Answer:
A: magnitude and direction
B: Force that the field exerts on a test charge
C: its magnitude and direction is the same.
D: electrostatic machine
two rollers that are driven by a motor that generates a rotation
Explanation:
Two wires carry current I1 = 73 A and I2 = 31 A in the opposite directions parallel to the x-axis at y1 = 3 cm and y2 = 13 cm. Where on the y-axis (in cm) is the magnetic field zero?
Answer:
The position on the y-axis where the magnetic field is zero is at y = 10 cm
Explanation:
The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by
B = μ₀i/2πR
Now, let y be the point where the magnetic fields of both wires are equal.
So, the magnetic field due to wire 1 carrying current i₁ = 73 A is
B₁ = μ₀i₁/2π(y - 3) and
the magnetic field due to wire 2 carrying current i₂ = 31 A is
B₂ = μ₀i₂/2π(13 - y)
At the point where the magnetic field is zero, B₁ = B₂. So,
μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)
cancelling out μ₀ and 2π, we have
i₁/(x - y) = i₂/(13 - y)
cross-multiplying, we have
(13 - y)i₁ = (y - 3)i₂
Substituting the values of i₁ and i₂, we have
(13 - y)73 = (y - 3)31
949 - 73y = 31y - 93
Collecting like terms, we have
949 + 93 = 73y + 31y
1042 = 104y
dividing through by 104, we have
y = 1042/104
y = 10.02 cm
y ≅ 10 cm
So, the position on the y-axis where the magnetic field is zero is at y = 10 cm
5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be changed to raise the wave speed to 180 m/s?
Answer:
The tension on string when the speed was raised is 134.53 N
Explanation:
Given;
Tension on the string, T = 120 N
initial speed of the transverse wave, v₁ = 170 m/s
final speed of the transverse wave, v₂ = 180 m/s
The speed of the wave is given as;
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
where;
μ is mass per unit length
[tex]v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}[/tex]
The final tension T₂ will be calculated as;
[tex]T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N[/tex]
Therefore, the tension on string when the speed was raised is 134.53 N
g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before interacting. From this information, find the time interval required for the strong interaction to occur.
Answer:
Time, [tex]t=1.07\times 10^{-23}\ s[/tex]
Explanation:
Given that,
When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel [tex]3.2\times 10^{-15}\ m[/tex] before interacting.
Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,
[tex]t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s[/tex]
So, the time interval is [tex]1.07\times 10^{-23}\ s[/tex]
g To decrease the intensity of the sound you are hearing from your speaker system by a factor of 36, you can
Answer:
Increase the distance by a factor of 6.
Explanation:
The intensity at a distance r is given by :
[tex]I=\dfrac{P}{4\pi r^2}[/tex]
Here,
P is power emitted
r is distance from source
It means that the intensity is inversely proportional to the distance from the source.
To decrease the intensity of the sound you are hearing from your speaker system by a factor of 36, we can increase the distance by a factor of 6. Hence, this is the required solution.
) Calculate current passing in an electrical circuit if you know that the voltage is 8 volts and the resistance is 10 ohms
Explanation:
Hey, there!
Here, In question given that,
potential difference (V)= 8V
resistance (R)= 10 ohm
Now,
According to the Ohm's law,
V= R×I { where I = current}
or, I = V/R
or, I = 8/10
Therefore, current is 4/5 A or 0.8 A.
(A= ampere = unit of current).
Hope it helps...
An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase the magnitude of the induced emf? A. Straighten the wire out to be flat B. Reduce the resistance of the wire of which the loop is made C. Turning the plane of the loop to be parallel to the magnetic field D. Reducing the diameter of the loop
Answer:
changing the magnetic field more rapidly
Explanation:
According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.
Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.
12. A concave lens has a focal length of 10 cm. An object 2.5 cm high is placed 30 cm from the lens. Determine the position and size of the image. (3)
Answer:
I think 9.5
Explanation:
............
The ability of a water strider to move along the surface of water without breaking the surface is due to:
Answer:
The ability of a water strider to move along the surface of water without breaking the surface is due to:
SURFACE TENSION
Explanation:
this is because Water molecules are more attracted to each other than they are to other materials, so they generate a force to stay together called surface tension. Which allows the strider to move without breaking the surface
Suppose you observed the equation for a traveling wave to be y(x, t) = A cos(kx − ????t), where its amplitude of oscillations was 0.15 m, its wavelength was two meters, and the period was 2/15 s. If a point on the wave at a specific time has a displacement of 0.12 m, what is the transverse speed of that point?
Answer:
15m/sExplanation:
The equation for a traveling wave as expressed as y(x, t) = A cos(kx − [tex]\omega[/tex]t) where An is the amplitude f oscillation, [tex]\omega[/tex] is the angular velocity and x is the horizontal displacement and y is the vertical displacement.
From the formula; [tex]k =\frac{2\pi x}{\lambda} \ and \ \omega = 2 \pi f[/tex] where;
[tex]\lambda \ is\ the \ wavelength \ and\ f \ is\ the\ frequency[/tex]
Before we can get the transverse speed, we need to get the frequency and the wavelength.
frequency = 1/period
Given period = 2/15 s
Frequency = [tex]\frac{1}{(2/15)}[/tex]
frequency = 1 * 15/2
frequency f = 15/2 Hertz
Given wavelength [tex]\lambda[/tex] = 2m
Transverse speed [tex]v = f \lambda[/tex]
[tex]v = 15/2 * 2\\\\v = 30/2\\\\v = 15m/s[/tex]
Hence, the transverse speed at that point is 15m/s
A navigational beacon in deep space broadcasts at a radio frequency of 50 MHz. A spaceship approaches the beacon with a relative velocity of 0.40c. What is the frequency of the beacon radio signal that is detected on the ship? Answer in MHz with no decimal places.
A) 55 MHz
B) 60 MHz
C) 66 MHz
D) 71 MHz
E) 76 MHz
Answer:
To first order f' = f (1 - u/c) where observer and source are separating
f' = (1 + .4) = 70 MHz so (d) would be correct
The next term would be 1/2 (u/c)^2 = 1/2 * ,4^2 = .08 and the correction would be .08 * 50 = 4 Mhz
Most approximations would use (d).
A car travels down the road for 535 m in 17.3 s. What is the velocity of the car in m/s and in km/h?
Answer:
30.92m/sExplanation:
[tex]Distance = 535m\\Time = 17.3s\\\\Velocity = \frac{Distane}{Time} \\\\V = \frac{535m}{17.3s} \\\\Velocity = 30.92m/s[/tex]
[tex]Distance = 535m\\\\535m \:to \: km=0.535km\\\\Time = 17.3s\\\\17.3s = 0.004805556hours\\\\Velocity = \frac{Distance}{Time}\\\\ V= \frac{0.535}{0.004805556} \\\\ V=111.329469472\\\\=111.33km/h[/tex]
Snell's Law: Light goes from material having a refractive index of n 1 into a material with refractive index n 2. If the refracted light is bent away from the normal, what can you conclude about the indices of refraction
Answer:
a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.
b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2
Explanation:
The expression for the angle of refraction is
n₁ sin θ₁ = n₂ sin θ₂
refractive index n₁ is for incident light and n₂ is for transmitted light.
We have two cases
a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.
b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2
A fireworks rocket is launched vertically upward at 40 m/s. At the peak of its trajectory, it explodes into two equal-mass fragments. One reaches the ground t1 = 2.71s after the explosion.When does the second reach the ground?t=?
Answer:
6.13 seconds
Explanation:
At the peak of the fireworks trajectory, the velocity of the firework would be zero. Using equation of motion, we have:
v² = u² + 2gh
0 = 40² - (2)(9.81)(h)
0 = 1600 - 19.62h
19.62h = 1600
h = 1600/19.62
h = 81.55 m
Now during the process of explosion, the two parts gained equal vertical momentum but in opposite directions.
We are told the first piece lands in a time of 2.71 s,
Using 3rd equation of motion, we have;
h = ut + ½gt²
81.55 = u(2.71) + ½(9.81 × 2.71²)
81.55 = 2.71u + 36.0228
2.71u = 81.55 - 36.0228
2.71u = 45.5272
u = 45.5272/2.71
u = 16.8 m/s
The time it takes a projectile to return back to its original launch point assuming the projectile was launched
vertically with speed u = 16.8 m/s is;
t = 2u/g
t = (2 × 16.8)/9.81
t = 3.43 s
Thus total time it takes the second mass to reach the ground = 3.43 + 2.71 = 6.13 seconds
Kinetic and
A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is holding the brick in place?
11 point)
O raise the board to a higher angle
O press down on the brick in a direction that is perpendicular to the board
O roughen up the texture of the wooden board
o lower the board so it's level with the ground
Answer:
raise the board to a higher angle
Explanation:
Static friction is the force opposite to the applied force.
Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.
So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.
Hence, the correct option is "raise the board to a higher angle".
How to do this question
Answer:
(a) 10 m/s
(b) 22.4 m/s
Explanation:
(a) Draw a free body diagram of the car when it is at the top of the loop. There are two forces: weight force mg pulling down, and normal force N pushing down.
Sum of forces in the centripetal direction (towards the center):
∑F = ma
mg + N = mv²/r
At minimum speed, the normal force is 0.
mg = mv²/r
g = v²/r
v = √(gr)
v = √(10 m/s² × 10.0 m)
v = 10 m/s
(b) Energy is conserved.
Initial kinetic energy + initial potential energy = final kinetic energy
½ mv₀² + mgh = ½ mv²
v₀² + 2gh = v²
(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²
v = 22.4 m/s
Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first dark fringes on either side of the central maximum is 4.70 mm . Part A What is the width of the slit
Answer:
The width of the slit is 0.4 mm (0.00040 m).
Explanation:
From the Young's interference expression, we have;
(λ ÷ d) = (Δy ÷ D)
where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.
Thus,
d = (Dλ) ÷ Δy
D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×[tex]10^{-9}[/tex] m)
d = (3.30 × 563 ×[tex]10^{-9}[/tex] ) ÷ (0.0047)
= 1.8579 × [tex]10^{-6}[/tex] ÷ 0.0047
= 0.0003951 m
d = 0.00040 m
The width of the slit is 0.4 mm (0.00040 m).
"How many wavelengths wide must a single slit be if, at a point 8o from the central maximum, there is a 72 rad phase difference between the top and bottom rays?"
Answer
82.3 wavelengths
A plastic balloon that has been rubbed with wool will stick to a wall.
a. Can you conclude that the wall is charged? If not, why not? If so, where does the charge come from?
b. Draw a series of charge diagrams showing how the balloon is held to the wall.
Answer:
Explanation:
When plastic balloon is rubbed with wool , charges are created on both balloon and silk in equal amount . Rubber balloon will acquire negative charge and silk will acquire positive charge .
Now when balloon is brought near a wall , there is induction of charge on the wall due to charge on the balloon . On the near surface of wall positive charge is produced and on the surface deep inside the wall negative charge is produced . The charge deep inside goes inside the earth but the positive charge near the surface of wall can not escape . It remains trapped by negative charge on the balloon .
hence there is mutual attraction between balloon and surface of wall is just like attraction between opposite charges . But once the ballon due to mutual attraction comes in contact with the wall , the charge on balloon and on wall neutralises each other and hence after some time the balloon falls off from the wall on the ground . It does not remain attracted to wall for ever . It happens due to neutralisation of charges on balloon and wall .
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram.
Answer:
a) k = 95.54 N / m, c = 19.55 , b) m₃ = 0.9078 kg
Explanation:
In a simple harmonic movement with friction, we can assume that this is provided by the speed
fr = -c v
when solving the system the angular value remains
w² = w₀² + (c / 2m)²
They give two conditions
1) m₁ = 1 kg
f₁ = 1.1 Hz
the angular velocity is related to frequency
w = 2π f₁
Let's find the angular velocity without friction is
w₂ = k / m₁
we substitute
(2π f₁)² = k / m₁ + (c / 2m₁)²
2) m₂ = 2 kg
f₂ = 0.8 Hz
(2π f₂)² = k / m₂ + (c / 2m₂)²
we have a system of two equations with two unknowns, so we can solve it
we solve (c / 2m)² is we equalize the expression
(2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁
k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)
k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)
a) Let's calculate
k = 4 π² (0.8² -1.1²) / (½ -1/1)
k = 39.4784 (1.21) / (-0.5)
k = 95.54 N / m
now we can find the constant of friction
(2π f₁) 2 = k / m₁ + (c / 2m₁)²
c2 = ((2π f₁)² - k / m₁) 4m₁²
c2 = (4ππ² f₁² - k / m₁) 4 m₁²
let's calculate
c² = (4π² 1,1² - 95,54 / 1) 4 1²
c² = (47.768885 - 95.54) 8
c² = -382.1689
c = 19.55
b) f₃ = 0.2 Hz
m₃ =?
(2πf₃)² = k / m₃ + (c / 2m₃) 2
we substitute the values
(4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²
1.579 = 95.54 / m₃ + 95.542225 / m₃²
let's call
x = 1 / m₃
x² = 1 / m₃²
- 1.579 + 95.54 x + 95.542225 x² = 0
60.5080 x² + 60.5080 x -1 = 0
x² + x - 1.65 10⁻² = 0
x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2
x = [1 ± 1.03] / 2
x₁ = 1.015 kg
x₂ = -0.015 kg
Since the mass must be positive we eliminate the second results
x₁ = 1 / m₃
m₃ = 1 / x₁
m₃ = 1 / 1.1015
A flatbed truck is supported by its four drive wheels, and is moving with an acceleration of 7.4 m/s2. For what value of the coefficient of static friction between the truck bed and a cabinet will the cabinet slip along the bed surface?
Answer:
The value is [tex]\mu = 0.76[/tex]
Explanation:
From the question we are told that
The acceleration is [tex]a = 7.4 \ m /s^2[/tex]
Generally the force by which the truck bed (truck) is moving with is mathematically represented as
[tex]F = ma[/tex]
Now for the truck cabinet to slip from the truck bed then the frictional force between the truck cabinet is equal the force by which the the truck bed is moving with that is
[tex]F_f = F[/tex]
Here [tex]F_f[/tex] is the frictional force which is mathematically represented as
[tex]F_f = \mu * m * g[/tex]
substituting into above equation
[tex]\mu * m * g = ma[/tex]
=> [tex]\mu = \frac{a}{g}[/tex]
substituting values
[tex]\mu = \frac{ 7.4 }{ 9.8}[/tex]
[tex]\mu = 0.76[/tex]
Peer assessment is a unique educational model. Think back to how you felt about peer assessment at the beginning of the term, and compare that to your feeling now. How have your feeling changed? Are you more comfortable with peer assessment? Have you learned something new while assessing your peer's work?
Answer:
In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.
for an answer to be complete,the units needs to be specified.why
Explanation:
unit is necessary to communicate values of the physical quantity for example can main to someone a particular length without using some sort of unit is impossible because a length cannot be described without a reference used to make sense of the value given
A fan rotating with an initial angular velocity of 1500 rev/min is switched off. In 2.5 seconds, the angular velocity decreases to 400 rev/min. Assuming the angular acceleration is constant, answer the following questions.
How many revolutions does the blade undergo during this time?
A) 10
B) 20
C) 100
D) 125
E) 1200
Answer:
The blade undergoes 40 revolutions, so neither of the given options is correct!
Explanation:
The revolutions can be found using the following equation:
[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
Where:
α is the angular acceleration
t is the time = 2.5 s
[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min
First, we need to find the angular acceleration:
[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]
Now, the revolutions that the blade undergo are:
[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]
Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!
I hope it helps you!
The Milky Way has a diameter (proper length) of about 1.2×105 light-years. According to an astronaut, how many years would it take to cross the Milky Way if the speed of the spacecraft is 0.890 c?
Answer:
t = 134834.31 years
Explanation:
First we find the speed of the ship:
v = 0.890 c
where,
v = speed of the ship = ?
c = speed of light = 3 x 10⁸ m/s
Therefore, using the values, we get:
v = (0.89)(3 x 10⁸ m/s)
v = 2.67 x 10⁸ m/s
Now, we find the distance in meters:
Distance = s = (1.2 x 10⁵ light years)(9.461 x 10¹⁵/1 light year)
s = 11.35 x 10²⁰ m
Now, for the time we use the following equation:
s = vt
t = s/v
t = (11.35 x 10²⁰ m)/(2.67 x 10⁸ m/s)
t = (4.25 x 10¹² s)(1 h/3600 s)(1 day/24 h)(1 year/365 days)
t = 134834.31 years
Explain how blood circulation takes place in humans?
Blood comes into the right atrium from the body, moves into the right ventricle and is pushed into the pulmonary arteries in the lungs. After picking up oxygen, the blood travels back to the heart through the pulmonary veins into the left atrium, to the left ventricle and out to the body's tissues through the aorta.
Hope it helps you
Mark my answer as brainlist
have a nice day
A bungee cord with a spring constant of 800 StartFraction N over m EndFraction stretches 6 meters at its greatest displacement. How much elastic potential energy does the bungee cord have? The bungee cord has J of elastic potential energy.
Explanation:
EE = ½ kx²
EE = ½ (800 N/m) (6 m)²
EE = 14,400 J
Answer:
14,400 J
Explanation:
Its the answer
how much is need to lift a load of 100n placed at a distance of 29 cm from fulcrum if effort is applied at 60cm from the fulcrum on opposite side of the load? calculate mechanical advantage and velocity ratio of the lever
Answer:
206.8965517 n
Explanation:
First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.
1. Why do only some people get addicted to
drugs?
Answer:
When drugs are taken in are body are brain release dopamine: which make us feel so pleasure and good, and for this some people are addicted to drugs which makes them feel good. on other hand damaging their health.
Open the sash half way up, take the beaker containing the dry ice / water out of the hood, and slowly move it from right in front of the hood all the way down to the floor. At what point do the fumes stop getting sucked up by the fume hood?
Answer:
The fumes stop getting sucked up by the fume hood once the beaker is pulled out of the hood.
Explain why the two plates of a capacitor are charged to the same magnitude when a battery is connected to the capacitor.
Answer:
This is because the same electron removed from the positively charged plate is what is taken to the negatively charged plate, maintaining the same amount of electron according to the conservation of charge in an electric circuit.
Explanation:
In any circuit, electrons are neither created nor destroyed according to the laws of conservation of charge, but are transferred from one point to another on the circuit. When the plates of a capacitor are connected to battery, the battery pushes the electron to move due to its potential difference. Electrons are then moved from the positive plate, at a steady rate, to the negative plate. The removal of electrons from the positive plate is what leaves it positively charged from deficiency of electrons, and the addition of electrons at the negatively charged plate is what leaves the plate negatively charge from excess of electrons. From this, we can see that the same electrons removed from the positively charged plate are taken to the negatively charged plate.