Answer:
the radius of bigger loop = 6 cm
Explanation:
given,
two concentric current loops
smaller loop radius = 3.6 cm
]current in smaller loop = 12 A
current in the bigger loop = 20 A
magnetic field at the center of loop = 0
Radius of the bigger loop = ?
[tex]B_t = B_1 + B_2[/tex]
[tex]0 = \dfrac{\mu_0I_1}{2R_1} +\dfrac{\mu_0I_2}{2R_2}[/tex]
now, on solving
[tex]\dfrac{I_1}{R_1} = \dfrac{I_2}{R_2}[/tex]
[tex]R_2 = I_2\dfrac{R_1}{I_1}[/tex]
= [tex]20\times \dfrac{3.6}{12}[/tex]
= 6 cm
hence, the radius of bigger loop = 6 cm
Hydroelectric dams use ------- to produce electricity. gravitational potential energy of falling water
nuclear energy
geothermal energy
fossil fuel energy
solar energy
Answer:
Gravitational potential energy
Explanation:
Hydroelectric dams are the power plants which generates electricity by using the energy of falling water from a great height.
A the water is stored in big reservoirs and at a great height, it contain lot of potential energy due to the height. As it falls downwards, the potential energy is converted into kinetic energy of the water. this kinetic energy of the falling water is used to run the turbine, and then the electric energy is generated.
So, in hydroelectric power stations, the potential energy of water is converted into the electric energy.
Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carrying a total charge of 4.95 μC .
Final answer:
To estimate the electric field near a uniformly charged wire, one must determine the linear charge density and integrate the contributions from infinitesimal charge elements along the wire, considering the symmetry that leads to the cancellation of horizontal components.
Explanation:
To estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire, we use the principle of superposition and integrate the contributions of the infinitesimally small charge elements along the wire.
First, we determine the linear charge density λ by dividing the total charge Q by the length L of the wire:
λ = Q / L
Then, we apply the formula for the electric field due to an infinitesimal charge element dQ at a distance r:
dE = (k * dQ) / r²
Where k is the Coulomb's constant (approximately 8.99 × 10⁹ Nm²/C²). Since we need to integrate over the length of the wire to find the total electric field, we use the linear charge density λ and a differential element of the wire's length dl:
dQ = λ * dl
The resulting integral, when evaluated, will give us the magnitude of the electric field E at the specified distance from the wire.
It is important to note that due to the symmetry of the problem, the horizontal components of the electric field due to charge elements on the wire will cancel out, leaving only the vertical components to contribute to the total electric field at the point of interest
If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in meters.
Answer:
Height, H = 25.04 meters
Explanation:
Initially the ball is at rest, u = 0
Time taken to fall to the ground, t = 2.261 s
Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :
[tex]H=ut+\dfrac{1}{2}at^2[/tex]
Here, a = g
[tex]H=\dfrac{1}{2}gt^2[/tex]
[tex]H=\dfrac{1}{2}\times 9.8\times (2.261)^2[/tex]
H = 25.04 meters
So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.
Final answer:
Using the equation of motion under gravity, it was calculated that a ball dropped and taking 2.261 s to hit the ground was released from a height of approximately 25.07 meters.
Explanation:
To find out from what height H a ball was released if it takes 2.261 s to fall to the ground, we use the equation of motion under gravity, which is H = 0.5 * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2 on the surface of the Earth) and t is the time in seconds.
Substituting the given values in the equation, we get: H = 0.5 * 9.8 * (2.261)^2. After calculating, we find that the height H approximately equals 25.07 meters.
Therefore, the ball was released from a height of approximately 25.07 meters.
A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 13.0 m/s when the hand is 2.30 m above the ground. How long is the ball in the air before it hits the ground?
Answer:
2.82 s
Explanation:
The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 is the starting position, 2.3 m in this case.
Vy0 is the starting speed, 13 m/s.
a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.
Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2
It will reach the ground when Y(t) = 0
0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2
-4.9 * t^2 + 13 * t + 2.3 = 0
Solving this equation electronically gives two results:
t1 = 2.82 s
t2 = -0.17 s
We disregard the negative solution. The ball spends 2.82 seconds in the air.
A charge located at (3,-5) in the x-y plane exerts an attractive force of 0.75N on another charge located at (2,7) (all distances are in cm) a) In what direction does this force point? The answer should be given as a counterclockwise angle with respect to the positive x-axis. b) Write down an expression for this force in terms of unit vectors i and (don't forget your units). c) If the charge at (3,-5) is +2pic, determine the magnitude and sign of the other charge.
Answer:
(a): [tex]247.77^\circ.[/tex]
(b): [tex]\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \dfrac{\hat i-12\hat j}{\sqrt{145}}.[/tex]
(c): Magnitude = [tex]0.604\ C.[/tex]
Sign = negative.
Explanation:
(a):
Given that first charge is located at (3,-5) and second charge is located at (2,7).
The electrostatic force between two charges acts long the line joining the two charges, therefore, the direction of electrostatic force of attraction between these two charges is along the position vector of charge at (2,7) with respect to the position of charge at (3,-5).
Assuming, [tex]\hat i,\ \hat j[/tex] are the unit vectors along positive x and y axes respectively.
Position vector of charge at (3,-5) with respect to origin is
[tex]\vec r_1 = 3\hat i+(-5)\hat j[/tex]
Position vector of charge at (2,7) with respect to origin is
[tex]\vec r_2 = 2\hat i+7\hat j[/tex]
The position vector of at (2,7) with respect to the position of charge at (3,-5) is
[tex]\vec r = \vec r_2-\vec r_1\\=(3\hat i-5\hat j)-(2\hat i+7\hat j)\\=1\hat i-12\hat j.[/tex]
If [tex]\theta[/tex] is the angle this position vector is making with the positive x axis then,
[tex]\rm \tan\theta =\dfrac{y\ component\ of\ (\hat i-12\hat j)}{x\ component\ of\ (\hat i-12\hat j}=\dfrac{-12}{1} = -12.\\\theta =\tan^{-1}(-12)=-85.23^\circ.[/tex]
The negative sign indicates that this position vector is [tex]85.23^\circ.[/tex] below the positive x axis, therefore, in counterclockwise direction from positive x axis, its direction = [tex]360^\circ-85.23^\circ = 247.77^\circ.[/tex]
The force is also along the same direction.
(b):
According to Coulomb's law, the expression of this force is given by
[tex]\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \hat r[/tex]
where,
[tex]q_1,\ q_2[/tex] are the charges.[tex]\vec r[/tex] = position vector of one charge with respect to another charge.[tex]\hat r[/tex] = unit vector along the direction of [tex]\vec r[/tex].[tex]k[/tex] = Coulomb's constant, which have value = [tex]\rm 9\times 10^9\ Nm^2/C^2[/tex].We have,
[tex]\vec r = \hat i-12\hat j\\\therefore |\vec r| = \sqrt{1^2+(-12)^2}=\sqrt{145}.\\\Rightarrow \hat r = \dfrac{\vec r}{|\vec r|} = \dfrac{\hat i-12\hat j}{\sqrt{145}}[/tex]
Putting this value, we get,
[tex]\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \dfrac{\hat i-12\hat j}{\sqrt{145}}[/tex]
(c):
If charge at (3,-5) is [tex]+2\ pC[/tex], then the magnitude of the force between the two charges is given by
[tex]F=k\ \dfrac{q_1q_2}{|\vec r|^2}[/tex]
where, we have,
[tex]F=0.75\ N\\|\vec r| = \sqrt{145}\ cm = \sqrt{145}\times 10^{-2}\ m.\\q_1 = +2\ pC = +2\times 10^{-12}\ C.[/tex]
Putting all these values,
[tex]0.75=9\times 10^9\times \dfrac{(+2\times 10^{-12})\times q_2}{(\sqrt{145}\times 10^{-2})^2}\\\Rightarrow q_2 = \dfrac{0.75\times(\sqrt{145}\times 10^{-2})^2}{9\times 10^9\times 2\times 10^{-12}} =0.604\ C.[/tex]
Since, the electric force between the two spheres is given to be attractive therefore this charge must be negative as the other charge is positive.
A fighterjet is launched from an aircraft carrier with the
aidof its own engines and a steam powered catapult. The thurst of
itsengine2.3*10^5N. In being launched from rest it moves through
adistance 87m and has KE 4.5*10^7J at lift off.What is the work
doneon the jet by the catapult.
Answer:
The work done on the jet is [tex]W_{jet} = 2.49\times 10^{7} J[/tex]
Given:
Force, F = [tex]2.3\times 10^{5} N[/tex]
Distance moved by the jet, x = 87 m
Kinetic Energy, KE = [tex]4.5\times 10^{7} J[/tex]
Solution:
Now, to calculate the work done on the fire fighter jet by the catapult, we find the difference between the KE of the jet at lift off and the work done by the engines.
This difference provides the amount of work done on the jet and is given by:
[tex]W_{jet} = KE - W_{engines}[/tex] (1)
Now, the work done by the engines is given by:
[tex]W_{engines} = Fx = 2.3\times 10^{5}\times 87 = 2.001\times 10^{7} J[/tex]
Now, using eqn (1):
[tex]W_{jet} = 4.5\times 10^{7} - 2.001\times 10^{7} = 2.49\times 10^{7} J[/tex]
A 1.8-kg block is released from rest at the top of a rough 45° inclined plane. As the block slides down the incline, its acceleration is 2.0 m/s, down the incline. Determine the magnitude of the force of friction acting on the block. a. 16.1 N b. 8.9 N c. 12.5 N d. 2.3 N e. 17.6N
Answer:
Ff= 8.9N:Force of friction acting on the block.
Explanation
Box kinetics: in x₁-y₁ :We apply the second law of Newton
∑Fx₁=ma : second law of Newton Formula (1)
Where:
∑Fx₁:algebraic sum of forces in the direction of x1, positive in the direction of movement of the block, which is down and negative in the direction opposite to the movement of the block
m: is the mass of the block
The x₁ axis coincides with the plane of sliding of the block, that is, the x₁ axis forms 45 degrees with the horizontal
The total weight (W) of the block is in the vertical direction and the tip of the vector down and its magnitude is calculated as follows:
W=m*g
Where:
m: is the mass of the block
g: is the acceleration due to gravity
Calculation of the weight in the direction x₁
Wx₁=Wcos45°= m*g*cos 45 Equation( 1)
Data
m=1.8kg
g=9.8 m/s²
a=2m/s
Wx₁=m*g*cos 45°= 1.8*9.8 *[tex]\frac{\sqrt{2} }{2}[/tex]= 12.47 N
Friction force (Ff ) calculation
We apply formula (1)
∑Fx₁=m*a
Wx₁ - Ff= m*a
Ff=Wx₁ - m*a
We replace data
Ff= 12.47 - 1.8*2 =8.87
Ff= 8.9N
Which of the following statements about electric field lines are true? (choose all that are true) a) They are only defined for positive charges.
b) They are always tangent to electric field vectors.
c) They are always perpendicular to charged surfaces.
d) They are a simple way to visualize the electric field vectors. e) None of the above.
Answer: b) TRUE and d) TRUE
Explanation: a) FALSE the electric field lines are used to represent the charges postives and negatives.
b) TRUE it is the definition of electric field lines , they are tangent to the electric field vector.
c) FALSE the electric field lines only for conductor are perpendicular to the surface in other any situation there is a tangencial electric field components.
d) TRUE since it is a way to describe and imagine the effect of vectorial fields.
e) FALSE
A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of O 100 m/s. 60 m/s. 50 m/s 80 m/s. 110 m/s.
Answer:
The initial velocity of the ball should be 50 m/s.
Explanation:
Since the trip of the ball shall consist of upward ascend and the downward descend and since the ascend and the descend of the ball is symmetrical we infer that the upward ascend of the ball shall last for a time of 5 seconds.
Now since the motion of ball is uniformly accelerated we can find the initial speed of the ball using first equation of kinematics as
[tex]v=u+gt[/tex]
where,
'v' is final velocity of the ball
'u' is initial velocity of the ball
'g' is acceleration due to gravity
't' is the time of motion
Now we know that the ball will continue to ascend until it's velocity becomes zero hence to using the above equation we can write
[tex]0=u-9.81\times 5\\\\\therefore u=9.81\times 5=49.05m/s\approx 50m/s[/tex]
The route followed by a hiker consists of three displacement vectors, X, Y and Z. Vector X is along a measured trail and is 1430m in a direction 42 degrees north of east. Vector Y is along a measured trail that goes 2200m due south. The hiker then follows vector Z and ends up back where they started, so that X + Y + Z = 0. Find the magnitude of Z and an angle that specifies its direction.
Answer:
magnitude : 1635.43 mAngle: 130°28'20'' north of eastExplanation:
First, we will find the Cartesian Representation of the [tex]\vec{X}[/tex] and [tex]\vec{Y}[/tex] vectors. We can do this, using the formula
[tex]\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]
where [tex]| \vec{A} |[/tex] its the magnitude of the vector and θ the angle. For [tex]\vec{X}[/tex] we have:
[tex]\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )[/tex]
[tex]\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )[/tex]
where the unit vector [tex]\hat{i}[/tex] points east, and [tex]\hat{j}[/tex] points north. Now, the [tex]\vec{Y}[/tex] will be:
[tex]\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )[/tex]
Now, taking the sum:
[tex]\vec{X} + \vec{Y} + \vec{Z} = 0[/tex]
This is
[tex]\vec{Z} = - \vec{X} - \vec{Y}[/tex]
[tex](Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )[/tex]
[tex](Z_x , Z_y) = ( \ - 1062.70 m \ , \ 2200 m \ - \ 956.86 m \ )[/tex]
[tex](Z_x , Z_y) = ( \ - 1062.70 m \ , \ 1243.14 m\ )[/tex]
Now, for the magnitude, we just have to take its length:
[tex]|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}[/tex]
[tex]|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}[/tex]
[tex]|\vec{Z}| = 1635.43 m[/tex]
For its angle, as the vector lays in the second quadrant, we can use:
[tex]\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m}) [/tex]
[tex]\theta = 180\° - arctan( -1.1720) [/tex]
[tex]\theta = 180\° - 45\°31'40'' [/tex]
[tex]\theta = 130\°28'20'' [/tex]
A barge is carrying a load of gravel along a river.
Itapproaches a low bridge, and the captain realizes that the top
ofthe pile of gravel is not going to make it under the bridge.
Thecaptain orders the crew to quickly shovel gravel from the pile
intothe water. Is this a good decision? Explain.
Answer:
Explanation:
Shoveling gravel into the water will increase the buoyancy of the barge, which will make it float higher. Without data it is hard to tell if it will raise the barge enough to be counterproductive, but in any case throwing away the payload is not a good idea. Adding some weights would make the barge float lower, and then maybe the plie can make it under the bridge.
Newton's second law: Rudolph the red nosed reindeer is pulling a 25 kg sled across the snow in a field. The coefficient of kinetic friction is .12 The rope that is pulling the sled is coming off at a 29 degree angle above the horizontal. Find the force in the rope when the acceleration is .12 m/s^2.
Answer:
[tex]F=39,68N[/tex]
Explanation:
Data:
Mass [tex]m=25 Kg[/tex]
Coefficient of kinetic friction [tex]\mu=0.12[/tex]
Angle = [tex]29^{0}[/tex]
Acceleration = [tex]0.12 \frac{m}{s^{2} }[/tex]
Solution:
By Newton's first law we know that for the x-axis:
[tex]F_{rope_x}-F_f=F_R[/tex] Where [tex]F_R[/tex] is the resulting force, and [tex]F_f[/tex] is the friction force.
And for the y-axis:
[tex]F_{rope_y}+N=W[/tex], where N is the normal force, and W is the weight of the sled.
We know that the resulting force's acceleration is [tex]0.12 \frac{m}{s^{2} }[/tex], and by using Newton's second law, we obtain:
[tex]F=m.a[/tex]
[tex]F_R=25Kg. 0.12\frac{m}{s^2} \\ F_R=3N[/tex] .
Now, the horizontal component of the force in the rope will be given by
[tex]F_{rope_x}=F_{rope}.cos(29^0)=F_R+F_f[/tex], since the resulting force is completely on the x-axis, and the friction opposes to the speed of the sled.
To obtain the friction force, we must know the normal force:
[tex]F_f=\mu. N[/tex]
Clearing N in the y-axis equation:
[tex]N=W-F_{rope_y}=W-F_{rope}.sin(29^0)[/tex]
So we can express the x-axis equation as follows:
[tex]F.cos(29^0)=F_R+\mu.(W-F_{rope}.sin(29^0))[/tex]
Finally, solving for F we get
[tex]F = (F_R + \mu. m.g) / (cos (29^0) + \mu.sin (29^0))[/tex]
[tex]F=39,68N[/tex]
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known, and identify its value. Then, identify the unknown and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long a time is the dolphin in the air? Neglect any effects resulting from his size or orientation.
(a) Initial velocity (v₀) = 0 m/s, Final velocity (v) = 15.0 m/s.
(b) The dolphin rises approximately 11.5 meters above the water.
(c) The dolphin is in the air for approximately 1.5 seconds.
(a)
Initial velocity (v₀) = 0 m/s (since the dolphin jumps straight up from the water, its initial velocity is zero)
Final velocity (v) = 15.0 m/s (given)
(b)
To solve for the height, we can use the kinematic equation that relates the final velocity (v), initial velocity (v₀), acceleration (a), and displacement (Δy):
v² = v₀² + 2aΔy
v² = v₀² + 2gh
v² = 2gh
Solving for h:
h = (v²) / (2g)
h = (15.0)² / (2 × 9.8)
h ≈ 11.5 m
Therefore, the dolphin rises approximately 11.5 meters above the water.
(c) To determine the time the dolphin is in the air, we can use the equation for time (t) derived from the kinematic equation:
Δy = v₀t + (1/2)at²
Δy = (1/2)at²
t = √((2Δy) / a)
t = √((2 × 11.5) / -9.8)
t ≈ 1.5 s
Therefore, the dolphin is in the air for approximately 1.5 seconds.
To know more about the velocity:
https://brainly.com/question/34025828
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The dolphin rises 8.64 meters above the water and is in the air for 1.33 seconds.
Explanation:(a) The knowns in this problem are:
Initial velocity of the dolphin: 13.0 m/sAcceleration due to gravity: -9.8 m/s²(b) To find the height above the water, we need to find the time taken by the dolphin to reach that height and use the equation:
Final velocity = Initial velocity + Acceleration × Time
Solving for time, we have:
Time = (Final velocity - Initial velocity) / Acceleration
Substituting the known values into the equation, we get:
Time = (0 m/s - 13.0 m/s) / -9.8 m/s² = 1.33 s
Next, we can use the formula for height:
Height = Initial velocity × Time + 0.5 × Acceleration × Time²
Substituting the known values, we have:
Height = 13.0 m/s × 1.33 s + 0.5 × -9.8 m/s² × (1.33 s)² = 8.64 m
(c) The time the dolphin is in the air is equal to the time calculated above, which is 1.33 seconds.
A box moving on a horizontal surface with an initial velocity of 20 m/s slows to a stop over a time period of 5.0 seconds due solely to the effects of friction. What is the coefficient of kinetic friction between the box and the ground?
Answer:
μ= 0.408 : coefficient of kinetic friction
Explanation:
Kinematic equation for the box:
[tex]a=\frac{v_{f} -v_{i} }{t}[/tex] Formula( 1)
a= acceleration
v_i= initial speed =0
v_f= final speed= 20 m/s
t= time= 5 s
We replace data in the formula (1):
[tex]a=\frac{0-20}{5}[/tex]
[tex]a= -\frac{20}{5}[/tex]
a= - 4m/s²
Box kinetics: We apply Newton's laws in x-y:
∑Fx=ma : second law of Newton
-Ff= ma Equation (1)
Ff is the friction force
Ff=μ*N Equation (2)
μ is the coefficient of kinetic friction
N is the normal force
Normal force calculation
∑Fy=0 : Newton's first law
N-W=0 W is the weight of the box
N=W= m*g : m is the mass of the box and g is the acceleration due to gravity
N=9.8*m
We replace N=9.8m in the equation (2)
Ff=μ*9.8*m
Coefficient of kinetic friction ( μ) calculation
We replace Ff=μ*9.8*m and a= -4m/s² in the equation (1):
-μ*9.8*m: -m*4 : We divide by -m on both sides of the equation
9.8*μ=4
μ=4 ÷ 9.8
μ= 0.408
Vesna Vulovic survived the longest fall on record without a parachute when her plane exploded and she fell 5 miles, 733 yards. What is this distance in meters?
Answer:
8717 meters.
Explanation:
We need to know the conversion factors. We know that:
1 mile = 1609.34 meters
1 yard = 0.9144 meters
This means that:
[tex]\frac{1609.34 meters}{1 mile}=1[/tex]
[tex]\frac{0.9144 meters}{1 yard}=1[/tex]
It is convenient to leave the units we want at the end in the numerator so the ones in the denominator cancel out with the ones we want to remove, as will be seen in the next step.
We will convert first the miles, then the yards, and add them up.
[tex]5miles=5miles\frac{1609.34 meters}{1 mile}=8046.7meters[/tex]
[tex]733yards=733yards\frac{0.9144 meters}{1 yard}=670.2552meters[/tex]
So total distance is the sum of these, 8717 meters.
Tarzan, in one tree, sights Jane in another tree. He grabs
theend of a vine with length 20 m that makes an angle of 45
degreeswith the vertical, steps off his tree limb, and swings down
andthen up to Jane's open arms. When he arrives, his vine makesan
angle of 30 degrees with the vertical. determine whetherhe gives
her a tender embrace or knocks her off her limb bycalculating
Tarzan's speed just before he reaches Jane. Youcan ignore air
resistace and the mass of the vine.
Answer:
He knocks her. V = 7.97m/s
Explanation:
Let A be Tarzan's starting position and B Jane's position (Tarzan's final position).
Since there is no air resistance, energy at position A must be equal to energy at B.
At position A, Tarzan's speed is zero and since the 45° of the vine is greater than the final 30°, Tarzan will have potential energy at point A.
[tex]E_{A}=m_{T}*g*L*(cos(30)-cos(45))[/tex]
At point B, it is the lowest point (between A and B), so it has no potential energy. It will only have kinetic energy (ideally zero, for Jane's sake, but we don't know).
[tex]E_{B}=\frac{m_{T}*V^{2}}{2}[/tex]
Because of energy conservation, we know that Ea=Eb, so:
[tex]m_{T}*g*L*(cos(30)-cos(45))=\frac{m_{T}*V^{2}}{2}[/tex] Solving for V:
[tex]V=\sqrt{2*g*L*(cos(30)-cos(45))}=7.97m/s[/tex]
The output of an ac generator connected to an RLC series combination has a frequency of 12 kHz and an amplitude of 28 V. If R = 4.0 Ohms, L = 30 μH, and C = 8 μF, find a. The impedance
b. The amplitude for current
c. The phase difference between the current and the emf of the generator
Please show all steps and units. Thank you.
Answer:
(a) 4.04 ohm
(b) 6.93 A
(c) 8.53°
Explanation:
f = 12 kHz = 12000 Hz
Vo = 28 V
R = 4 ohm
L = 30 micro Henry = 30 x 10^-6 H
C = 8 micro Farad = 8 x 10^-6 F
(a) Let Z be the impedance
[tex]X_{L} = 2\pi fL=2\times3.14\times12000\times30\times10^{-6}= 2.26 ohm[/tex]
[tex]X_{c} = \frac{1}{2\pi fC}=\frac{1}{2\times3.14\times12000\times8\times10^{-6}}= 1.66 ohm[/tex]
[tex]Z = \sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{4^{2}+\left ( 2.26-1.66 \right )^{2}}[/tex]
Z = 4.04 Ohm
(b) Let Io be the amplitude of current
[tex]I_{o}=\frac{V_{o}}{Z}[/tex]
[tex]I_{o}=\frac{28}{4.04}[/tex]
Io = 6.93 A
(c) Let the phase difference is Ф
[tex]tan\phi = \frac{X_{L}-X_{C}}{R}[/tex]
[tex]tan\phi = \frac{2.26-1.66}{4}[/tex]
tan Ф =0.15
Ф = 8.53°
A ball is dropped from an upper floor, some unknown distance above your apartment. As you look out of your window, which is 1.50 m tall, you observe that it takes the ball 0.210 s to traverse the length of the window. 1)Determine how high above the top of your window the ball was dropped. Ignore the effects of air resistance. (Express your answer to three significant figures.)
Answer:1.902 m
Explanation:
Given
height of apartment=1.5 m
It takes 0.21 sec to reach the bottom from apartment
So
[tex]s=u_1t+\frac{gt^2}{2}[/tex]
[tex]1.5=u_1\times 0.21+\frac{9.81\times 0.21^2}{2}[/tex]
[tex]u_1=6.11 m/s[/tex]
i.e. if ball is dropped from top its velocity at window is 6.11 m/s
So height of upper floor above window
[tex]v^2-u^2=2as[/tex]
where s= height of upper floor above window
here u=0
[tex]6.11^2=2\times 9.81\times s[/tex]
[tex]s=\frac{6.11^2}{2\times 9.81}[/tex]
s=1.902 m
An airplane flies 200 km due west from city A to city B and then 275 km in the direction of 26.0° north of west from city B to city C. (a) In straight-line distance, how far is city C from city A?
Relative to city A, in what direction is city C?
(c) Why is the answer only approximately correct?
The straight-line distance from city A to city C is approximately 458.80 km. The direction from city A to city C is 14.9° north of west. The approximation is due to ignoring Earth's curvature.
Explanation:To solve this problem, we will use the laws of vector addition as airplanes movements are vector quantities as it involves both magnitude (distance) and direction. Initially, the airplane flies 200 km west. Then, it changes its direction 26.0° north of west and flies an additional 275 km.
When we decompose the second segment of the flight into its westward and northward components, we can use trigonometric principles. The horizontal (westward) distance is 275 km * cos(26.0°) = 245.56 km. Adding that to the 200 km the airplane flew initially gives us a total westward distance of 445.56 km. The vertical (northward) distance is 275 km * sin(26.0°) = 118.40 km.
To find the straight-line distance between City A and City C, we apply the Pythagorean theorem: √[(445.56 km)² + (118.4 km)²] = 458.80 km. To find the direction, we take the arctan of the northward distance over the westward distance: arctan(118.4 km / 445.56 km) = 14.9° north of west.
It's only approximately correct because we ignore the curvature of the Earth, which would slightly modify the distances and angles involved.
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The straight-line distance from city A to city C is approximately 460 km and the direction is 15° north of west. The answer is only an approximation because we've considered a flat plane, ignoring the spherical nature of the Earth.
Explanation:The question is asking for the linear distance and direction from city A to city C, which can be solved through vector addition and trigonometric calculations in physics. The displacement from A to B is 200 km west. The displacement from B to C is 275 km, 26.0° north of west. These can be seen as components of a right triangle, and by applying the Pythagorean theorem, the straight-line distance, or the hypotenuse, can be calculated.
First, we derive the north and west components of the 275 km using cosine and sine respectively as the motion is angled. The westward motion is 275cos(26) = 244 km. Add this to the 200 km westward motion to get a total westward motion of 444 km. The northward motion is 275sin(26) = 121 km.
Then, we apply Pythagorean theorem. √(444^2 + 121^2) ≈ 460 km. This is the straight-line distance from city A to city C.
For the direction, we find the angle, using tan^(-1)(121/444) ≈ 15° north of west.
The answer is only approximately correct because in real situation we have to take into account the spherical nature of the Earth, but here we've considered a flat plane for simplicity.
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If you are playing soccer and you always hit the left goal post instead of scoring, are you: a. accurate? b. precise? c. both? or d. neither?
Answer:
b. precise
Explanation:
Accuracy means how close a measurement is to the true value (in our example the true value is score a goal )
Precision refers how close two or more measurements to each other.
So if the playing soccer hit always the left goal post, he is not accurate (e doesn't score any goals ) but he is very precise
A hot-air balloon is descending at a rate of 2.3 m/s when a pas- senger drops a camera. If the camera is 41 m above the ground when it is dropped, (a) how much time does it take for the cam- era to reach the ground, and (b) what is its velocity just before it lands? Let upward be the positive direction for this problem.
Answer:
a) time taken = 2.66 s
b) v = 28.34
Explanation:
given,
rate of descending = 2.3 m/s
height of camera above ground = 41 m
using equation of motion
[tex]h = u t + \dfrac{1}{2}gt^2[/tex]
[tex]41 =2.3t + \dfrac{1}{2}\times 9.8\times t^2[/tex]
4.9 t² + 2.3 t - 41 =0
t = 2.66 ,-3.13
time taken = 2.66 s
b) v² = u² + 2 g h
v² = 0 + 2× 9.8 × 41
v = 28.34
A hot air ballo0n is ascending straight up at a constant
speedof 7.0 m/s. When the balloon is 12.0 m above the ground, agun
fires a pellet straight up from ground level with an initialspped
of 30.0 m/s. Along the paths of the ballon and thepellet, there are
two places where each of them has the altitude atthe same time. How
far above ground level are theseplaces?
Answer: The two places altitudes are: 16.17 m and 40.67 m
Explanation:
Hi!
Lets call z to the vertical direction (z= is ground) . Then the positions of the balloon and the pellet, using the values of the velocities we are given, are:
[tex]z_b =\text{balloon position}\\z_p=\text{pellet position}\\z_b=(7\frac{m}{s})t\\z_p=30\frac{m}{s}(t-t_0)-\frac{g}{2}(t-t_0)^2\\g=9.8\frac{m}{s^2}[/tex]
How do we know the value of t₀? This is the time when the pellet is fired. At this time the pellet position is zero: its initial position. To calculate it we know that the pellet is fired when the ballon is in z = 12m. Then:
[tex]t_0=\frac{12}{7}s[/tex]
We need to know the when the z values of balloon and pellet is the same:
[tex]z_b=z_p\\(7\frac{m}{s})t =30\frac{m}{s}(t-\frac{12}{7}s)-\frac{g}{2}(t-\frac{12}{7}s)^2[/tex]
We need to find the roots of the quadratic equation. They are:
[tex]t_1=2.31s\\t_2=5.81[/tex]
To know the altitude where the to objects meet, we replace the time values:
[tex]z_1=16,17m\\z_2=40,67m[/tex]
A person walks 27.0° north of east for 4.00 km. How far due north and how far due east would she have to walk to arrive at the same location?
Answer:
Along North, 1.816 km
Along east , 3.564 km
Explanation:
d = 4 km 27° North of east
As we need to find the distance traveled due north and the distance traveled due east, they are the components along north and along east. The component along north is the vertical component and the distance traveled along east is the horizontal component.
Distance traveled due North
y = d Sin 27 = 4 Sin 27 = 1.816 km
Distance traveled due east
x = d cos 27 = 4 cos 27 = 3.564 km
Jack and Jill ran up the hill at 2.8 m/s . The horizontal component of Jill's velocity vector was 2.1 m/s . What was the angle of the hill?What was the vertical component of Jill's velocity?
Answer:[tex]\theta =41.409 ^{\circ}[/tex]
Explanation:
Given
Jack and Jill ran up the hill at 2.8 m/s
Horizontal component of Jill's velocity vector was 2.1 m/s
Let [tex]\theta [/tex]is the angle made by Jill's velocity with it's horizontal component
Therefore
[tex]2.8cos\theta =2.1[/tex]
[tex]cos\theta =\frac{2.1}{2.8}[/tex]
[tex]cos\theta =0.75[/tex]
[tex]\theta =41.409 ^{\circ}[/tex]
Vertical velocity is given by
[tex]V_y=2.8sin41.11=1.85 m/s[/tex]
The question can be solved using the Pythagorean theorem to calculate the vertical velocity component and apply trigonometry to find the angle of the hill. Remember the conversion from radians to degrees.
Explanation:The question asks about the angle of the hill that Jack and Jill climbed, as well as the vertical component of Jill’s velocity. We know that Jill ran up the hill at a velocity of 2.8 m/s and the horizontal component of her velocity was 2.1 m/s. These two components form a right angle triangle where the hypotenuse is the total velocity (2.8 m/s), one side is the horizontal velocity (2.1 m/s), and the other side, which we're finding, is the vertical component of the velocity.
We can find the angle of the hill, θ, using the tangent function of trigonometry. Tan θ = opposite side / adjacent side. In this case, the 'opposite side' is the vertical velocity component we're after, and the 'adjacent side' is the horizontal component (2.1 m/s). To find the vertical velocity component, you can use the Pythagorean theorem, which states that: (Hypotenuse)² = (Adjacent Side)² + (Opposite Side)² or (2.8 m/s)² = (2.1 m/s)² + (vertical velocity)².
Once you solve for the vertical velocity through the Pythagorean theorem, you then insert it into the tangent equation to solve for the angle θ. Remember that when you use the arctangent function on a calculator to find the angle, the answer will likely be in radians, so you might need to convert them into degrees.
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A bicyclist starts from rest and accelerates at a rate of 2.3 m/s^2 until it reaches a speed of 23 m/s. It then slows down at a constant rate of 1.0 m/s^2 until it stops. How much time elapses from start to stop? We assume an answer in seconds
Answer:
33 seconds.
Explanation:
The equation for speed with constant acceleration at time t its:
[tex]V(t) \ = \ V_0 \ + \ a \ t[/tex]
where [tex]V_0[/tex] is the initial speed, and a its the acceleration.
First half of the problemStarting at rest, the initial speed will be zero, so
[tex]V_0 = 0[/tex]
the final speed is
[tex]V(t_{f1}) = 23 \frac{m}{s}[/tex]
and the acceleration is
[tex]a = 2.3 \frac{m}{s^2}[/tex].
Taking all this together, we got
[tex]V(t_{f1}) = 23 \frac{m}{s} = 0 + 2.3 \ \frac{m}{s^2} t_{f1}[/tex]
[tex]23 \frac{m}{s} = 2.3 \ \frac{m}{s^2} t_{f1}[/tex]
[tex]\frac{23 \frac{m}{s}}{2.3 \ \frac{m}{s^2}} = t_{f1}[/tex]
[tex]10 s = t_{f1}[/tex]
So, for the first half of the problem we got a time of 10 seconds.
Second half of the problemNow, the initial speed will be
[tex]V_0 = 23 \frac{m}{s}[/tex],
the acceleration
[tex]a=-1.0 \frac{m}{s^2}[/tex],
with a minus sign cause its slowing down, the final speed will be
[tex]V(t_{f2}) = 0[/tex]
Taking all together:
[tex]V(t_{f2}) = 0 = 23 \frac{m}{s} - 1.0 \frac{m}{s^2} t_{f2}[/tex]
[tex] 23 \frac{m}{s} = 1.0 \frac{m}{s^2} t_{f2}[/tex]
[tex] \frac{23 \frac{m}{s}}{1.0 \frac{m}{s^2}} = t_{f2}[/tex]
[tex] 23 s = t_{f2}[/tex]
So, for the first half of the problem we got a time of 23 seconds.
Total time[tex]t_total = t_{f1} + t_{f2} = 33 \ s[/tex]
If a beam passes from a material with a refractive index of 1.3 into a material with a refractive index of 1.7 at an angle of 25 degrees (from normal), what is the angle of refraction of the beam? Is the beam bent towards normal or away from it? Sketch a diagram of this problem with rays, angles, and the interface labeled.
Answer:
18.86° , it will bend towards normal.
Explanation:
For refraction,
Using Snell's law as:
[tex]n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}[/tex]
Where,
[tex]{\theta_i}[/tex] is the angle of incidence ( 25.0° )
[tex]{\theta_r}[/tex] is the angle of refraction ( ? )
[tex]{n_r}[/tex] is the refractive index of the refraction medium (n=1.7)
[tex]{n_i}[/tex] is the refractive index of the incidence medium ( n=1.3)
Hence,
[tex]1.3\times {sin\ 25.0^0}={1.7}\times{sin\theta_r}[/tex]
Angle of refraction = [tex]sin^{-1}0.3232[/tex] = 18.86°
Since, the light ray is travelling from a material with a refractive index of 1.3 into a material with a refractive index of 1.7 or lighter to denser medium, it will bend towards normal.
The diagram is shown below:
A steel rope has a 25 mm diameter and density of 5000 kg/m^3. The largest tension it can withstand is 320,000 N. What is the longest length of of this rope in m that can hang without breaking?
Answer:
longest length is 13304.05 m
Explanation:
given data
diameter = 25 mm
so radius r = 12.5 × [tex]10^{-3}[/tex] m
density = 5000 kg/m³
tension = 320000 N
to find out
longest length of of this rope
solution
we know that here tension is express as
T = m × g
and m is mass and g is acceleration due to gravity
here mass = density × volume
and volume = π×r²×l
here r is radius and l is length
so T = density c volume × g
put here value
320000 = 5000 × π×12.5² × [tex]10^{-3}[/tex] × l × 9.8
solve it we get length l
l = 13304.05
so longest length is 13304.05 m
A Horizontal rifel is fired at a bull's-eye. The muzzle
speedof the bullet is 670 m/s. The barrel is pointed directly at
thecenter of the bull's-eye, but the bullet strikes the
target0.025m below the center. What is the horizontal distance
betweenthe end of the rifel and the bull's eye?
Answer:
The horizontal distance is 478.38 m
Solution:
As per the question:
Initial Speed of the bullet in horizontal direction, [tex]v_{x} = 670 m/s[/tex]
Initial vertical velocity of the bullet, [tex]v_{y} = 0 m/s[/tex]
Vertical distance, y = 0.025 m
Now, for the horizontal distance, 'x':
We first calculate time, t:
[tex]y = v_{y}t - \frac{1}{2}gt^{2}[/tex]
(since, motion is vertically downwards under the action of 'g')
[tex]0.025 = 0 - \frac{1}{2}\times 9.8t^{2}[/tex]
[tex]t = \sqrt{0.05}{9.8} = 0.0714 s[/tex]
Now, the horizontal distance, x:
[tex]x = v_{x}t + \frac{1}{2}a_{x}t^{2}[/tex]
[tex]x = v_{x}t + \frac{1}{2}0.t^{2}[/tex]
(since, the horizontal acceleration will always be 0)
[tex]x = 670\times 0.714 = 478.38 m[/tex]
An aircraft is at a standstill. It accelerates to 140kts in 28 seconds. The aircraft weighs 28,000 lbs. How many feet of runway was used?
Answer:
runway use is 3307.8 feet
Explanation:
given data
velocity = 140 kts = 140 × 0.5144 m/s = 72.016 m/s
time = 28 seconds
weight = 28000 lbs
to find out
How many feet of runway was used
solution
we will use here first equation of motion for find acceleration
v = u + at ..............1
here v is velocity given and u is initial velocity that is 0 and a is acceleration and t is time
put here value in equation 1
72.016 = 0 + a(28)
a = 2.572 m/s²
and
now apply third equation of motion
s = ut + 0.5×a×t² .......................2
here s is distance and u is initial speed and t is time and a is acceleration
put here all value in equation 2
s = 0 + 0.5×2.572×28²
s = 1008.24 m = 3307.8 ft
so runway use is 3307.8 feet
To find the runway distance used by an aircraft accelerating to 140 knots in 28 seconds, we convert knots to feet per second, find the acceleration, and then apply the kinematic formula for distance to obtain approximately 3308 feet.
Explanation:To calculate how many feet of runway an aircraft used to accelerate to 140 knots in 28 seconds, we need to convert the speed to consistent units and use the kinematic equations for uniformly accelerated motion. Knots need to be converted to feet per second (since the answer is required in feet). Since 1 knot = 1.68781 feet per second, 140 knots is equivalent to 236.293 feet per second. Now, we can use the formula for distance d when given initial velocity vi, final velocity vf, and acceleration a:
d = (vi + vf) / 2 * t
Here, as the plane is starting from a standstill, vi is 0, vf is 236.293 feet/second, and t is 28 seconds. We first need to find the acceleration:
a = (vf - vi) / t = 236.293 / 28 = 8.439 feet/second2
Now we can calculate the distance:
d = (0 + 236.293) / 2 * 28 = 3308.1 feet
The aircraft used approximately 3308 feet of runway.
A force of magnitude 33.73 lb directed toward the right is exerted on an object. What other force must be applied to the object so that the resultant force is zero?
An additional force of 33.73 lb must be applied to the left to cancel out the initial force and create a resultant force of zero, according to Newton's first law of motion.
Explanation:To ensure that the resultant force on the object is zero, another force of equal magnitude but in the opposite direction must be applied. In this case, since a force of 33.73 lb is directed toward the right, an additional force must be applied to the left with the same magnitude, i.e., 33.73 lb to the left. This balances out the forces acting upon the object, creating a state of equilibrium where the sum of the forces equals zero, thus satisfying Newton's first law of motion.
Now, considering the provided example where a man applies a force of +50 N, it's clear that the force of friction opposing the motion must also be 50 N but in the negative direction (-50 N) for the forces to cancel each other and fulfill the condition of Newton’s first law where Fnet = 0.
In the context of three forces acting on an object, discussing the necessary condition for an object to move straight down, the forces in the horizontal plane (to the right and left) must be balanced. This means that force A and force B, which act in opposite directions, must be equal in magnitude.