If we someday put a spacecraft in low Jupiter orbit, its orbital speed will be greater than that for an Earth satellite.
What is orbital speed?The orbital speed of an astronomical body or object in a gravitationally bound system is its speed with respect to the most massive body's center of mass, or the barycenter.
Orbital speed is ≈ √(gR) for low orbit. As the radius of Jupiter is 11 times that of Earth and the free fall acceleration near its surface is 2.5 times that on earth, the orbital speed of a spacecraft in low Jupiter orbit will be greater than that for an Earth satellite.
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The figure below shows a motorcycle leaving the end of a ramp with a speed of 39.0 m/s and following the curved path shown. At the peak of the path, a maximum height h above the top of the ramp, the motorcycle's speed is 37.1 m/s. What is the maximum height h? Ignore friction and air resistance. (Enter your answer in m.)
_____m
The maximum height h of the curved path is 7.38 m.
Maximum height of the curved path
Apply the following kinematic equation;
v² = u² - 2gh
where;
v is the final velocity of the motorcycleu is initial velocity of the motorcycleh is the maximum height(u² - v²)/2g = h
(39² - 37.1²)/(2 x 9.8) = h
7.38 m = h
Thus, the maximum height h of the curved path is 7.38 m.
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When a scientific calculator shows the quantity below, what does it mean?
1.5E8
. What is the potential energy of a 0.40 kg ball at a height of 9.2 m?
Explanation:
gravitational potential energy = mgh (must be in S.I. unit)
m= 0.4 kg ; g= 10m/s (gravitational acceleration occurs); h=9.2 m
hence mgh=0.4×10×9.2= 36.8J
unit for energy is joules and since the variables are in S.I. unit, we can use Joules as the final unit for measurement
Answer:
36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s²
Explanation:
Potential Energy = Mass x Gravitational Acceleration x Height
It can be expressed as [tex]\displaystyle{PE = mgh}[/tex] where PE is potential energy, m is mass, g is gravitational acceleration and h is height.
In this case, we know mass is 0.40 kg and height of 9.2 m as well as gravitational acceleration is defined to be 9.8 m/s² (You can also define g = 10 m/s²)
Therefore, substitute given information in formula:
[tex]\displaystyle{PE=0.40 \ \times \ 9.8 \ \times 9.2 }\\\\\displaystyle{PE=36.06 \ J}[/tex]
With g = 10 m/s², you'll get:
[tex]\displaystyle{PE = 0.40 \ \times 10 \ \times 9.2}\\\\\displaystyle{PE = 36.8 \ J}[/tex]
Note that J is for joule unit.
Therefore, the answer is 36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s² - both work.
do you think an Electromagnet can be used for separating plastic bags from a garbage heap? explain
Answer:
No
Explanation:
Plastic bags are not magnetic materials, only magnetic materials (such as iron) can be attracted by the magnet.
Hope this helps.
How do I connect with my higher self?
Answer:
Create space
Watch your breath
Watch your thoughts
Be gentle with yourself
Affirm what you want
Explanation:
an 8 kilogram ball is fired horizontally form a 1 times 10^3 kilogram cannon initially at rest after having been fired the momentum of the ball is 2.4 time 10^3 kilogram meters per second east calculate the magnitude of the cannons velocity after the ball is fired
The magnitude of the cannons velocity after the ball is fired will be 2.4m/s.
To find the answer, we have to know more about the law of conservation of linear momentum.
How to find the magnitude of the cannons velocity after the ball is fired?The law of conservation of linear momentum states that, the initial momentum of a system will be equal to the final momentum.Here, then initial momentum of the system will be equal to zero, since the initial velocity of the cannon is equal to zero.Given that,[tex]M_B=8kg\\M_C=1*10^3 kg\\P_f=2.4*10^3 kgm/s.\\[/tex]
When the ball is fired from the canon, then the cannon will have a recoil velocity in the opposite direction of motion of the ball.Since it has a final momentum towards east, the recoil momentum will be in the west.Thus, the velocity of the cannon after when the ball is fired will be,[tex]P_f=M_CV_C\\V_C=\frac{P_f}{M_C}=\frac{2.4*10^3}{1*10^3} =2.4m/s \\west[/tex]
Thus, we can conclude that, the magnitude of the cannons velocity after the ball is fired is 2.4m/s.
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After the ball is shot, the cannon will move at a speed of 2.4 m/s.
We must learn more about the rule of conservation of linear momentum in order to locate the solution.
How can I determine the cannon's post-ball velocity magnitude?
According to the law of conservation of linear momentum, a system's starting and ultimate velocities are equal.Since the cannon's initial velocity is 0 in this case, the system's initial momentum will be equal to zero as well.We have,[tex]m=8kg\\M=1000kg\\P_f=2.4*10^3kgm/s[/tex]
When a cannon fires a ball, the cannon will recoil quickly and in the opposite direction of the ball's motion.Given that it is now moving eastward, the recoil momentum is towards the west.As a result, when the ball is fired, the cannon's velocity will be,[tex]P_f=MV\\V=\frac{P_f}{M} =2.4m/s[/tex]
Thus, we can infer that the cannon's maximum speed once the ball is fired is 2.4 m/s.
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A cart of mass 0.5 kg sits at rest on a table on which it can roll without friction. It is attached to an unstretched spring. You give the mass a push with a constant force over a distance of 5 cm in the direction that compresses the spring, after which the mass starts undergoing simple harmonic motion with a frequency of 0.5 complete oscillations per second and an amplitude of 15 cm.
A) What is the spring constant of the spring?
B) How fast was the cart moving at the instant when you finished pushing it?
C) What force did you exert on the cart?
(A) The spring constant of the spring is 4.94 N.
(B) The speed of the cart after pushing it is 0.47 m/s.
(C) The force applied to the cart is 0.75 N.
Spring constant
ω = √k/m
where;
ω is angular frequencyk is spring constantm is mass0.5 rev/s = 0.5(2π) rad/s = π rad/s = 3.142 rad/s
ω² = k/m
k = mω²
k = 0.5 x (3.142)²
k = 4.94 N/m
Energy stored in the springE = ¹/₂kA²
where;
A is amplitude
E = ¹/₂(4.94)(0.15)²
E = 0.056 J
Speed of the cartE = ¹/₂mv²
2E = mv²
v² = 2E/m
v² = (2 x 0.056)/(0.5)
v² = 0.224
v = √0.224
v = 0.47 m/s
Force exerted on the cartE = ¹/₂FA
2E = FA
F = 2E/A
F = (2 x 0.056)/(0.15)
F = 0.75 N
Thus, the spring constant of the spring is 4.94 N. The speed of the cart after pushing it is 0.47 m/s. The force applied to the cart is 0.75 N.
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A 15.0-kg child descends a slide 2.40 m high and reaches the bottom with a speed of 1.10 m/s .
How much thermal energy due to friction was generated in this process?
Express your answer to three significant figures and include the appropriate units.
The thermal energy that is generated due to friction is 344J.
What is the thermal energy?Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.
Thus, the kinetic energy is given as;
KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J
PE = mgh = 15.0-kg * 9.8 m/s^2 * 2.40 m = 352.8 J
The thermal energy is; 352.8 J - 9.1 J = 344J
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What is the speed of light while traveling through (a) a vacuum, (b) air at 30°C, and (c) air at 0°C?
The speed of light through a transparent substance is 2.00 × 108 m/s. What is the substance?
The wavelength of light from a monochromatic source is measured to be 6.80 × 10−7 m. (a) What is the frequency of this light? (b) What color would you observe?
What is the energy of a photon of red light with a frequency of 4.3 × 1014 Hz?
(1) The speed of light through vacuum is 3 x 10⁸ m/s.
(2) The speed of light in air at 30⁰C is 350 m/s.
(3) The speed of light in air at 0⁰C is 330 m/s.
(4) The transparent substance is glass.
(5) The color of light observed is red.
(6) The energy of a photon of red light is 2.85 x 10⁻¹⁹ J.
Speed of light
The speed of light through vacuum is 3 x 10⁸ m/s.
The speed of light in air at 30⁰C is 350 m/s.
The speed of light in air at 0⁰C is 330 m/s
Refractive index of the substancen = speed of light in vacuum/speed of light in the substance
n = ( 3 x 10⁸ m/s) / ( 2 x 10⁸ m/s)
n = 1.5
The refractive index of glass is 1.5, thus, the transparent substance is glass.
Frequency of lightf = v/λ
f = (3 x 10⁸) / (6.8 x 10⁻⁷)
f = 4.41 x 10¹⁴ Hz
The color of light observed is red based on the frequency and wavelength.
Energy of the photonE = hf
E = (6.626 x 10⁻³⁴)(4.3 x 10¹⁴)
E = 2.85 x 10⁻¹⁹ J
Thus, the speed of light through vacuum is 3 x 10⁸ m/s.
The speed of light in air at 30⁰C is 350 m/s.
The speed of light in air at 0⁰C is 330 m/s.
The transparent substance is glass.
The color of light observed is red.
The energy of a photon of red light is 2.85 x 10⁻¹⁹ J.
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Motorcycle safety helmet extend the time of collision hence decreasing,
a:chance of collision
b:force acting
c: velocity
c:Impulse
Answer:
D. Impulse
Explanation: Hope this helps
According to the "Law of Increasing Opportunity Costs," what would be the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning?
A. Food or Drink.
B. Money or income.
C. Sleep or rest.
C. The opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.
What is law of opportunity cost?
The law of increasing opportunity cost is an economic principle that describes how opportunity costs increase as resources are applied.
As the student gives up his sleep or night rest in the place of his exam preparation, we say that the opportunity cost is the sleep or rest.
Thus, the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.
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using a weston cadmium cell of 1.0283 v and a standard resistance of .1ohm a potentiometer was adjusted so that 1.0183 m was equivalent to the emf of the cell: when a certain direct current was flowing through the standard resistance, the voltage across it corresponds to 150 cm. what was the value of current
Answer:
Explanation:
the value is -8 cm
What is the approximate uncertainty in the area of a circle of radius 4.3×104cm ?
The approximate uncertainty in the area of a circle is 5.4%.
Area of the circle
A = πr²
where;
r is radius of the circleA = π(4.3 x 10⁴)²
A = 5.81 x 10⁹ cm²
Let the error in measurement = 1 x 10⁴ cm
Error in Area measurementA = π(1 x 10⁴)² = 3.14 x 10⁸ cm²
Uncertainty in measurement% = (error/actual area) x 100%
% = ( 3.14 x 10⁸) / (5.81 x 10⁹) x 100%
% = 5.4 %
Thus, the approximate uncertainty in the area of a circle is 5.4%.
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converting 67 m•s¹ to km•h¹
Answer:
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A 0.80-kg mass attached to the end of a string swings in a vertical circle (radius = 2.0 m). When the mass is at the highest point of the circle, the speed of the mass is 9.0 m/s. What is the magnitude of the force of the string on the mass at this position?
Answer:
Approximately [tex]25\; {\rm N}[/tex] assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
This mass is in a circular motion of radius [tex]r[/tex]. Hence, when the velocity of the mass is [tex]v[/tex], the acceleration of this mass should be [tex](v^{2} / r)[/tex]. The net force on this mass should be [tex](\text{net force}) = (m\, v^{2}) / r[/tex] towards the center of the circle.
When this [tex]m = 0.80\; {\rm kg}[/tex] mass is at the top of this circle, both gravitational pull and the force of the string (tension) point downwards. Hence, the net force on this mass would be:
[tex](\text{net force}) = (\text{weight}) + (\text{tension})[/tex].
Thus:
[tex]\begin{aligned} (\text{tension}) &= (\text{net force}) -(\text{weight})\\ &= \frac{m\, v^{2}}{r} - m\, g \\ &= m\, \left(\frac{v^{2}}{r} - g\right) \\ &= 0.80\; {\rm kg}\times \left(\frac{(9.0\; {\rm m\cdot s^{-1}})^{2}}{2.0\; {\rm m}} - 9.81\; {\rm m\cdot s^{-2}}\right) \\ &\approx 25\; {\rm kg \cdot m\cdot s^{-2}} \\ &= 25\; {\rm N}\end{aligned}[/tex].
*Look at attachment for photo of object**
An object, whose mass is 0.520 kg, is attached to a spring with a force constant of 106 N/m. The object rests upon a frictionless, horizontal surface (shown in the figure below).
The object is pulled to the right a distance A = 0.150 m from its equilibrium position (the vertical dashed line) and held motionless. The object is then released from rest.
(a) At the instant of release, what is the magnitude of the spring force (in N) acting upon the object?
__________ N
(b) At that very instant, what is the magnitude of the object's acceleration (in m/s2)?
________ m/s2
(c)
In what direction does the acceleration vector point at the instant of release?
- The direction is not defined (i.e., the acceleration is zero).
- Toward the equilibrium position (i.e., to the left in the figure).
- Away from the equilibrium position (i.e., to the right in the figure).
- You cannot tell without more information.
A. The magnitude of the spring force (in N) acting upon the object is 15.9 N
B. The magnitude of the object's acceleration (in m/s²) is 30.58 m/s²
C. The direction of the acceleration vector points toward the equilibrium position (i.e., to the left in the figure).
A. How to determine the force Extension (e) = 0.150 mSpring constant (K) = 106 N/mForce (F) = ?F = Ke
F = 106 × 0.15
F = 15.9 N
B. How to determine the accelerationMass (m) = 0.52 KgForce (F) = 15. 9 NAcceleration (a) =?F = ma
Divide both sides by m
a = F / m
a = 15.9 / 0.52
a = 30.58 m/s²
C. How to determine the direction of the acceleration vectorConsidering the diagram, we can see that the spring was pulled away from the equilibrium point.
Thus, when the spring is released, it will move toward the equilibrium point. This is also true about the acceleration.
Therefore, we can conclude that the direction of the acceleration vector is towards the equilibrium point.
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An object with a mass of m = 3.85 kg is suspended at rest between the ceiling and the floor by two thin vertical ropes.
The magnitude of the tension in the lower rope is 12.8 N. Calculate the magnitude of the tension in the upper rope.
The tension in the upper rope is 50.53 N.
Tension in the upper rope
The tension in the upper rope is calculated as follows;
T(up) = T(dn) + mg
where;
T(dn) is the tension in the lower ropemg is the weight of the objectT(up) = 12.8 N + (3.85 x 9.8) N
T(up) = 50.53 N
Thus, the tension in the upper rope is 50.53 N.
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The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×10^3 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10^-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×10^5 m around the planet Glob. Calculate the speed of the satellite.
(a) The initial speed of the rock as it left the astronaut's hand is 19.35 m/s.
(b) The speed of the satellite is 50.24 m/s.
Acceleration due to gravity of the planet
g = GM/R²
where;
M is mass of the planetR is radius of the planetg = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(63200)²
g = 0.13 m/s²
Initial speed of the rockv² = u² - 2gh
where;
v is final velocityu is initial velocityat maximum height, v = 0
u² = 2gh
u = √2gh
u = √(2 x 0.13 x 1,440)
u = 19.35 m/s
Speed of the satellitev = √GM/r
M is mass of the planet Globr is the total distance from the center of the planet Globr = radius of planet Glob + radius of the satellite
r = 63200 m + 145,000 m = 208,200 m
v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(208,200)]
v = 50.24 m/s
Thus, the initial speed of the rock as it left the astronaut's hand is 19.35 m/s.
The speed of the satellite is 50.24 m/s.
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A bowling ball of mass m = 1.1 kg is resting on a spring compressed by a distance d = 0.35 m when the spring is released. At the moment the spring reaches its equilibrium point, the ball is launched from the spring into the air in projectile motion at an angle of θ = 39° measured from the horizontal. It is observed that the ball reaches a maximum height of h = 4.4 m, measured from the initial position of the ball. Let the gravitational potential energy be zero at the initial height of the bowling ball.
a) what is the spring constant, k, in newtons per meter?
(I got that the speed of the ball after the launch is 14.76)
The spring constant, k, in newtons per meter is 1,955.9 N/m.
Speed of the ball after the launchh = v²sin²θ/2g
v = √[(2gh)/sin²θ]
v = √[(2 x 9.8 x 4.4)/ (sin 39)²]
v = 14.76 m/s
Energy of the ball at topE = K.E + P.E
E = ¹/₂m(v cosθ)² + mgh
E = ¹/₂(1.1)(14.76 cos39)² + (1.1 x 9.8 x 4.4)
E = 119.8 J
Spring constantE = ¹/₂kx²
k = 2E/x²
k = (2 x 119.8)/(0.35²)
k = 1,955.9 N/m
Thus, the spring constant, k, in newtons per meter is 1,955.9 N/m.
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in the absence of friction, the output power of a winding engine is 100kw but thus is reduced by friction to 90kw . how much oil initially at 120° is required per second to to keep the Temperature of the bearing down to 70°C ? specific heat capacity of oil is 2100 j/kg°C.
please I need it now Bosses ♀️
Answer:
Explanation:
The angular speed of the rotor is 200 rad/s.
The torque needed to be transmitted by the engine is 180 Nm.
The power of the rotor required to transmit energy to apply a torque τ to rotate a motor with angular speed ω,
P=τω
=180×200W
=36kW
A proton is moved from the negative to the positive plate of a parallel-plate arrangement. The plates are 1.50 cm apart, and the electric field is uniform with a magnitude of 1 500 N/C.
What is the proton’s potential energy change?
What is the potential difference between the plates?
What is the potential difference between the negative plate and a point midway between the plates?
If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate?
(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
(b) The potential difference between the negative plate and a point midway between the plates is 11.25 V.
(c) The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
Potential energy of the proton
U = qΔV
where;
q is charge of the protonΔV is potential differenceU = q(Ed)
U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)
U = 3.6 x 10⁻¹⁸ J
Potential difference between the negative plate and a point midwayΔV = E(0.5d)
ΔV = 0.5Ed
ΔV = 0.5 (1500)(1.5 x 10⁻²)
ΔV = 11.25 V
Speed of the protonU = ¹/₂mv²
U = mv²
v² = 2U/m
where;
m is mass of proton = 1.67 x 10⁻²⁷ kgv² = (2 x 3.6 x 10⁻¹⁸) / ( 1.67 x 10⁻²⁷)
v² = 4.311 x 10⁹
v = √(4.311 x 10⁹)
v = 6.57 x 10⁴ m/s
Thus, the proton’s potential energy change is 3.6 x 10⁻¹⁸ J.
The potential difference between the negative plate and a point midway between the plates is 11.25 V.
The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.
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A force of 100 newtons is applied to a box at an angle of 36° with the horizontal. If the mass of the box is 25 kilograms, what is the horizontal
acceleration of the box?
OA
1.52 meters/second²
OB. 3.24 meters/second²
OC. 5.48 meters/second²
O D.
6.87 meters/second²
OE
7.15 meters/second²
Answer:
See below
Explanation:
I will assume the force is in a DOWNWARD direction ( I believe it makes no answer difference)
Horizontal component is then 100 cos 36° =80.9 N
F = ma
80.9 = 25 kg *a
a = 3.24 m/s^2
Answer:
See image
Explanation:
Plato
In physics, when a baseball player catches a ball, which one of newtons laws is it an example of?
A. 1st law
B. 2nd law
C. 3rd law
A traffic light hangs from a pole as shown in (Figure 1). The uniform aluminum pole AB is 7.60 m long and has a mass of 10.0 kg . The mass of the traffic light is 23.5 kg. 1.) Determine the tension in the horizontal massless cable CD. 2) Determine the vertical component of the force exerted by the pivot A on the aluminum pole. 3.) Determine the horizontal component of the force exerted by the pivot A on the aluminum pole.
There is a 446N force in the horizontal massless cable CD. The vertical component of the pivot A's force on the aluminum pole, which is 328.3N. The force that the pivot A applies to the aluminum pole has a horizontal component of 446N.
We need to be aware of the force in order to discover the solution.
How can I determine the tension in the CD's horizontal massless cable?The free body diagram of the masses must be drawn in order to determine the tension in the horizontal cable CD.To obtain the tension on CD in the free body diagram, let's balance all the vertical and horizontal forces. [tex]TH-mg\frac{l}{2}cos\alpha -Mglcos\alpha =0\\T=\frac{glcos\alpha (\frac{m}{2}+M )}{h} \\T=446N[/tex]where, H=3.8m, l=7.6m, m=10kg, M=23.5 kg and alpha= 37 degree,
How to calculate the force the pivot A exerted on aluminum's vertical and horizontal components?The overall force acting vertically is,[tex]F_V-mg-Mg=0\\F_v=328.2N\\[/tex]
The overall force acting horizontally is,[tex]F_H=T=446N[/tex]
In light of this, we may say that the tension in the horizontal massless cable CD and the horizontal component of the force applied by the pivot A to the aluminum pole are identical and each exert 466N of force, whereas the vertical component of that force is 328.3N.
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The image shows a wheel that’s wound up and released. The wheel moves up and down as shown. Identify the position of the wheel when its potential energy is greatest.
The highest point of the wheel is the position of the wheel when its potential energy is greatest.
At what position of the wheel potential energy is greatest?The position of the wheel when its potential energy is greatest when it is at the highest point because potential energy depends on the height of an object. If the object is at more height then it has more potential energy and vice versa.
So we can conclude that the highest point of the wheel is the position of the wheel when its potential energy is greatest.
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Jane, looking for Tarzan, is running at top speed (6.0 m/s ) and grabs a vine hanging vertically from a tall tree in the jungle.
How high can she swing upward?
Express your answer to two significant figures and include the appropriate units.
Does the length of the vine affect your answer?
(a) The maximum height reached by Jane is 1.8 m.
(b) The length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.
Maximum height Jane can swing
apply the principle of conservation of energy;
P.E = K.E
mgh = ¹/₂mv²
h = v²/2g
where;
v is speed of janeg is acceleration due to gravityh = (6²)/(2 x 9.8)
h = 1.84 m
Time of motion of JaneAssuming Jane to be in simple harmonic motion, the time of motion is calculated as;
T = 2π√(L/g)
where;
L is the length of the vineThus, the length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.
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White light is spread out into its spectral components by a diffraction grating. If the grating has 1975 lines per centimeter, at what angle does red light of wavelength 640 nm appear in first-order spectrum? (Assume that the light is incident normally on the grating.)
The angle of the red light is mathematically given as
[tex]\theta = 7.24 \textdegree[/tex]
What angle does red light of wavelength 640 nm appear in the first-order spectrum?Generally, the equation for the grating element is mathematically given as
d= 1 / N
Therefore
d= 1/1965
d= 5.089 * 10^{-6} m
Generally, the equation for the differential formula is mathematically given as
[tex]d sin \theta = m\lambda[/tex]
Therefore
[tex]sin \theta = \lambda / d[/tex]
[tex]sin \theta= (640 * 10 ^ {-9} m)/(5.089 * 10 ^ {-6} m)[/tex]
[tex]\theta = 7.24 \textdegree[/tex]
In conclusion, The angle of the red light
[tex]\theta = 7.24 \textdegree[/tex]
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An ideal toroidal solenoid (Figure 1) has inner radius r1 = 15.1 cm and outer radius r2 = 18.3 cm. The solenoid has 180 turns and carries a current of 8.40 A.
a) What is the magnitude of the magnetic field at 12.6 cm from the center of the torus?
b) What is the magnitude of the magnetic field at 16.3 cm from the center of the torus?
c) What is the magnitude of the magnetic field at 20.7 cm from the center of the torus?
The solution for the three questions is mathematically given as
Parts A and C are both zeros.
Part B B = 0.001855Tesla
What is the magnitude of the magnetic field at 12.6 cm from the center of the torus?Parts A and C are both zeros.
For component A, the magnetic field is zero since 12.6 cm is still inside the toroidal solenoid. Part C has no magnetic field since it is 20.7 cm outside of the toroidal solenoid.
Generally, the equation for magnetic field is mathematically given as
B = (mu_0*N*I)/(2*pi*r)
Therefore
B = ((4*pi*10^{-7})*180*8.40)/(2*pi*0.163)
B = 0.001855Tesla
In conclusion, the magnitude of the magnetic field at 16.3 cm from the center of the torus is
B = 0.001855Tesla
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Answer:
Part A & C: 0 T
Part B: 1.855*10^-3 T
Explanation:
The formula that models the magnetic field of a toroid is B=μ0*N*I/2π*r .
Note: Toroids keep their magnetic field rotating within the coils that carry current.
Part A & C: Thus the B field magnitude 12.6 cm & 20.7 cm away from the center is 0 T.
Part B: [tex]B=\frac{(4\pi *10x^{-7} )(180)(8.4 A)}{(2\pi )(16.3*10x^{-2} m)}[/tex]
B=1.855*10^-3 T
An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.
a) Find the magnitude of the magnetic field this electron produces at the point A .
b) Find the magnitude of the magnetic field this electron produces at the point B .
c) Find the magnitude of the magnetic field this electron produces at the point C .
d) Find the magnitude of the magnetic field this electron produces at the point D
Hi there!
We can use Biot-Savart's Law for a moving particle:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }[/tex]
B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)
q = charge of particle (1.6 × 10⁻¹⁹ C)
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
r = distance from particle (2.10 μm)
There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }[/tex]
Where 'θ' is the angle between the velocity and radius vectors.
a)
To find the angle between the velocity and radius vector, we find the complementary angle:
θ = 90° - 60° = 30°
Plugging 'θ' into the equation along with our other values:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }[/tex]
[tex]B = \boxed{7.07 *10^{-10} T}[/tex]
b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.
c)
In this instance, the radius vector and the velocity vector are perpendicular so
'θ' = 90°.
[tex]B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}[/tex]
d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.
Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.
[tex]\boxed{B = 0 T}[/tex]
A 21.0 kg uniform beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 66.0° what is the tension in the cable?
Since upward forces must be equal to the downward forces, the tension in the cable is 225.3 N
What are the two conditions for equilibrium ?The two conditions are;
Sum of the upward forces must be equal to the sum of the downward forces.The sum of the clockwise moment must be equal to the sum of the anticlockwise moment.The given parameters are;
Mass m = 21 kgangle θ = 66°Tsinθ = mg
Tsin 66 = 21 x 9.8
T = 205.8 / 0.914
T = 225.3 N
Therefore, the tension in the cable is 225.3 N
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