interact with the poles of the bar magnet?
Answer: Yes
Explanation:
Answer:
Yes. The behavior of the bar magnet and the needle is similar to that in part A.
Explanation:
plato
1) A rock of mass 1.50kg is released from rest at a height of 30m. Ignore air resistance and
calculate;
(a) its speed half way down
(b) its speed on reaching the ground
(c) its total mechanical energy
(i) at a height of 30m
(ii) its kinetic energy half way through
(iii) its potential energy on striking the ground
Here's link to the answer:
bit.[tex]^{}[/tex]ly/3a8Nt8n
What is the average speed for a runner who finished the 100 meter dash in 9.8 seconds?
A. 0.98 m/s
B. 10.2 m/s
C. 1.02 m/s
D. 980 m/s
Answer:
10.2
the average speed of the runner is 10.2m/s.
Can someone please help me with this one
Answer:
im pretty sure the answer is A. up but im not 100% sure
What is the speed of a wave with a period of 25 sec/cycle and a wavelength of 15 m/cycle?please help me !
Someone tryna help a homie out
Answer: thats hard...
Explanation:1 c
2 a
3 b
4 d
6 b
7 a
8 b
i tried
The speed of sound is calculated to be 343m/s. If a sound wave has a frequency of 20Hz, what is the wavelength?
A trumpeter plays at a sound level of 75dB. three equally loud trumpet players join in. what is the new sound level?
a problem solving method that involves trying all possible solutions until one works is using_____.
Answer:
trial and error.
Explanation:
a problem solving method that involves trying all possible solutions until one works is using trail and error.
The 10-lb block A attains a velocity of 2ft/s in 5 seconds, starting from rest. Determine the tension in the cord and the coefficient of kinetic friction between block A and the horizontal plane. Neglect the weight of the pulley. Block B has a weight of 8 lb. Please work on this by using the principle of linear impulse-momentum
Answer:
Explanation:
Let T be the tension in the cord.
Impulse by cord = change in momentum of block A .
T x 5s = 10 ( 2 -0) = 20
T = 4 poundal .
acceleration of block B = 2 / 5 = 0.4 m /s²
Net force applied on A = m ( g + a ) where m is mass of block B , a is acceleration of block B .
= 8 ( 32 + .4 ) = 259.2 poundal
Frictional force on block A = 259.2 - 4 = 255.2 poundal
μ x 10 x 32 = 255.2
320μ = 255.2
μ =0 .8 .
5. What is the elevation along the shoreline (sea level)?
O ft
1 ft
10 ft
100 ft
When a wave of light strikes a diffraction screen (or grating) it will cause the wave to pass through each slit, causing the emerging light to radiate outward as if from a new source. These newly formed waves create alternating patterns of constructive and destructive interference, which can be observed when projected onto a screen. Select the variables listed below that determine the angle at which these patterns occur (you might need to select more than one answer):
Answer:
The answer is "the angle depends on wavelength and distance of slits".
Explanation:
[tex]d \sin \theta = m\lambda[/tex] as well as angular, of mth order darkened border is indicated only by angular position, the [tex]\theta[/tex] mth order bright border generated by a light wavelength [tex]\lambda[/tex] , and use the grating of the segregation slit.
[tex]d \sin \theta = ( m + \frac{1}{2}) \lambda[/tex]
Therefore, [tex]\theta[/tex] depends on d and [tex]\lambda[/tex] for a specific order.
do u ever think that how are u living cause we could not even be here and God but made us but had did it all started I believe in god btw I'm just saying how
Answer:
What is the question
Explanation:
lol
An electric motor operates a pump that irri-
gates a farmer's crop by pumping 10 000 L of
water a vertical distance of 8.0 m into a field
each hour. The motor has an operating resis-
tance of 22.0 2 and is connected across a
110-V source.
a. What current does it draw?
b. How efficient is the motor?
Answer:
a. 5A
b. 39.60%
Explanation:
The computation is shown below:
a. The current does it draw is
= v ÷ R
= 110v ÷ 22
= 5A
b. Now the efficiency of the motor is
n = mgh ÷ vlt
= (10,000 × 9.8 × 8) ÷ (5 × 3600 × 110)
= 784000 J ÷ 1,980,000
= 39.60%
hence, the above formulas are applied & the same is relevant
Two 6 ohm resistors in parallel gives an equivalent resistance of
1/Req = 1/R1 + 1/R2 + 1/R3...
3 ohm
6 ohms
9 ohms
12 ohms
Answer:
3 ohms
Explanation:
6×6/6+6 =3 .............
What is the mass of 1.000 L of seawater? kg
Answer:
1000 L= 1000 kg
1.000 L= 1.000 kg
Explanation:
It will be the same because L and kg have the same mass
Choose the statement(s) that are FALSE. A. An object will undergo constant acceleration if it is in equilibrium. B. Subject to the same net force, a larger mass will accelerate at a greater rate than a smaller mass. C. If the vector sum of the forces on an object is not zero, the object is in equilibrium. D. A net force of 125 N acts horizontally on an object that weighs 125 N. Therefore, the object does not accelerate. E. If an object is at rest, then there are no forces acting on the object. F. None of these statements are false.
Answer:
A, B, C, D and E
Explanation:
A. An object will undergo constant acceleration if it is in equilibrium.
This is false because an object in equilibrium has no acceleration.
B. Subject to the same net force, a larger mass will accelerate at a greater rate than a smaller mass.
This is false because, since F = ma and a = F/m where a is acceleration and m = mass. Since F is constant, a ∝ 1/m. So acceleration is inversely proportional to mass. So, the smaller mass would accelerate at a greater rate than the larger mass.
C. If the vector sum of the forces on an object is not zero, the object is in equilibrium.
This is false because an object can only be in equilibrium if the vector sum of the forces acting on it is zero.
D. A net force of 125 N acts horizontally on an object that weighs 125 N. Therefore, the object does not accelerate.
This is false because, the weight acts in the vertical direction and has no effect in the horizontal direction. The only forces in the horizontal direction are the frictional force and the 125 N horizontally applied force.
The objects accelerates if the 125 N applied force is greater than the frictional force and since frictional force = μN = μW where μ = coefficient of friction, N = normal force and W = weight of the object. Since μ < 1, f < W. So f < 125 N. So, the object accelerates.
E. If an object is at rest, then there are no forces acting on the object.
This is false because, an object might be at rest when the net force acting on it is zero not only when no force acts on it.
There is a current of 0.83 A through a light bulb in a 120.0v circuit. What is the
resistance of the light bulb?
Answer:
144.6ohms
Explanation:
v=IR
R =V/I=120/0.83
R=144.6
11) A tank of kerosene with density of 750 kg/m3 has a syphon used to remove the fluid that then exits into the local atmosphere, with pressure of 101 kPa. The pressure above the kerosene in the tank is 120 kPa absolute. The syphon tube has a diameter of 2 cm, exits the tank rising to 10 cm above the level of the kerosene and then drops down to 15 cm below the level of the kerosene where it exits into the atmospheric pressure. Calculate the exit velocity from the tube.
Answer:
[tex]7.32\ \text{m/s}[/tex]
Explanation:
[tex]v_1[/tex] = Velocity at initial point = 0
[tex]P_1[/tex] = Pressure in tank = 120 kPa
[tex]P_2[/tex] = Pressure at outlet = 101 kPa
[tex]\rho[/tex] = Density of kerosene = [tex]750\ \text{kg/m}^3[/tex]
[tex]Z_1[/tex] = Tank height = 15 cm
[tex]Z_2[/tex] = Height of pipe exit = 0
[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
From Bernoulli's equation we have
[tex]\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+Z_2\\\Rightarrow \dfrac{P_1}{\rho g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}\\\Rightarrow v_2=\sqrt{2g(\dfrac{P_1}{\rho g}+Z_1-\dfrac{P_2}{\rho g})}\\\Rightarrow v_2=\sqrt{2\times 9.81(\dfrac{120\times 10^3}{750\times 9.81}+0.15-\dfrac{101\times 10^3}{750\times 9.81})}\\\Rightarrow v_2=7.32\ \text{m/s}[/tex]
The exit velocity from the tube is [tex]7.32\ \text{m/s}[/tex].