Answer:
you will get huge electricity bills ............
the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic field for that wave at P
Answer:
[tex]6.63\times 10^8\ N/C[/tex]
Explanation:
Given that,
The magnitude of magnetic field, B = 2.21
We need to find the magnitude of the electric field. Let it is E. So,
[tex]\dfrac{E}{B}=c\\\\E=Bc[/tex]
Put all the values,
[tex]E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C[/tex]
So, the magnitude of the electric field is equal to [tex]6.63\times 10^8\ N/C[/tex].
ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on its circular path is 6.0 m/s. What is its kinetic energy at an instant when the string makes an angle of 50 degree with the vertical
Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J
Gradual shifting or movement of a time series to relatively higher or lower values over a longer period of time is called _____.
Answer:
Gradual shifting of a time series to relatively higher or lower values over a long period of time is called a Trend.
Imagine you were given a converging lens and a meter stick and sent outside on a sunny day. In a few sentences, describe a method to measure, as accurately as possible, the focal length of the lens using only the lens, a meter stick, and your outside surroundings. Explain your reasoning
Answer:
the Sun is the object that is at a very great distance and the focus point of the sun's shape is the distance to the magnesia, this image is equal to the seal distance
Explanation:
The method for measuring the focal length of a lens is based on the use of the constructor's equation
[tex]\frac{1}{f } = \frac{1}{p} + \frac{1}{q}[/tex]
where q and q are the distance to the object and the image respectively, f is the focal length.
If we place the object very far away (infinity) the equation remains
[tex]\frac{1}{f} = \frac{1}{q}[/tex]
Therefore with this we can devise a means for measuring the Sun is the object that is at a very great distance and the focus point of the sun's shape is the distance to the magnesia, this image is equal to the seal distance
A box that is sliding across the floor experiences a net force of 10.0 N. If the box has a mass of 1.50 kg, what is the resulting acceleration of the box g
Answer:
a = 6.67 m/s²
Explanation:
F = 10.0 N
m = 1.50 kg
a = ?
F = ma
10.0 = (1.50)a
6.67 = a
The angular velocity of an object is given by the following equation: ω(t)=(5rads3)t2\omega\left(t\right)=\left(5\frac{rad}{s^3}\right)t^2ω(t)=(5s3rad)t2 What is the angular displacement of the object (in rad) between t = 2 s and t = 4 s?
Answer:
The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.
Explanation:
The angular velocity of the object ([tex]\omega[/tex]), in radians per second, is given by the following expression:
[tex]\omega(t) = 5\cdot t^{2}[/tex] (1)
Where [tex]t[/tex] is the time, measured in seconds.
The change in the angular displacement ([tex]\Delta \theta[/tex]), in radians, is found by means of the following definite integral:
[tex]\Delta \theta = \int\limits^{4}_{2} {5\cdot t^{2}} \, dt[/tex] (2)
Then we proceed to integrate on the function in time:
[tex]\Delta \theta = \frac{5}{3}\cdot (4^{2}-2^{2})[/tex]
[tex]\Delta \theta = 20\,rad[/tex]
The angular displacement of the object between [tex]t = 2\,s[/tex] and [tex]t = 4\,s[/tex] is 20 radians.
A metre rule is used to measure the length of a piece of string in a certain experiment. It is found to be 20 cm long to the nearest millimeter. How should thisresult be recorded in a table of results? a. 0.2000m b. 0.200m c. 0.20m d. 0.2m
Answer:
C
Explanation:
20 cm = 0.2m
since uncertainty is 0.1 cm (0.001 m), should be recorded to same number of decimal place as uncertainty
therefore it's 0.200m
What is life like in a cave camp? Do you think you would like to experience this? Why or why not?
Answer:
There's no risk of animals or bad weather interfering with your campsite, either. You don't even really need a tent. A sleeping pad, sleeping bag and a mindful eye to pick up everything you brought in is all you really need to enjoy overnight caving. Do your research
Explanation:
A cylindrical tank with radius 7 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing (in m/min)?
Answer:
0.013 m/min
Explanation:
Applying,
dV/dt = (dh/dt)(dV/dh)............. Equation 1
Where
V = πr²h................ Equation 2
Where V = volume of the tank, r = radius, h = height.
dV/dh = πr²............ Equation 3
Substitute equation 3 into equation 1
dV/dt = πr²(dh/dt)
From the question,
Given: dV/dt = 2 m³/min, r = 7 m, π = 3.14
Substitute these values into equation 3
2 = (3.14)(7²)(dh/dt)
dh/dt = 2/(3.14×7²)
dh/dt = 0.013 m/min
A flat coil of wire is placed in a uniform magnetic field that is in the y-direction.
The magnetic flux through the coil is maximum if the coil is:_________.
(a) in the XY plane
(b) in either the XY or the YZ plane
(c) in the XZ plane
(d) in any orientation, because it is constant.
Answer:
The correct answer is c
Explanation:
Flow is defined by
Ф = B . A
bold letters indicate vectors.
The magnetic field is directed to the y axis, The area of the coil is represented by a vector normal to the plane of the coil, so to have a flux
i.i = j.j = k.k = 1
and the tori scalar products are zero
a) If the coil must be in the xy plane so that its normal vector is in the Z axis, so there is no flux
b) if the coil is in the plane yz the normal veto is in the x axis, so the flux is zero
C) If the coil is in XZ, the normal vector points in the y direction, usually the scalar product is one and there is a flux in this configuration
The correct answer is c
Would this pressure difference be greater or smaller if the scuba diver were in seawater (density 1050 kg/m3 ) and went to the same depth you calculated in question D1, took and held his breath, and then returned to the surface
Answer:
Greater.
Explanation:
This pressure difference will be greater if the scuba diver were in seawater and went to the same depth because the seawater have salts which increases the density of water as compared to freshwater. Salt in water increases the density which automatically increases the pressure on the diver so that's why we can say that the pressure will be increases for the scuba diver in seawater as compared to freshwater.
Which of the following categories of motion is mutually exclusive with each of the others? A. Translational motion B. Rectilinear motion C. Rotational motion D. Curvilinear motion
Answer:
C. Rotational motion
Explanation:
The kinematics of rotational motion describes the relationships between the angle of rotation, angular velocity, angular acceleration, and time. It only describes motion—it does not include any forces or masses that may affect rotation (these are part of dynamics). Recall the kinematics equation for linear motion: v = v+at (constant a).
Rotational motion is mutually exclusive with each of the others. Hence, option (C) is correct.
What is Rotational motion?"The motion of an object around a circular route, in a fixed orbit, is referred to as rotational motion."
Rotational motion dynamics are identical to linear or translational dynamics in every way. The motion equations for linear motion share many similarities with the equations for the mechanics of rotating objects. Rotational motion only takes stiff bodies into account. A massed object that maintains a rigid shape is referred to as a rigid body.
What is Curvilinear motion?Curvilinear motion is the movement of an object along a curved route. Example: A stone hurled at an angle into the air.
The motion of a moving particle that follows a predetermined or known curve is referred to as curvilinear motion. Two coordinate systems—one for planar motion and the other for cylindrical motion—are used to examine this type of motion.
Learn more about motion here:
https://brainly.com/question/22810476
#SPJ2
The cable lifting an elevator is wrapped around a 1.1m diameter cylinder that is turned by the elevator's motor. The elevator is moving upward at a speed of 2.6 ms. It then slows to a stop, while the cylinder turns one complete revolution. How long does it take for the elevator to stop? Express your answer to two significant figures and include the appropriate units.
Answer:
t = 2.7 s
Explanation:
This is a kinematics problem.
How the elevator reduces its speed to zero by a distance equal to the length of the cylinder
y = 2π r = 2 π d / 2
y = π d
y = π 1.1
y = 3,456 m
now we can look for the acceleration of the system
v² = v₀² - 2 a y
0 = v₀² - 2 a y
a = v₀² / 2y
a = 2.6² / 2 3.456
a = 0.978 m / s²
now let's calculate the time
v = v₀ - a t
0 = v₀ - at
t = v₀ / a
t = 2.6 /0.978
t = 2.658 s
ask for the result with two significant figures
t = 2.7 s
A small object A, electrically charged, creates an electric field. At a point P located 0.250 m directly north of A, the field has a value of 40.0 N/C directed to the south. If a second object B with the same charge as A is placed at 0.250m south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P?
Answer:
E_total = - 50 N / A
Explanation:
The electric field is a vector magnitude whereby
E_total = Eₐ + E_b
where the bold letters indicate vectors, in this case the charges of the two objects A and B are the same and they are on the same line
E_total = - E_a - E_b
the electric field for a point charge is
E_a = [tex]k \ \frac{q_a}{r_a^2 }[/tex]
qₐ= Eₐ rₐ² / k
indicates that Eₐ = 40.0 N / C
qₐ = 40.0 0.250²/9 10⁹
qₐ = 2.777 10⁻¹⁰ C
indicates that the charge of the two points is the same
qₐ = q_b
E_total = - k qₐ / rₐ² - k qₐ / (2 rₐ)²
E_total = [tex]-k \ \frac{q_a}{r_a^2} \ ( 1 + \frac{1}{4} )[/tex]
we calculate
E_total = - 40.0 (5/4)
E_total = - 50 N / A
When the insulation resistance between a motor winding and the motor frame is tested, the value obtained is 1.0 megohm (106 Ω). How much current passes through the insulation of the motor if the test voltage is 1000 V?
Answer:
0.001 A
Explanation:
Applying,
V = IR.............. Equation 1
Where V = Voltage of the motor, I = current, R = resistance
make I the subject of the equation
I = V/R.............. Equation 2
From the question,
Given: V = 1000 V, R = 1 MΩ = 10⁶ Ω
Substitute these values into equation 2
I = 1000/10⁶
I = 10⁻³ A
I = 0.001 A
A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the vehicles get stuck together. What is their velocity after the collision? A. 2.9 m/s east B. 1.6 m/s east m C. 2.6 m/s east D. 1.8 m/s east
Answer:
Explanation:
This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is
[tex][m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a[/tex] Filling in:
[tex][1200(4.5)+2100(0)]=[(1200+2100)v][/tex] which simplifies to
5400 + 0 = 3300v
so v = 1.6 m/s to the east, choice B
brainly A person's eye lens is 2.9 cm away from the retina. This lens has a near point of 25 cm and a far point at infinity. What must the focal length of this lens be in order for an object placed at the near point of the eye to focus on the retina
Answer: The focal length of the lens is 2.60 cm
Explanation:
The equation for lens formula follows:
[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]
where,
f = focal length = ? cm
v = image distance = 2.9 cm
u = Object distance = -25 cm
Putting values in above equation, we get:
[tex]\frac{1}{f}=\frac{1}{2.9}-\frac{1}{(-25)}\\\\\frac{1}{f}=\frac{1}{2.9}+\frac{1}{(25)}\\\\\frac{1}{f}=\frac{25+2.9}{2.9\times 25}\\\\f=\frac{72.5}{27.9}=2.60cm[/tex]
Hence, the focal length of the lens is 2.60 cm
If earth is compressed to the volume of moon, its acceleration due to gravity
* i. decreases
ii. remains same as before
iii. increases
iv. none of these
Answer:
increase
Explanation:
hope it will help you:)
State whether plastic is biodegradable or non-biodegradable ? Give reasons for your answer.
Answer:
non biodegradable
Explanation:
It is non biodegradable because plastic cannot dispose off easily ..
If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. What will its speed be, in meters per second, when it reaches the positive plate in this case
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
g Consider a mass-spring system where the spring constant is 5 N/m and the mass on the spring is 0.5 kg. The mass is moved a distance of -0.9 m from its equilibrium position. How much work is done by the spring
Answer:
The work done by the spring is 2.025 J
Explanation:
Given;
mass on the spring, m = 0.5 kg
spring constant, k = 5 N/m
extension of the spring, x = 0.9 m
The work done by the spring is calculated as;
[tex]W = \frac{1}{2} kx^2\\\\W = \frac{1}{2} \times 5 \times (0.9)^2\\\\W = 2.025 \ J[/tex]
Therefore, the work done by the spring is 2.025 J
The table below describes some features of methods used to generate electricity. Name method 4.
Answer:
Hydroelectricity
Explanation:
Because of flooding of water, we can assume that the electricity was generated by Water which is known as Hydroelectricity.
We can presume that the energy was produced by water because of the flooding of the water, which is a process known as hydroelectricity.
What is hydroelectricity?Hydroelectric power, often known as hydropower, is the name given to electricity generated by turbines that turn the potential energy of falling or swiftly running water into mechanical energy. As of 2019, hydropower accounted for more than 18% of the world's total power generation capacity, giving it the most frequently used renewable power source in the early 21st century.
When water is used to produce energy, it is first gathered or stored at a higher altitude and then transported through extensive pipelines or tunnels (called pen stocks) to a lower level; the difference between these two altitudes is referred to as the head. The falling water triggers the rotation of turbines at the bottom of its descent through the pipes.
To know more about Hydroelectricity:
https://brainly.com/question/15228003
#SPJ2
I need help with this physics question.
The acceleration will increase by 61.3%.
Explanation:
The centripetal acceleration [tex]a_c[/tex] is given by
[tex]a_c = \dfrac{v^2}{r}[/tex]
If the velocity of the object increases by 27.0%, then its new velocity v' becomes
[tex]v' = 1.270v[/tex]
The new centripetal acceleration [tex]a'_c[/tex] becomes
[tex]a'_c = \dfrac{(1.270v)^2}{r} = 1.613 \left(\dfrac{v^2}{r} \right)[/tex]
[tex]\:\:\:\:\:\:\:\:\:= 1.613a_c[/tex]
A seesaw made of a plank of mass 10.0 kg and length 3.00 m is balanced on a fulcrum 1.00 m from one end of the plank. A 20.0-kg mass rests on the end of the plank nearest the fulcrum. What mass must be on the other end if the plank remains balanced?
Answer:
7.5 kg
Explanation:
We are given that
[tex]m_1=10 kg[/tex]
Length of plank, l=3 m
Distance of fulcrum from one end of the plank=1 m
[tex]m_2=20 kg[/tex]
We have to find the mass must be on the other end if the plank remains balanced.
Let m be the mass must be on the other end if the plank remains balanced.
In balance condition
[tex]20\times 1=10\times (1.5-1)+m\times (1.5+0.5)[/tex]
[tex]20=10(0.5)+2m[/tex]
[tex]20=5+2m[/tex]
[tex]2m=20-5=15[/tex]
[tex]\implies m=\frac{15}{2}[/tex]
[tex]m=7.5 kg[/tex]
Hence, mass 7.5 kg must be on the other end if the plank remains balanced.
Answer:
The mass at the other end is 7.5 kg.
Explanation:
Let the mass is m.
Take the moments about the fulcrum.
20 x 1 = 10 x 0.5 + m x 2
20 = 5 + 2 m
2 m = 15
m = 7.5 kg
Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by strings with the same tension.a. Compare the final kinetic energies of pucks A and B. b. Compare the final speeds of pucks A and B.
Answer:
(a) 1 : 2
(b) same
Explanation:
Let the mass of puck A is m and the mass of puck B is 2 m.
initial speed for both the pucks is same as u and the distance is same for both is s.
let the tension is T for same.
The kinetic energy is given by
[tex]K = 0.5 mv^2[/tex]
(a) As the speed is same, so the kinetic energy depends on the mass.
So, kinetic energy of A : Kinetic energy of B = m : 2m = 1 : 2
(b) A the distance s same so the final velocities are also same.
(a) The kinetic energy of puck B is 2 times the kinetic energy of puck A.
(b) The final speed of both the puck A and B are same.
Let the mass of puck A is m and the mass of puck B is 2 m.
Initial speed for both the pucks is same as u and the distance is same for both is s.
Let the tension is T for same.
Then, the kinetic energy is given as,
[tex]KE = \dfrac{1}{2}mv^{2}[/tex]
(a)
As the speed is same, so the kinetic energy depends on the mass.
Then,
[tex]\dfrac{KE_{A}}{KE_{B}} = \dfrac{1/2 \times mv^{2}}{1/2 \times (2m)v^{2}}\\\\\\\dfrac{KE_{A}}{KE_{B}} =\dfrac{1}{2}[/tex]
So, kinetic energy of A : Kinetic energy of B = 1 : 2.
Thus, we can conclude that the kinetic energy of puck B is 2 times the kinetic energy of puck A.
(b)
The final speed for the puck is given as,
v = s/t
here, s is the distance covered.
Since, both pucks are pulled the same distance across frictionless ice. Then, the final speed of each puck is also same.
Thus, we can conclude that the final speed of both the puck A and B are same.
learn more about the kinetic energy here:
https://brainly.com/question/12669551
A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of
ax = 5.10 m/s2,
while the one on the back gives an acceleration component in the y direction of
ay = 7.30 m/s2.
The engines turn off after firing for 670 s, at which point the spacecraft has velocity components of
vx = 3670 m/s and vy = 4378 m/s.
What was the magnitude and the direction of the spacecraft's initial velocity before the engines were turned on? Express the magnitude as m/s and the direction as an angle measured counterclockwise from the +x axis.
magnitude m/s
direction ° counterclockwise from the +x-axis
Answer:
a) v = 517.99 m / s, b) θ = 296.3º
Explanation:
This is an exercise in kinematics, we are going to solve each axis independently
X axis
the acceleration is aₓ = 5.10 1 / S², they are on for t = 670 s and reaches a speed of vₓ= 3670 m / s, let's use the relation
vₓ = v₀ₓ + aₓ t
v₀ₓ = vₓ - aₓ t
v₀ₓ = 3670 - 5.10 670
v₀ₓ = 253 m / s
Y axis
the acceleration is ay = 7.30 m / s², with a velocity of 4378 m / s after
t = 670 s
v_y = v_{oy} + a_y t
v_{oy} = v_y - a_y t
v_oy} = 4378 - 7.30 670
v_{oy} = -513 m / s
to find the velocity modulus we use the Pythagorean theorem
v = [tex]\sqrt{v_o_x^2 + v_o_y^2}[/tex]
v = [tex]\sqrt{253^2 +513^2}[/tex]
v = 517.99 m / s
to find the direction we use trigonometry
tan θ ’= [tex]\frac{v_o_y}{v_o_x}[/tex]
θ'= tan⁻¹ [tex]\frac{voy}{voy}[/tex]
θ'= tan⁻¹ (-513/253)
tea '= -63.7
the negative sign indicates that it is below the ax axis, in the fourth quadrant
to give this angle from the positive side of the axis ax
θ = 360 - θ
θ = 360 - 63.7
θ = 296.3º
Two cylindrical resistors are made from copper. The first one is of length L and of radius r . The 2nd resistor is of length 6L and of radius 2r. The ratio of these two resistances R1/R2 is:
Answer:
[tex]R1/R2=\frac{2}{3}[/tex]
Explanation:
From the question we are told that:
1st's Length [tex]l=L[/tex]
1st's radius [tex]r=r[/tex]
2nd's Length [tex]l=6L[/tex]
2nd's radius [tex]r=2r[/tex]
Generally the equation for Resistance R is mathematically given by
[tex]R=\frac{\rho L}{\pi r^2}[/tex]
Therefore
[tex]R_1=\frac{\rho L}{\pi r^2}[/tex]
And
[tex]R_2=\frac{\rho 6L}{\pi (2r)^2}[/tex]
Therefore
[tex]R1/R2=\frac{\frac{\rho L}{\pi r^2}}{\frac{\rho 6L}{\pi (2r)^2}}[/tex]
[tex]R1/R2=\frac{2}{3}[/tex]
Write one advantage of MKS system over CGS system.
The diagram here shows an image being formed by a convex lens. Compared to the object at right, the image at left is-
larger and upright.
smaller and upright.
smaller and upside down.
larger and upside down.
Answer:
Smaller and upside down
Explanation:
To answer the question, we must recognise that the characteristics of the image formed by a convex lens depends on the position of the object from the lens.
From the diagram given in the question above, the following data were obtained:
1. The image is smaller than the object.
2. The image is inverted i.e upside down.
3. The image is closer to the lens
4. The image between 2f and f
Now, considering the options given in question above, the correct answer to the question is:
The image is smaller and upside down.
An object accelerates from rest, and after traveling 145 m it has a speed of 420 m/s. What was the acceleration of the object?
I am not sure how to calculate acceleration without being given the time directly.
Explanation:
Here,we've been given that,
Initial velocity (u) = 0 m/s (as it starts from rest)Distance (s) = 145 mFinal velocity (v) = 420 m/sWe've to find the acceleration of the object. By using the third equation of motion,
→ v² - u² = 2as
→ (420)² - (0)² = 2 × a × 145
→ 176400 - 0 = 290a
→ 176400 = 290a
→ 176400 ÷ 290 = a
→ 608.275862 m/s² = a
If you know initial speed and final speed, you can find the average speed. Then, knowing distance, you can find the time.
KimYurii posted the first answer to this question.
That answer is well organized, well presented, elegant and correct, and it deserves to be awarded "Brainliest" and several merit badges.
My problem is that I can never remember all the different formulas. I guess I had to work with so many uvum in all the Physics, Geometry, and Calculus classes that I took, I filled up all the memory slots with formulas, and over the years they all eventually merged into a big glob of goo. Now, the only formulas I can remember are the ones I had to use as an Electrical Engineer.
When I see this kind of question, I can only remember one or two simple formulas, and I reason it out like this:
Starting speed . . . zero
Ending speed . . . 420 m/s
Formula: Average speed . . . (1/2)·(0 + 420) = 210 m/s
Distance covered . . . 145 m
Formula: Time taken = (distance) / (average speed) = (145/210) second
(Now you have the time.)
Formula: Distance = (1/2)·(acceleration)·(time²)
145 m = (1/2)·(acceleration)·(145/210 sec)²
Acceleration = 290 m / (145/210 s)²
Acceleration = 608.28 m/s²