Answer:
[tex]4.61\times 10^{-5}\ \text{T}[/tex]
[tex]1.05\times 10^{-6}\ \text{Nm}[/tex]
Explanation:
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi10^{-7}\ \text{H/m}[/tex]
[tex]r_l[/tex] = Radius of loop = 15 cm
[tex]I_l[/tex] = Current in loop = 11 A
[tex]r_c[/tex] = Radius of coil = 0.76 cm
N = Number of turns of coil = 66
[tex]I_c[/tex] = Current in coil = 1.9 A
Magnetic field is given by
[tex]B=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}\times 11}{2\times 0.15}\\\Rightarrow B=4.61\times 10^{-5}\ \text{T}[/tex]
Magnitude of magnetic field produced by the loop at its center is [tex]4.61\times 10^{-5}\ \text{T}[/tex].
Torque is given by
[tex]\tau=BI_c\pi r_c^2N\sin90^{\circ}\\\Rightarrow \tau=4.61\times 10^{-5}\times 1.9\times \pi\times (0.76\times 10^{-2})^2\times 66\sin90^{\circ}\\\Rightarrow \tau=1.05\times 10^{-6}\ \text{Nm}[/tex]
Magnitude of torque on the coil due to the loop is [tex]1.05\times 10^{-6}\ \text{Nm}[/tex]
Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of mA = 16.7 kg and an initial velocity of = 7.26 m/s, due east. Object B, however, has a mass of mB = 29.3 kg and an initial velocity of = 4.39 m/s, due north. Find the (a) magnitude and (b) direction of the total momentum of the two-object system after the collision.
Answer:
a) v = 3,843 m / s, b) 46.7º North- East
Explanation:
Moment is a vector quantity, so one of the best ways to solve this problem is to solve each component separately.
The system is formed by the two vehicles so that the moment is preserved during the crash
Direction to the East
initial instant. Before the crash
p₀ = mₐ vₐ₀
final insttne. After the crash
p_f = (mₐ + m_b) vₓ
p₀ = p_f
mₐ vₐ₀ = (mₐ + m_b) vₓ
vₓ = [tex]\frac{m_a}{m_a + m_b} \ v_{ao}[/tex]
let's calculate
vₓ = [tex]\frac{16.7}{16.7 + 29.3} \ 7.26[/tex]
vₓ = 2,636 m / s
direction north
initial p₀ = m_b v_{bo}
final p_f = (mₐ + m_b) v_y
p₀ = p_f
m_b v_{bo} = (mₐ + m_b) v_y
v_y = [tex]\frac{m_b}{m_a+m_b} \ v_{bo}[/tex]
let's calculate
v_y = [tex]\frac{29.3}{16.7 + 29.3} \ 4.39[/tex]
v_y = 2.796 m / s
the final speed of the two two vehicles is
v = (2,636 i ^ + 2,796 j ^) m / s
a) the magnitude of the velocity
let's use the Pythagorean theorem
v = [tex]\sqrt{v_x^2 + v_y^2}[/tex]
v = [tex]\sqrt{2.636^2 + 2.796^2}[/tex]
v = 3,843 m / s
b) let's use trigonometry to find the direction
tan θ = v_y / vₓ
θ = tan⁻¹ v_y / vₓ
θ = tan⁻¹ (2,796 / 2,636)
θ = 46.7º
This direction is 46.7º North East
Describe the formation of the land, the atmosphere, and the oceans of earth
Which of the following is an example of heat transfer by conduction?
A. Heat is transferred to the air above a candle flame.
B. Heat is transferred to the soil on a sunny day.
c. Heat is transferred to your hand from a warm cup.
D. Heat is transferred to the air from a warm lightbulb.
Answer:
option c
Heat is transferred to your hand from a warm cup
conduction is the process of transferring of heat from one material to another when they are in contact
hope it helps
an object is acted by force of 22 newtons to the right and a force of 13 newtons to left
Answer:
Explanation:
I'm going to guess that you want the net force.
These two forces are acting in opposite directions. Therefore the forces are subtracted in effect.
F = F1 - F2
F1 = 22N to the right
F2 = 13N to the left
F = 22 - 13 = 9 N to the right.
Specifying the direction is very important. Forces do have directions and you must specify what that is. Otherwise, the question should be marked incorrect.
PLEASE HELP NO LINKS NEED HELP FAST
Use the scenario to answer the question.
An astronomer discovers a new galaxy using a telescope. The astronomer wants to investigate how the galaxy is moving relative to the Milky Way galaxy.
In one or two sentences, make a hypothesis about the movement of the galaxy and explain at least one way to test the hypothesis.
Answer:
The galaxies outside of our own are moving away from us, and the ones that are farthest away are moving the fastest. This means that no matter what galaxy you happen to be in, all the other galaxies are moving away from you
Explanation:
The hypothesis about the movement of the galaxy is that galaxies are moving far from each other continuously.
What is the milky way galaxy?The milky way galaxy is a galaxy that contains over a hundred billion stars and it also includes our solar system. Its name describes its appearance when viewed from the earth. All the individual stars in the whole sky are a portion of the Milky Way Galaxy, the term "Milky Way" is because of the band of light.
The astronomer has discovered a new galaxy which means our universe is continuously expanding. This is because the universe encloses everything that exists.
Galaxies are moving in space and since the universe space is continuously expanding so the galaxies continuously move from each other. The farther the galaxy is from the milky way which is an observable part, the faster will be moving the galaxy and the closer the galaxy is to the milky way, the slower will be movement of the galaxy.
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The rollercoaster is near the
bottom of the hill after the first big
drop. Which best describes the
potential and kinetic energy?
A. It has mostly kinetic energy.
B. It has mostly potential energy.
C. The kinetic energy is decreasing.
D. The potential energy is about 50%
and increasing.
Answer:
A. It has mostly kinetic energy
Explanation:
Kinetic energy refers to movement. Potential energy refers to height. In this case, the big drop just got over. So, when the coaster is at the bottom, it has more kinetic than potential energy . Potential energy is still present but kinetic is more at the bottom.
The steering wheel of a car has a radius of 0.19 m, and the steering wheel of a truck has a radius of 0.25 m. The same force is applied in the same direction to each steering wheel. What is the ratio of the torque produced by this force in the truck to the torque produced in the car
Answer:
[tex]\frac{T_t}{T_c} = 1.32[/tex]
Explanation:
The torque applied on an object can be calculated by the following formula:
[tex]T = Fr[/tex]
where,
T = Torque
F = Applied Force
r = radius of the wheel
For car wheel:
[tex]T_c = Fr_c\\[/tex]
For truck wheel:
[tex]T_t = Fr_t[/tex]
Dividing both:
[tex]\frac{T_t}{T_c} = \frac{Fr_t}{Fr_c}[/tex]
for the same force applied on both wheels:
[tex]\frac{T_t}{T_c} = \frac{r_t}{r_c} \\[/tex]
where,
rt = radius of the truck steering wheel = 0.25 m
rc = radius of the car steering wheel = 0.19 m
Therefore,
[tex]\frac{T_t}{T_c} = \frac{0.25\ m}{0.19\ m} \\[/tex]
[tex]\frac{T_t}{T_c} = 1.32[/tex]
A rod that is 96.0 cm long is made of glass that has an index of refraction equal to 1.60. The rod has its ends ground to convex spherical surfaces that have radii equal to 8.00 cm and 19.1 cm. An object is in air on the long axis of the rod 19.9 cm from the end that has the 19.1-cm radius.
(a) Find the image distance due to refraction at the 19.1-cm radius surface.
(b) Find the position of the final image due to refraction at both surfaces.
(c) Is the final image real or virtual?
Please help me someone !
Answer:
The object is moving at constant speed.
Explanation:
The spaces between the dots are equal.
TIME REMAINING
15:56:15
A plant root is an example of
Type here to search
Answer:
h
Explanation:
what happens during subduction
Answer:
Subduction , Latin for "carried under," is a term used for a specific type of plate interaction. It happens when one lithospheric plate meets another—that is, in convergent zones —and the denser plate sinks down into the mantle.
Fluid mechanics questions and answers
Answer:
Fluid mechanics is considered one of the toughest subdisciplines within mechanical and aerospace engineering. It is unique from almost any other field an undergraduate engineer will encounter. It requires viewing physics in a new light, and that's not always an easy jump to make.
Which of the following is NOT true about
newspaper weather maps?
A. They report the temperature of the area in degrees
Fahrenheit.
B. They let you know how much precipitation to expect.
C. They provide more detail than weather service maps.
D. They tell you about the warm and cold air fronts.
Two points on a progressive wave are out of phase by 0.41 rad.
What is this phase difference?
[1 mark]
A 23° [e]
B 47° [e]
C 74° [e]
D 148° [e]
The phase difference between two points on a progressive wave are out of phase by 0.41 rad is 23°.
What is an equation?An equation is an expression that shows the relationship between two or more numbers and variables.
The two points on a progressive wave are out of phase by 0.41 rad.
Hence, Phase difference = 0.41 rad
But:
Rad to degree = (rad * 180/π)°
Hence:
0.41 rad = (0.41 rad * 180/π) = 23°
The phase difference between two points on a progressive wave are out of phase by 0.41 rad is 23°.
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1.
is the rate that velocity changes
O Acceleration
O
Time
O
Distance
O
Mass
Answer:
a) acceleration
Explanation:
Acceleration is, by definition, the change of an object's velocity.
A wave has a frequency of 2 Hz. Find its period
Which of the following is true of the
thermocline layer of the ocean?
A. rapidly decreases in temperature
B. warmest and least dense of the ocean layers
C. is the bottom layer of the ocean
D. is the top layer of the ocean
Answer:
d is trueeeeeeeeeeeeeeeee
Humans impact the Earth in good AND bad ways.
A) True
B) False
Answer:
True
Explanation:
yes we can see that we are helping animals but we create pollution which is very bad
stainless steel, tell us about its properties and what should be taken into account when using it?
g A velocity selector consists of crossed electric and magnetic fields. The electric field has a magnitude of 480 N/C and is in the negative z direction. What should the magnetic field (magnitude and direction) be to select a proton moving in the negative x direction with a velocity of 3.50 cross times 10 to the power of 5 m/s to go un-deflected
Answer:
B = 1.37 mT
Explanation:
Given that,
The magnitude of the electric field, E = 480 N/C
The speed of the proton, [tex]v=3.50 \times 10^5\ m/s[/tex]
We need to find the magnitude of the magnetic field. In a velocity selector, the electric field is balanced by the magnetic field. So,
[tex]qE=qvB[/tex]
Where
B is the magnetic field
[tex]B=\dfrac{E}{v}\\\\B=\dfrac{480}{3.5\times 10^5}\\\\B=1.37\times 10^{-3}\ T\\\\or\\\\B =1.37\ mT[/tex]
So, the magnetic field is equal to 1.37 mT.
A turntable, with a mass of 1.5 kg and diameter of 20 cm, rotates at 70 rpm on frictionless bearings. Two 540 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick to it.
Required:
What is the turntable's angular speed, in rpm, just after this event?
Answer:
The turntable's angular speed after the event is 28.687 revolutions per minute.
Explanation:
The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:
[tex]I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}[/tex] (1)
Where:
[tex]I_{T}[/tex] - Moment of inertia of the turntable, in kilogram-square meters.
[tex]r[/tex] - Distance of the block regarding the center of the turntable, in meters.
[tex]m[/tex] - Mass of the object, in kilograms.
[tex]\omega_{o}[/tex] - Initial angular speed of the turntable, in radians per second.
[tex]\omega_{f}[/tex] - Final angular speed of the turntable-objects system, in radians per second.
In addition, the momentum of inertia of the turntable is determined by following formula:
[tex]I_{T} = \frac{1}{2}\cdot M\cdot r^{2}[/tex] (2)
Where [tex]M[/tex] is the mass of the turntable, in kilograms.
If we know that [tex]\omega_{o} \approx 7.330\,\frac{rad}{s}[/tex], [tex]M = 1.5\,kg[/tex], [tex]m = 0.54\,kg[/tex] and [tex]r = 0.1\,m[/tex], then the angular speed of the turntable after the event is:
[tex]I_{T} = \frac{1}{2}\cdot M\cdot r^{2}[/tex]
[tex]I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}[/tex]
[tex]I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}[/tex]
[tex]\omega_{f} = \frac{I_{T}\cdot \omega_{o}}{2\cdot r^{2}\cdot m +I_{T}}[/tex]
[tex]\omega_{T} = 3.004\,\frac{rad}{s}[/tex] ([tex]28.687\,\frac{rev}{min}[/tex])
The turntable's angular speed after the event is 28.687 revolutions per minute.
Can anybody help in number 6? <3
Answer:
5.4 will be the weight in illustrate form
It is almost as if each outer planet is a solar system in its own right.
True or False
A charged particle (charge 1.6x10-19 C and mass 1.67x10-27 kg) is initially moving with a velocity of 2x105 m/s and then moves into a region having a magnetic field and an electric field (6x104 V/m). The direction of initial velocity is perpendicular to the electric field and magnetic field. If the charged particle keep moving in the original direction without being deflected, what is the magnitude of the magnetic field
Answer:
[tex]0.3\ \text{T}[/tex]
Explanation:
q = Charge = [tex]1.6\times 10^{-19}\ \text{C}[/tex]
m = Mass of particle = [tex]1.67\times 10^{-27}\ \text{kg}[/tex]
v = Velocity = [tex]2\times 10^5\ \text{m/s}[/tex]
E = Electric field = [tex]6\times 10^4\ \text{V/m}[/tex]
B = Magnetic field
Magnetic field is given by
[tex]B=\dfrac{E}{v}\\\Rightarrow B=\dfrac{6\times 10^4}{2\times 10^5}\\\Rightarrow B=0.3\ \text{T}[/tex]
The magnitude of magnetic field is [tex]0.3\ \text{T}[/tex]
Select all the correct answers.
Which statements about our solar system are false?
Our solar system is made up of the Sun and other objects that orbit the Sun.
The Sun is the only star in our solar system.
Dwarf planets have several other bodies in their path orbiting the Sun just as they do.
The Kuiper Belt is between Uranus and Neptune.
A large number of irregularly shaped comets are located in a vast ring between the orbits of Mars and Jupiter.
D and E are false.
A, B, and C are true.
Answer:
- Our solar system is made up of the Sun and other objects that orbit the Sun.
- The Sun is the only star in our solar system.
- Dwarf planets have several other bodies in their path orbiting the Sun just as they do.
Explanation:
plato
A 101 kg basketball player crouches down 0.380 m while waiting to jump. After exerting a force on the floor through this 0.380 m, his feet leave the floor and his center of gravity rises 0.920 m above its normal standing erect position. (a) Using energy considerations, calculate his velocity (in m/s) when he leaves the floor. m/s (b) What average force (in N) did he exert on the floor
Answer:
[tex]4.25\ \text{m/s}[/tex]
[tex]3391.22\ \text{N}[/tex]
Explanation:
y = Height of compression = 0.38 m
m = Mass of basketball player = 101 kg
h = Height of center of gravity after jump = 0.92 m
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Energy balance of the system is given by
[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.92}\\\Rightarrow v=4.25\ \text{m/s}[/tex]
The velocity of the player when he leaves the floor is [tex]4.25\ \text{m/s}[/tex]
[tex]Fy=mgy+\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{mgy+\dfrac{1}{2}mv^2}{y}\\\Rightarrow F=\dfrac{101\times 9.81\times 0.38+\dfrac{1}{2}\times 101\times 4.25^2}{0.38}\\\Rightarrow F=3391.22\ \text{N}[/tex]
The force exerted on the floor is [tex]3391.22\ \text{N}[/tex].
A student hangs a block from a light string that is attached to a massive pulley of unknown radius R, as shown in the figure. The student allows the block to fall from rest to the floor. Which two of the following sets of data that could be measured or determined should the student use together to determine the final angular velocity of the pulley just before the block hits the floor? Select two answers. Justify your selections.
Answer:
The mass of the block, the distance of the block above the floor, and the time it takes the block to reach the floor, because these quantities can be used to determine the acceleration of the block.
The radius and the mass of the pulley, because these quantities can be used together to determine the rotational inertia of the pulley.
Explanation:
If the motion starts from rest, the initial angular velocity will be zero and the final angular velocity can be determined with the product of angular acceleration and time of motion of the pulley.
Angular velocity is defined as the change in the angular displacement per change in time of motion. This can be expressed mathematically as follows;
[tex]\omega = \frac{\Delta \theta}{\Delta t} = vr[/tex]
where;
Ф is the angular displacementt is the time of the motionv is the linear velocityr is the radius of the circular path.In a circular motion that starts from rest and ends with final velocity, the equation is given as;
[tex]\omega_f =\omega_i + \alpha t[/tex]
Where;
[tex]\omega_f[/tex] is the final angular velocity[tex]\omega_i[/tex] is the initial angular velocity[tex]\alpha[/tex] is the angular accelerationThus, if the motion starts from rest, the initial angular velocity will be zero and the final angular velocity can be determined with the product of angular acceleration and time of motion of the pulley.
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A belt of negligible mass passes between cylinders A and B and is pulled to the right with a force P. Cylinders A and B weigh, respectively, 5 and 20 lb. The shaft of cylinder A is free to slide in a vertical slot and the coefficients of friction between the belt and each of the cylinders are µs = 0.50 and µk = 0.40. For P = 3.6 lb,
Determine:
(a) Whether slipping occurs between the belt and either cylinder,
(b) The angular acceleration of each cylinder.
Slipping doesn't occur between the belt and cylinder B because the force of static friction is greater than force exerted on cylinder B.
Given the following data:
Mass A = 5 lb to kg = 2.27 kg.
Mass B = 20 lb to kg = 9.02 kg.
Force = 3.6 lb to N = 16.02 Newton.
How to calculate angular acceleration.In order to calculate the angular acceleration of each cylinder, we would take moment about the two cylinders.
For cylinder A:
[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_A\alpha _A = F_A (0.1)\\\\(\frac{m_Ar^2}{2}) \alpha _A = F_A (0.1)\\\\(\frac{2.27 \times 0.1^2}{2}) \alpha _A = F_A (0.1)\\\\0.1F_A=0.01135\alpha _A\\\\F_A=\frac{0.01135\alpha _A}{0.1} \\\\F_A= 0.1135\alpha _A[/tex]
For cylinder B:
[tex]\sum M_G=\sum (M_G)_{eff}\\\\I_B\alpha _B = F_B (0.2)\\\\(\frac{m_Br^2}{2}) \frac{\alpha _A}{2} = F_B (0.2)\\\\(\frac{9.02 \times 0.1^2}{4}) \alpha _A = F_B (0.2)\\\\0.1F_B=0.02255\alpha _A\\\\F_B=\frac{0.02255\alpha _A}{0.2} \\\\F_B= 0.1128\alpha _A[/tex]
For the belt, we have
[tex]\sum F_A =0\\\\P-F_B-F_A=0\\\\16.02-0.1128\alpha _A-0.1135\alpha _A=0\\\\16.02=0.2263\alpha _A\\\\\alpha _A=\frac{16.02}{0.2263} \\\\\alpha _A=70.79 \;rad/s^2[/tex]
Also, we would determine the angular acceleration of cylinder B:
[tex]0.1\alpha _A=0.2\alpha _B\\\\0.1 \times 70.79 = 0.2\alpha _B\\\\7.079= 0.2\alpha _B\\\\\alpha _B=\frac{7.079}{0.2} \\\\\alpha _B=35.40\;rad/s^2[/tex]
Next, we would calculate the forces acting on the cylinders:
[tex]F_A = 0.1135\alpha _A\\\\F_A = 0.1135 \times 70.79\\\\F_A = 8.04 \;Newton[/tex]
[tex]F_B = P-F_A\\\\F_B = 16.02 - 8.04\\\\F_B = 7.98\;Newton[/tex]
Next, we would determine the force of static friction:
[tex]F_s = \mu_s N = \mu_s m_B g\\\\F_s = 0.50 \times 9.02 \times 9.8\\\\F_s=44.198\;Newton[/tex]
From the above calculation, we can deduce that the force of static friction is greater than force exerted on cylinder B. Therefore, slipping doesn't occur between the belt and cylinder B.
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A 1.8 kg book has been dropped from the top of the football stadium. Its speed is 4.8 m/s when it is 2.9 m above the ground. How high is the stadium?
Answer:
the height of the stadium is 4 m
Explanation:
The computation of the height of the stadium is shown below:
but before that total mechanic energy should be determined
E = PE + KE
where
PE = mgh
and, KE = 1 ÷2 mv^2
Now
E = mgh + 1 ÷2 mv^2
= (1.8) (9.8) (2.9) + 1 ÷ 2 (1.8) (4.8)^2
= 71.9J
= 72J
Now the height of the stadium is
TE = mgh
72 = (1.8) × (9.8) × h
So, h = 4 m
Hence, the height of the stadium is 4 m
What is the equation for frequency?
a. number of cycles +unit of time
b. number of cycles - unit of time
c. number of cycles ×unit of time
d. number of cycles/ unit of time
Answer:
d
Explanation: