A 70-kg circus performer is fired from a cannon that is elevated at an angle of 37° above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by 3.8 m from their unstrained length. He takes 4.1 s to travel between the launch point (where he is free from the bands) and the net into which he is shot. Assume the launch and landing points are at the same height and do not neglect the change in height during stretching.
Answers
The above is not complete. Couldn't find the it online but i have found a similar question which would help you solve the above
Answer:
Explanation:
From the question we are given that
Mass = 70kg
Angle of elevation = 40°
Length of stretch = 3.00 m
Height of performer above floor = Height of the net into the into which he is shot
Time traveled = 2.14 s
Distance traveled = 26.8 m
First obtain the horizontal velocity = [tex]\frac{Distance}{time} =\frac{26.8}{2.14} = 12.523 \ m/s[/tex]
To obtain the initial we would divide the final velocity by the cosine of the angle of elevation
Initial velocity = [tex]\frac{12.523}{cos\ 40} = 16.348m/s[/tex]
Next is to obtain the initial kinetic energy
This is equal to = [tex]\frac{1}{2} mv^2= \frac{1}{2} *70*16.348^2 = 9354 \ Joules[/tex]
Looking at the diagram in the second uploaded the cannon raised the performer before releasing him is [tex]30sin(40) = 19284 \ m[/tex]
So the potential energy given by the cannon is = mgh [tex]=70 *9.80*1.9284 = 1323 \ Joules[/tex]
Hence the total energy the band gives the performer i.e the total energy stored in the band is = 9354 +1323 = 10677 Joules
To obtain the Spring constant we would use the stored energy formula
i.e Stored Energy [tex]=\frac{1}{2}kx^2[/tex]
And we have calculated the stored energy as 10677
Substituting
[tex]10677 = \frac{1}{2} *k * 3.00^2[/tex]
[tex]k = 2373 \ N/m[/tex]
The circus performer's stunt includes aspects of elastic potential energy and projectile motion. The performer is launched due to the elastic potential energy in the cannon bands and then follows the laws of projectile motion. The performer's mass does not influence the time of flight.
Explanation:This problem demonstrates the principles of projectile motion and elastic potential energy applied to a real-world situation. The 70-kg circus performer is propelled out of the cannon because of the elastic potential energy stored in the stretched bands. Once launched, the performer's motion is an example of projectile motion.
In projectile motion, the horizontal and vertical motions are independent of each other. Therefore, it's possible to find the horizontal and vertical components of the performer's velocity separately and use them to construct the overall motion. The angle of 37° is used to calculate these components.
While the performer is in flight, the vertical motion is under constant acceleration due to gravity, while the horizontal motion has a constant velocity because air resistance is negligible. This allows us to calculate various aspects of the motion, such as the time it takes for the performer to land in the net.
Lastly, the performer's mass does not influence the time of flight in a vacuum. This illustrates an important principle of physics known as the independence of motion.
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Related Questions
A steel bucket contains 4 liters of water at 128C. An electric immersion heater rated at 1400 Watts is placed in the * Indicates an advanced level in solving. Problems 61 62 CHAPTER 1 Introduction bucket. Determine how long it will take for water to heat to 708C. Assume that the empty bucket weighs 1.1 kg. The specific heat of steel is 0.46 kJ/(kg 8C). Use an average specific heat of water of 4.18 kJ/(kg 8C). Disregard any heat loss to the surroundings.
Answers
Answer:
The time is 0.713 sec.
Explanation:
Given that,
Weight of water = 4 liters
Initial temperature = 128°C
Power = 1400 Watts
Final temperature = 708°C
Weight = 1.1 kg
Specific heat of steel = 0.46 kJ/kg°C
Specific heat of water = 4.18 kJ/kg°C
We need to calculate the heat gained by bucket
Using formula of heat
[tex]Q_{b}=mc\Delta T[/tex]
Put the value into the formula
[tex]Q_{b}=1.1\times0.46\times(70-12)[/tex]
[tex]Q_{b}=29.348\ kJ[/tex]
We need to calculate the heat gained by water
Using formula of heat
[tex]Q_{w}=mc\Delta T[/tex]
Put the value into the formula
[tex]Q_{w}=4\times4.18\times(70-12)[/tex]
[tex]Q_{w}=969.76\ kJ[/tex]
We need to calculate the total heat
Using formula of heat
[tex]Q=Q_{b}+Q_{w}[/tex]
Put the value into the formula
[tex]Q=29.348+969.76[/tex]
[tex]Q=999.108\ kJ[/tex]
We need to calculate the time
Using formula of time
[tex]t=\dfrac{Q}{P}[/tex]
Put the value into the formula
[tex]t=\dfrac{999.108}{1400}[/tex]
[tex]t=0.713\ sec[/tex]
Hence, The time is 0.713 sec.
The space between two concentric conducting spherical shells of radii b = 1.70 cm and a = 1.20 cm is filled with a substance of dielectric constant ? = 27.0. A potential difference V = 64.5 V is applied across the inner and outer shells.
(a) Determine the capacitance of the device.
nF
(b) Determine the free charge q on the inner shell.
nC
(c) Determine the charge q' induced along the surface of the inner shell.
nC
Answers
Answer:
a) C = 1.065 * 10^-10 F
b) 7.775 * 10^-9
c) 7.444 * 10^-9 C
Explanation:
A spherical capacitor, with inner radius of a = 1.2 cm and outer radius
of b = 1.7 cm is filled with a dielectric material with dielectric constant of
K = 27 and connected to a potential difference of V = 64.5 V.
(a) The capacitance of a filled air spherical capacitor is given by equation :
C = 4*π*∈o*(a*b/b-a)
if the capacitor is filled with a material with dielectric constant K, we need
to modify the capacitance as ∈o ---->k∈o , thus:
C = 4*π*∈o*(a*b/b-a)
substitute with the given values to get:
C = 4*π*(27)*(8.84*10^-12)[(1.2*10^-2)*(1.7*10^-2)/(1.7*10^-2)-(1.2*10^-2)*]
C = 1.065 * 10^-10 F
(b) The charge on the capacitor is given by q = CV, substitute to get:
q = (1.065 * 10^-10)*64.5 V
= 7.775 * 10^-9
(c) The induced charge on the dielectric material is given by equation as:
q' = q(1-1/k)
substitute with the given values to get:
q' = (7.775 * 10^-9)*(1-1/27)
= 7.444 * 10^-9 C
note:
calculation maybe wrong but method is correct. thanks
The capacitance is 3.36 nF, the free charge is 216.72 nC and the induced charge is zero.
Given information:
Radius, a = 0.017 m
b = 0.012 m
Potential difference, V = 64.5 V
(a)
The capacitance is given by:
C = (4πε₀ / (1/b - 1/a))
C = (8.85*10⁻²×3.14×4)/(1/0.012-1/0.017)
C = (4π(8.85 x 10^-12) / 293.3)
C = 3.36 nF
Hence, the capacitance is 3.36 nF.
(b)
The free charge can be calculated from the relation of charge, capacitance, and voltage:
q = CV
q = 3.36×64.5
q= 216.72 nC
Hence, the free charge is 216.72 nC.
(c)
The induced charge is given by:
q' = q - C × V
q' = 0 nC
Hence, the charge is 0 nC.
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Oxygen (O 2 ) is confined to a cubic container, 15 cm on a side, at a temperature of 300 K. Compare the average kinetic energy of a molecule of the gas to the change in its gravitational potential energy if it falls from the top of the container to the bottom.
Answers
Answer: Average kinetic energy is greater than gravitational potential energy.
Explanation: The average kinetic energy formulae for a molecule of a gas relative to temperature (in Kelvin) is given below as
E = 3/2 (KT).
Where E = average kinetic energy =?
K = boltzman constant = 1.381×10^-23 m²kg/s²k
T = temperature = 300 k
By substituting the parameters in the formulae, we have that
E = 3/2 ×( 1.381×10^-23 × 300)
E = 1.5 × 4.143×10^-21
E = 6.2145×10^-21 J
To get the gravitational potential energy, we use the fact that
Gravitational potential energy = gravitational energy at the top - gravitational energy at bottom.
At the top, the height of cube is h= 15cm = 0.15m, g = acceleration dude to gravity = 9.8m/s², m = mass of molecule of oxygen = 1.661×10^-27 kg
Gravitational potential energy = mgh = 1.661×10^-27× 9.8 × 0.15 = 2.442×10^-27 J
At the bottom, height is zero, hence gravitational potential energy is also zero.
Hence the final gravitational potential energy = potential energy at top - potential energy at bottom =
2.442×10^-27 - 0 = 2.442×10^-27 J.
Gravitational potential energy = 2.442×10^-27 J
Average kinetic energy = 6.2145×10^-21 J
As we can see that the average kinetic energy is bigger than the gravitational potential energy.
Compare the average kinetic energy of oxygen molecules at 300 K to the change in their gravitational potential energy if they fall from the top to the bottom of a container.
Oxygen (O2) molecules at 300 K have an average kinetic energy of 3kT, where k is Boltzmann's constant. If an oxygen molecule falls from the top of a container to the bottom, its change in gravitational potential energy is mgΔh, where m is the mass of the molecule, g is the acceleration due to gravity, and Δh is the height change.
Air enters a horizontal, well-insulated nozzle operating at steady state at 12 bar, 500 K, with a velocity of 50 m/s and exits at 7 bar, 440 K. The mass flow rate is 1 kg/s. Determine the net force, in N, exerted by the air on the duct in the direction offlow.Answer:
Answers
Answer:
The net force is 300.8 N
Explanation:
∆H = Cp(T1 - T2) = 1/2(V2^2 - V1^2)
Cp is the heat capacity of air at constant pressure = 1005 J/kg
T1 is initial temperature of air = 500 K
T2 is the exit temperature of air = 440 K
V1 is the initial velocity of air = 50 m/s
V2 is the exit velocity of air
1005(500 - 440) = 1/2(V2^2 - 50^2)
60,300×2 = V2^2 - 2,500
V2^2 = 120,600 + 2,500 = 123,100
V2 = sqrt(123,100) = 350.8 m/s
Net force (F) = mass flow rate × change in velocity = 1 kg/s × (350.8 - 50)m/s = 1 kg/s × 300.8 m/s = 300.8 kgm/s^2 = 300.8 N
The net force exerted by the air on the duct in the direction of flow is approximately [tex]\( 0.00293 \) N[/tex].
Given:
- Inlet conditions (state 1):
- [tex]\( P_1 = 12 \) bar = \( 12 \times 10^5 \) Pa[/tex]
- [tex]\( T_1 = 500 \) K[/tex]
- [tex]\( v_1 = 50 \) m/s[/tex]
- Exit conditions (state 2):
- [tex]\( P_2 = 7 \) bar = \( 7 \times 10^5 \) Pa[/tex]
- [tex]\( T_2 = 440 \) K[/tex]
- Mass flow rate, [tex]\( \dot{m} = 1 \) kg/s[/tex]
1. Calculate specific volumes:
[tex]\( v_1 = \frac{RT_1}{P_1} = \frac{287 \times 500}{12 \times 10^5} \approx 0.02379 \) m^3/kg[/tex]
[tex]\( v_2 = \frac{RT_2}{P_2} = \frac{287 \times 440}{7 \times 10^5} \approx 0.02086 \) m^3/kg[/tex]
2. Calculate the change in specific volume:
[tex]\( \Delta v = v_2 - v_1 = 0.02086 - 0.02379 = -0.00293 \) m^3/kg[/tex]
3. Calculate the net force in the direction of flow:
[tex]\( F = \dot{m} \cdot \Delta v = 1 \cdot (-0.00293) = -0.00293 \) N[/tex]
Since the force is negative, it indicates that the force is acting opposite to the direction of flow.
Therefore, the net force exerted by the air on the duct in the direction of flow is approximately [tex]\( 0.00293 \) N[/tex].
A system of two cylinders fixed to each other is free to rotate about a frictionless axis through the common center of the cylinders and perpendicular to the page. A rope wrapped around the cylinder of radius 2.50 m exerts a force of 4.49 N to the right on the cylinder. A rope wrapped around the cylinder of radius 1.14 m exerts a force of 9.13 N downward on the cylinder. What is the magnitude of the net torque acting on the cylinders about the rotation axis? Answer in three decimal places.
Answers
Torque is the force's twisting action about the axis of rotation magnitude of the net torque acting on the cylinders about the rotation axis will be 331.402 Nm.
What is torque?Torque is the force's twisting action about the axis of rotation. Torque is the term used to describe the instant of force. It is the rotational equivalent of force. Torque is a force that acts in a turn or twist.
The amount of torque is equal to force multiplied by the perpendicular distance between the point of application of force and the axis of rotation.
In the first situation, force is delivered in the right direction, hence torque is applied upwards.
The value of torque for case 1
[tex]\rm \tau_1=4.49\times 2.50\\\\\rm \tau_1=11.25 Nm[/tex]
The value of torque for case 2
[tex]\rm \tau_2=91.3\times 1.14\\\\\rm \tau_2=104.08[/tex]
The resultant value of torque will be;
[tex]\rm \tau =\sqrt{(11.25)^2+(41.90)^2} \\\\\ \rm \tau =331.402 Nm[/tex]
Hence the magnitude of the net torque acting on the cylinders about the rotation axis will be 331.402 Nm.
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Final answer:
The magnitude of the net torque acting on the system of two cylinders is 21.633 N⋅m.
Explanation:
The question asks us to calculate the net torque on a system of two cylinders with ropes exerting forces at different radii. Torque (τ) is calculated by the product of the radius (r), the force (F), and the sine of the angle (θ) between them, τ = rFsinθ. In this case, as both forces are perpendicular to the radii, sinθ = 1, simplifying the torque to τ = rF.
For the cylinder with radius 2.50 m, the torque exerted by the 4.49 N force is τ = 2.50 m * 4.49 N = 11.225 N⋅m. For the cylinder with radius 1.14 m, the torque from the 9.13 N force is τ = 1.14 m * 9.13 N = 10.4082 N⋅m. Since both torques are acting to rotate the system in the same direction (counterclockwise), we can sum them to find the net torque.
The magnitude of the net torque is 11.225 N⋅m + 10.4082 N⋅m = 21.6332 N⋅m, or to three decimal places, 21.633 N⋅m.
Three identical resistors, when connected in series, transform electrical energy into thermal energy at a rate of 12 W (4.0 W per resistor). Part A Determine the power consumed by the resistors when connected in parallel to the same potential difference. Express your answer with the appropriate units.
Answers
Answer:
108 Watts
Explanation:
The total circuit resistance when the resistors are connected in series is
R + R + R = 3R
When he resistors are connected in parallel, the resistance reduces from 3R in the series circuit to become;
[tex]\frac{1}{R} + \frac{1}{R} + \frac{1}{R}[/tex]
= [tex]\frac{R}{3}[/tex] Ω
[tex]Power = \frac{V^{2}}{R}[/tex]
The voltage supply was given to be constant for both the series and parallel circuits. This implies that V² is constant and power is inversely proportional to resistance.
Therefore;
Power for the parallel connected circuit = [tex]\frac{3R}{\frac{R}{3} } * 12 W[/tex]
= 9 × 12 W = 108 Watts
Answer:
Explanation:
Potential difference, V and let each resistance, R
Resistors are in series, total resistance, Rₓ = R1 + R2 + R3
= R + R + R
= 3R
Power, P = V²/Rₓ
12 = V²/3R
V²/R = 36
Resistors are in parallel, total resistance, 1/Rₓ = 1/R1 + 1/R2 + 1/R3
Rₓ = R/3
P = V²/Rₓ
P = V²/(R/3)
P = 3(V²/R)
= 3(36)
= 108 W.
Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the same time by pushing on box A with a horizontal pushing force FP = 8.9 N. Here, A has a mass mA = 10.2 kg and B has a mass mB = 7.0 kg. The contact force between the two boxes is FC. The coefficient of kinetic friction between the boxes and the floor is 0.04. (Assume FP acts in the +x direction.)
Answers
Answer:
Explanation:
The force of friction acting on the system
= .04 x 9.8 ( 10.2 + 7 )
= 6.74 N
Net force = 8.9 - 6.74
= 2.16 N
Acceleration in the system
= 2.16 / ( 10.2 + 7 )
= .12558 m / s ²
Contact force between boxes = FP
Considering force on box A
Net force = 8.9 - FP
Applying Newton's law on box A
8.9 - FP = 10.2 x .12558
= 1.28
FP = 8.9 - 1.28
= 7.62 N
A wire is formed into a circle having a diameter of 10.3 cm and is placed in a uniform magnetic field of 2.98 mT. The wire carries a current of 5.00 A. Find the maximum torque on the wire.
Answers
Answer:
T(max) = 1.17 × 10⁻⁴Nm
= 117μNm
Explanation:
T = BIA sinθ
A = area enclosed
θ = angle between normal plane
for max. torque θ = 90, (sin90° =1)
T = BIA sin90°
T= BI (πd/4)
T = [tex]T_m_a_x = \frac{1}{4} (2.98 * 10^-^3)(5)(\pi )\\T_m_a_x = 1.17 * 10^-^4Nm\\T_m_a_x = 117UNm[/tex]
A flat circular loop of radius 0.10 m is rotating in a uniform magnetic field of 0.20 T. Find the magnetic flux through the loop when the plane of the loop and the magnetic field vector are perpendicular.'
Answers
Answer:
[tex]\phi=6.28\times10^{-3}\;\;weber[/tex]
Explanation:
Given,
Magnetic field [tex]B=0.2\;\;T\\[/tex]
Radius [tex]r=0.1\;\;m[/tex]
The angle between the area vector and magnetic field is 0 degree, because the direction of area vector is always perpendicular to the plane.
[tex]\phi=BAcos\theta\\\phi=o.2\times\pi (0.1)^2\times cos0^o\\\phi=0.00628\;\;weber\\\phi=6.28\times10^{-3}\;\;weber[/tex]
The magnetic flux through a flat circular loop rotating in a uniform magnetic field is zero when the plane of the loop and the magnetic field vector are perpendicular.
Explanation:To find the magnetic flux through the loop when the plane of the loop and the magnetic field are perpendicular, we will need to use the formula for magnetic flux:
Φm = BA cos θ
where Φm is the magnetic flux through the surface, B is the magnitude of the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field direction and the normal (perpendicular) to the surface. In this case, if the loop and the magnetic field are perpendicular, θ = 90°, and cos 90° = 0. Hence, the magnetic flux will be zero.
However, if the provided problem included a different angle, we would adjust the equation to accommodate that. For example, if the field were parallel to the loop (θ = 0), then cos θ would equal 1 and the magnetic flux would just be the product of B and A (A = πr²)
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after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF capacitor from a camera flash unit retains a voltage of 150 V when an unwary student removes it from the camera. If the student accidentally touches the two terminals with his hands, and if the resistance of his body between his hands is 1.8 kΩ, for how long will the current across his chest exceed the danger level of 50 mA?
Answers
Answer:
The time is [tex]110.16\times10^{-3}\ sec[/tex]
Explanation:
Given that,
Capacitor = 120 μF
Voltage = 150 V
Resistance = 1.8 kΩ
Current = 50 mA
We need to calculate the discharge current
Using formula of discharge current
[tex]i_{0}=\dfrac{V_{0}}{R}[/tex]
Put the value into the formula
[tex]i_{0}=\dfrac{150}{1.8\times10^{3}}[/tex]
[tex]i_{0}=83.3\times10^{-3}\ A[/tex]
We need to calculate the time
Using formula of current
[tex]i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}[/tex]
Put the value into the formula
[tex]50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}[/tex]
[tex]\dfrac{50}{83.3}=e^{\frac{-t}{RC}}[/tex]
[tex]\dfrac{-t}{RC}=ln(0.600)[/tex]
[tex]t=0.51\times1.8\times10^{3}\times120\times10^{-6}[/tex]
[tex]t=110.16\times10^{-3}\ sec[/tex]
Hence, The time is [tex]110.16\times10^{-3}\ sec[/tex]
The current across the student's chest will exceed the danger level for approximately 216 seconds.
Explanation:The current that flows through the student's chest when he accidentally touches the terminals of the 120 mF capacitor can be calculated using Ohm's law: I = V/R, where I is the current, V is the voltage, and R is the resistance. In this case, the voltage is 150 V and the resistance of the student's body is 1.8 kΩ. Plugging in these values, we can find the current.
I = (150 V)/(1.8 kΩ) = 0.0833 A
To find how long the current will exceed the danger level of 50 mA, we need to calculate the time it takes for the charge on the capacitor to dissipate through the student's body. Since the current is constant, we can use the formula Q = I*t, where Q is the charge and t is the time. Rearranging the formula, we can solve for t.
t = Q/I = (120 mF * 150 V) / (0.0833 A) = 216 s
Therefore, the current will exceed the danger level of 50 mA for approximately 216 seconds.
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A 6861 kg car traveling at 48 m/s is at the foot of a hill that rises 142 m in 2.3 km. At the top of the hill, the speed of the car is 10 m/s. The acceleration of gravity is 9.81 m/s 2 . Assuming constant acceleration, find the average power delivered by the car’s engine, neglecting any internal frictional losses.
Answers
Answer: P = 25050.8w
Explanation:
total energy at top = K.E + P.E
= (1/2)(6861)(100) + 6861(9.81)(142)
total energy at bottom
= (1/2)(6861)(48)^2
work done = energy at top - energy at bottom
average velocity = (48+10)/2
time = 2300/average velocity
power = work done/time
plus potential) at the base and the top; is the energy input from the engine
the ascent time is the average speed, (top + bottom) / 2; divided by the 2.3 km distance
energy / time equals power
a certain engine has a second-law efficiency of 85%. During each cycle, it absorbs 480J of heat form a reservoir at 300C and dumps 300J of heat to a cold termperature reservoir.What is the temperature of the cold reservoir?
Answers
Final answer:
The temperature of the cold reservoir for an engine with a second-law efficiency of 85% that absorbs 480J of heat from a hot reservoir at 300C and dumps 300J into the cold reservoir is 358.125 K.
Explanation:
The question asks for the temperature of the cold reservoir for an engine with a second-law efficiency of 85% that absorbs 480J of heat from a hot reservoir at 300C and dumps 300J into the cold reservoir.
The second-law efficiency η of a heat engine is defined as the ratio of the work output W to the heat input Qh at the high temperature, while for a Carnot engine, the efficiency can also be related to the temperatures of the hot (Th) and cold (Tc) reservoirs as:
η = 1 - (Tc/Th)
Given the heat absorbed (Qh = 480 J) and the heat rejected (Qc = 300 J), we can calculate the work done (W = Qh - Qc) which is 180 J here. We know that Th is the temperature of the hot reservoir in kelvin, which we obtain by converting 300C to kelvin (Th = 573 K). Note that 0 degrees Celsius is equivalent to 273 K.
Using the given second-law efficiency:
η = W / Qh = 180 J / 480 J = 0.375
For a Carnot engine:
η = 1 - (Tc/Th)
0.375 = 1 - (Tc/573 K)
Tc = 573 K * (1 - 0.375)
Tc = 358.125 K
The temperature of the cold reservoir for this engine is therefore 358.125 K.
Two objects have the same size and shape, but one is much heavier than the other. When they are dropped simultaneously from a tower, they reach the ground at the same time (assuming that there is no air resistance), but the heavier one has a greater
speed
acceleration
none of the above
all of these
Answers
Answer:
None of the above.
The correct answer would be momentum
Answer:
Momentum (None of the above)Explanation:
The two objects free-fall at the same rate of acceleration, thus giving them the same speed when they hit the ground. The heavier object however has more momentum since momentum takes into account both the speed and the mass of the object (p=m*v).
A projectile is fired with initial speed vo at an angle of 45o above the horizontal. Assume no air resistance.
i: During the flight, the x-component of the projectile's momentum remains constant.
ii: During the flight, the y-component of the projectile's momentum remains constant.
a) i is True and ii is False
b) i is True and ii is True
c) i is False and ii is True
d) i is False and ii is False
Answers
Answer:
The correct answer is a
Explanation:
At projectile launch speeds are
X axis vₓ = v₀ = cte
Y axis [tex]v_{y}[/tex] = v_{oy} –gt
The moment is defined as
p = mv
For the x axis
pₓ = mvₓ = m v₀ₓ
As the speed is constant the moment is constant
For the y axis
p_{y} = m v_{y} = m (v_{oy} –gt) = m v_{oy} - m (gt)
Speed changes over time, so the moment also changes over time
Let's examine the answer
i True
ii False. The moment changes with time
The correct answer is a
In projectile motion without air resistance, the horizontal component of momentum remains constant while the vertical component changes due to gravity. A projectile with initial velocity 2i + j will have a final velocity of 2i - 2j before striking the ground. A projectile with velocity 2i + 3j m/s is ascending towards its maximum height.
Explanation:The student's question regards the momentum components of a projectile fired at a 45-degree angle assuming no air resistance. The correct answer to the student's multiple-choice question is that statement (i) is True and (ii) is False. This is because, in projectile motion, the horizontal component of momentum, which is dependent on the horizontal component of velocity (Vx), remains constant due to the absence of horizontal forces acting on the projectile (assuming no air resistance). On the other hand, the vertical component of momentum changes because gravity acts in the vertical direction, affecting the vertical component of velocity (Vy) over time.
Considering the examples provided, the velocity of a projectile with an initial velocity of 2i + j (in terms of unit vectors i and j) before striking the ground is 2i - 2j, considering that gravity (g = 10 m/s²) acts downward, affecting only the vertical component of velocity. For the second example, a projectile with a velocity of 2i + 3j m/s could be either ascending or descending, but since the vertical component of velocity is still positive, it suggests that the projectile is ascending to the maximum height. So, the answer would be option (d).
The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion. Assume that the North American continent can be represented by a slab of rock 4700 km on a side and 25 km deep and that the rock has an average mass density of 2880 kg/m3 . The continent is moving at the rate of about 3.2 cm/year. What is the mass of the continent
Answers
The mass of the North American continent is calculated by multiplying the volume of the continent by the density of the rock it is made of. The volume is determined based on the given dimensions, converted to meters. Multiplying the resulting volume by the rock's density provides the estimated mass of 1.59712e21 kilograms.
Explanation:In order to find the mass of a continent, we need to multiply the volume of the continent by the density of the rock that it's made of. First, we need to convert the dimensions of the North American continent into meters from kilometers since the density is in kg/m³. That's 4700km x 1000m/km =4.7e6m for length & width, and 25km x 1000m/km = 2.5e4m for depth. Thus, the volume of the continent becomes 4.7e6m x 4.7e6m x 2.5e4m = 5.54e17m³.
Now, we multiply by the rock's density, which is 2880 kg/m³. So, mass = volume * density = 5.54e17 m³ x 2880 kg/m³ = 1.59712e21 kg. So, the theoretical mass of the North American continent is approximately 1.59712e21 kilograms considering its simplified shape as a slab of rock.
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A horizontal power line carries a current of 7250 A from south to north. Earth's magnetic field (65.7 µT) is directed toward the north and is inclined downward at 70.0° to the horizontal. Find the (a) magnitude and (b) direction of the magnetic force on 140 m of the line due to Earth's field.
Answers
Answer:
a) [tex]||\vec F_{B}||=62.664\,N[/tex], b) From east to west.
Explanation:
Vectorially, the magnetic force can be calculed by the following formula:
[tex]\vec F_{B} = i\cdot \vec l\, \times \, \vec B[/tex]
The cross product is:
[tex]\vec F_{B} = \left|\begin{array}{ccc}i&j&k\\1015000\,A\cdot m&0\,A\cdot m&0\,A\cdot m\\22.471\times 10^{-6}\,T&-61.738\times 10^{-6}\,T&0\,T\end{array}\right|[/tex]
[tex]\vec F_{B} = - 62.664\,N\cdot k[/tex]
a) The magnitude of the magnetic force is:
[tex]||\vec F_{B}||=62.664\,N[/tex]
b) The direction of the magnetic force is:
From east to west.
Two charged particles are separated by 10 cm. suppose the charge on each particle is doubled. By what factor does the electric force between the particles change?
Answers
Answer:
The electric force increases by a factor of 4.
Explanation:
The electric force between two charges [tex]q_1[/tex] and [tex]q_2[/tex] separated a distance d can be calculated using Coulomb's Law:
[tex]F=\frac{kq_1q_2}{d^2}[/tex]
where [tex]k=9\times10^9Nm^2/C^2[/tex] is the Coulomb constant.
If the value of each charge is doubled, then we will have a force between them which is:
[tex]F'=\frac{k(2q_1)(2q_2)}{d^2}=4\frac{kq_1q_2}{d^2}=4F[/tex]
So the new force is 4 times larger than the original force.
Doubling the charge on each particle increases the electric force between them by a factor of 4.
Explanation:The force between two charged particles is given by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. So, if we denote the electric force as F, the charges as q1 and q2, and the distance as r, we can write Coulomb's law as F = k* q1*q2/r^2, where k is a constant.
Now if you double the charges (q1 and q2 become 2q1 and 2q2), and use these values in the formula, we get Fnew = k*(2q1) *(2q2)/r^2 = 4 * k*q1*q2/r^2 = 4F.
So, by doubling the charge on each particle, the electric force between them is multiplied by the factor of 4. So, the force increases fourfold.
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When tension is applied to a metal wire of length L , it stretches by Δ L . If the same tension is applied to a metal wire of the same material with the same cross-sectional area but of length 2 L , by how much will it stretch?
Answers
Answer:
The metal wire will stretch by [tex]2 \delta L[/tex]
Explanation:
[tex]T = \frac{kA \delta L}{L}[/tex]......................................(1)
Where T = Tension applied
ΔL = Extension
L = length
k = constant
T₁ = T₂ = T
A₁ = A₂ =A
L₁ = L
L₂ = 2L
(ΔL)₁ = ΔL
(ΔL)₂ = ?
From equation (1)
[tex]TL/kA = \delta L[/tex].....................(2)
[tex]TL/kA = (\delta L) ........................(3)\\ 2TL/kA = (\delta L)_{2} ........................(4)[/tex]
Divide (4) by (3)
[tex]\frac{(\delta L)_{2} }{\delta L} =\frac{\frac{2TL}{kA} }{\frac{TL}{kA} } \\\frac{(\delta L)_{2} }{\delta L} = 2\\ (\delta L)_{2} = 2\delta L[/tex]
If the starting length is twice, the extension is doubled as well.
Given that;
Length of metal wire = L
Stretch = ΔL
So,
It will stretched to a length of 2L.
Although when tension is continuous, the length of the extension is proportional to the size of the initials.
As a result, if the starting length is doubled, the extension will be twice as well.
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Why do you think it would be more practical to use an electromagnet to move scrap metal than to use a permanent magnet?
Answers
Using an electromagnet to lift scrap metal is advantageous because it can be turned off to release the metal, and its strength can be adjusted to handle different loads.
Using an electromagnet to move scrap metal is more practical than using a permanent magnet for several reasons. Firstly, an electromagnet can be turned on and off, useful when you want to release the metal after lifting it. This is not possible with a permanent magnet, which would require a physical effort to detach the metal pieces. Secondly, the strength of an electromagnet can be adjusted by controlling the electric current. This allows for strong magnetic effects that can be finely tuned for the weight and type of scrap being lifted. Lastly, industrial electromagnets can be designed to lift thousands of pounds of metallic waste, which might not be feasible with the size and strength of a permanent magnet.
However, there are limits to how strong electromagnets can be made, mainly due to coil resistance leading to overheating. In cases where extremely strong magnetic fields are necessary, such as in particle accelerators, superconducting magnets may be employed, although these also have their limits, since superconducting properties can be destroyed by excessively strong magnetic fields.
Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 7.50 cm apart. You see an object jump from side to side by 0.95 degree as you blink back and forth between your eyes. How far away is the object?
Answers
Answer:
Distance of the object from eye is approx 4.52 m
Explanation:
As we know that the object subtend a small angle on both the eyes which is given as
[tex]\theta = 0.995 degree[/tex]
now we know that the distance between two eyes is given as
d = 7.50 cm
so we have
[tex]angle = \frac{arc}{Radius}[/tex]
so here the radius is same as the distance from eye while arc is the distance between two eyes
so we have
[tex]0.95 \frac{\pi}{180} = \frac{7.50}{R}[/tex]
[tex]R = 452 cm[/tex]
A 84.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 36.0 m/s. If both are initially at rest and if the ice is frictionless, how far (in m) does the player recoil in the time it takes the puck to reach the goal 24.0 m away? (Enter the magnitude.)
Answers
Answer: 3333333222135790075
Explanation:Set term u equal to initial velocity for simplicity
Set V equal to final velocity for simplicity
2
To begin this problem, one must look at the system to have multiple stages. These being before and after hitting the puck. In these first few steps, we look at BEFORE the human hits the puck
3
This collision is elastic because the puck and the human do not join together after interaction
4
Because the initial velocity of both the puck and the human are both 0, the terms on the left of the equal sign become 0
5
Solving for the final velocity of the human gives this formula. This number should be negative as the negative indicates the direction he is going (left)
The final velocity of the puck is already given in the problem
6
Because the ice is frictionless, the final velocity before hitting the puck is equal to the initial velocity after hitting the puck
Now we begin to look at the system AFTER the puck has been hit
7
Using the formula for final position allows us to solve for time it takes the puck to travel the distance given
8
9
Solve for time
10
We can now use the formula for the final position of the human to solve for the final answer
11
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Plugging in formulas from steps 5 and 9 gives the final answer
Again, this number should be negative as the negative sign denotes the direction the human is going. Because the problem does not ask for snijndij hinu9nub hvtj c v7 yf jhmb tfgnb nb fyhgbv
The distance traveled by the player ( recoil ) in the time the puck reaches the goal is 0.043m.
What is law of conservation of linear momentum?According to the law of conservation of momentum, the sum of the momentum of the object before and after the collision must be equal.
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
where m₁ and m₂ is the mass of the objects, u₁ and u₂ are initial speed while v₁ & v₂ is final speed.
Given the initial velocity of the player is u₁ = 0 and the puck is u₂ = 0
The mass of the player m₁ = 84 Kg
The mass of the puck, m₂ = 0.150 Kg
The final velocity of the puck, v₂ = 36 m/s
From the law of conservation of momentum, find the velocity of the player:
m₁ u₁ + m₂ u₂ =m₁ v₁ + m₂v₂
84 × 0 + m ×0 = 84 × v + 0.150 × 36
v = - 0.064 m/s
A negative sign shows the player and puck moving in the opposite direction.
Now, we calculte the time taken for the puck to trach the goal:
Time = Distance/ Velocity
t = 24/36 = 0.667 sec
Next, we calculate the distance traveled by the player( recoil ) in the time of 0.667 seconds.
Distance = Velocity× time
S = 0.064 ×0.667
S = 0.043m
Therefore, the distance covered by the player in the time the puck reaches the goal is 0.043m.
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Tarik winds a small paper tube uniformly with 189 turns 189 turns of thin wire to form a solenoid. The tube's diameter is 7.99 mm 7.99 mm and its length is 2.19 cm 2.19 cm . What is the inductance, in microhenrys, of Tarik's solenoid?
Answers
Answer:
102.8 μH
Explanation:
The (self) inductance of a coil based on its own geometry is given as
L = (μ₀N²A)/l
where
μ₀ = magnetic constant = (4π × 10⁻⁷) H/m
N = number of turns = 189
A = Cross sectional Area = (πD²/4) = (π×0.00799²/4) = 0.00005014 m²
l = length of the solenoid = 2.19 cm = 0.0219 m
L = (4π × 10⁻⁷ × 189² × 0.00005014)/0.0219
L = 0.0001028131 H = (1.028 × 10⁻⁴) H = (102.8 × 10⁻⁶) H = 102.8 μH
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Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer connection.17 pointsb. Calculate the efficiency of the transformer in this connection when it is supplying its rated load at unity power factor. 17 points
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The High Voltage Rating for Auto - Transformer is 86kV
The Low Voltage Rating for Auto - Transformer is 78kV
The MVA rating is 268.75[tex]MVA[/tex]
b
The efficiency is 99.4%
Explanation:
From the question we are given are given that
The transformer has Mega Volt Amp rating of 25MVA
The frequency is 60-Hz
Voltage rating 8.0kV : 78kV
The short circuit test gives : 453kV,321A,77.5kW
The open circuit test gives : 8.0kV, 39.6A, 86.2kW
This can be represented on a diagram shown on the second uploaded image
From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the Low Voltage Rating for Auto - Transformer is 78kV
Now to obtain the current flowing through the 8kV coil in the Auto-transformer we have
[tex]\frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}[/tex]
The volt will cancel each other
[tex]\frac{25*10^6}{8*10^3} = 3125\ A[/tex]
Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.
[tex]MVA \ rating = (86*10^3)(3125) =268.75[/tex]
We need to understand that Iron losses is due to open circuit test which has power = 86.2kW
While copper loss is due to short circuit test which has power = 77.5kW
The the current flowing through the secondary coil [tex]I_2[/tex] as shown in the circuit diagram can be obtained as
[tex]I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321[/tex]
Now the efficiency can be obtained as thus
[tex]\frac{(operational \ MVA )*(Power factor \pf))}{(operational\ MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}[/tex]
=99.941%
Two long, parallel wires are separated by a distance of 3.30 cm. The force per unit length that each wire exerts on the other is 5.00×10−5 N/m, and the wires repel each other. The current in one wire is 0.620 A.
a. What is the current in the second wire?
b. Are the two currents in the same direction or in opposite directions?
Answers
Answer with Explanation:
We are given that
Distance between two parallel long wires=r=3.3 cm=[tex]\frac{3.3}{100}=0.033m[/tex]
1 m=100 cm
[tex]\frac{F}{l}=5\times 10^{-5} N/m[/tex]
[tex]I_1=0.62 A[/tex]
a.We have to find the current in the second wire.
We know that
[tex]\frac{F}{l}=\frac{2\mu_0I_1I_2}{4\pi r}[/tex]
Using the formula
[tex]5\times 10^{-5}=\frac{2\times 10^{-7}\times 0.62\times I_2}{0.033}[/tex]
Where [tex]\frac{\mu_0}{4\pi}=10^{-7}[/tex]
[tex]I_2=\frac{5\times 10^{-5}\times 0.033}{2\times 10^{-7}\times 0.62}[/tex]
[tex]I_2=13.3 A[/tex]
Hence, the current in the second wire=13.3 A
b.We are given that the wires repel each other.When the current carrying in the wires in opposite direction then, the wires repel to each other.
Hence,the two currents in opposite directions.
What is the phase angle of an AC series circuit that is constructed of a 14.5-Ω resistor along with 16.5-Ω inductive reactance and 9.41-Ω capacitive reactance?
Answers
Answer: cosθ = 0.7531
Explanation: the phase angle cosθ is given as
cosθ = R/Z
Where R = resistive reactance = 14.5 ohms
Z = impeadance = √R^2 +(Xl - Xc)^2
Where Xl = inductive reactance = 16.5 ohms and Xc= capacitive reactance = 9.41 ohms
By substituting the parameters, we have that
Z = √14.5^2 + (16.5^2 - 9.41^2)
Z = √210.25 + (272.25 - 88.5481)
Z = √210.25 + 183.7019
Z = √393.9519
Z = 19.85 ohms
Z = 19.85 ohms, R = 14.5 ohms
cosθ = R/Z = 14.5/19.85
cosθ = 0.7531
26.06°
Explanation:Given an RLC circuit [a circuit containing a capacitor, inductor and resistor], the phase angle (Φ), which is the difference in phase between the voltage and the current in the circuit, is given by;
Φ = tan⁻¹ [ [tex]\frac{X_{L} - X_{C}}{R}[/tex]] --------------------------(i)
Where;
[tex]X_{L}[/tex] = inductive reactance of the circuit
[tex]X_{C}[/tex] = capacitive reactance of the circuit
R = resistance of the circuit
From the question;
[tex]X_{L}[/tex] = 16.5 Ω
[tex]X_{C}[/tex] = 9.41 Ω
R = 14.5 Ω
Substitute these values into equation (i) as follows;
Φ = tan⁻¹ [ [tex]\frac{16.5 - 9.41}{14.5}[/tex]]
Φ = tan⁻¹ [ [tex]\frac{7.09}{14.5}[/tex]]
Φ = tan⁻¹ [ 0.4890]
Φ = 26.06°
Therefore the phase angle of the AC series circuit is 26.06°
A cylinder with moment of inertia 41.8 kg*m^2 rotates with angular velocity 2.27 rad/s on a frictionless vertical axle. A second cylinder, with moment of inertia 38.0 kg*m^2, initially not rotating, drops onto the first cylinder and remains in contact. Since the surfaces are rough, the two eventually reach the same angular velocity. Calculate the final angular velocity.
Answers
Explanation:
When any body rotates about its axis , the angular momentum of the body remains constant .
Thus the product of moment of inertia and its angular velocity remains constant .
In first case
The moment of inertia of cylinder = 41.8 kg m²
and Angular velocity = 2.27 rad/s
Thus angular momentum L₁ = 41.8 x 2.27 N-m s
In the second case
The moment of inertia = 41.8 + 38.0 = 79.8 kg m²
Suppose the angular velocity = ω
Thus angular momentum L₂ = 79.8 x ω
But according to principle of conservation of momentum
L₁ = L₂
41.8 x 2.27 = 79.8 x ω
Thus ω = [tex]\frac{41.8x2.27}{79.8}[/tex]
ω = 1.19 rad/s
Answer:
Final Angular Velocity=[tex]\omega[/tex]=1.189 rad/s
Explanation:
Given Data:
Moment of Inertia of 1st cyclinder=[tex]I_1=41.8 kg/m^{2}[/tex]
Angular Velocity of 1st cyclinder=[tex]\omega_1[/tex]=2.27 rad/s
Moment of Inertia of After contact=[tex]I_2=(41.8+38) kg/m^{2}=79.8 kg/m^{2}[/tex]
Required:
Final Angular velocity =[tex]\omega[/tex]=?
Formula:
Angula Momentum=L=[tex]I\omega[/tex]
Solution:
According to the conservation of angular momentum:
[tex]L_1=L_2[/tex]
[tex]I_1 \omega_1=I_2\omega\\41.8*2.27=(41.8+38)*\omega\\\omega=\frac{41.8*2.27}{79.8}\\\omega= 1.189\ rad/s[/tex]
Final Angular Velocity=[tex]\omega[/tex]=1.189 rad/s
The maximum allowed power dissipation for a 23.9 Ω 23.9 Ω resistor is stated to be 10.0 W . 10.0 W. Calculate the largest current that this resistor can take safely without burning out.
Answers
Answer:
0.65 A.
Explanation:
Given:
Pmax = 10 W
R = 23.9 Ω
Formula for calculating power,
P = I × V
= I^2 × R
I^2 = 10/23.9
I = 0.65 A.
Tarik winds a small paper tube uniformly with 183 turns 183 turns of thin wire to form a solenoid. The tube's diameter is 9.49 mm 9.49 mm and its length is 2.09 cm 2.09 cm . What is the inductance, in microhenrys, of Tarik's solenoid?
Answers
143μH
Explanation:The inductance (L) of a coil wire (e.g solenoid) is given by;
L = μ₀N²A / l --------------(i)
Where;
l = the length of the solenoid
A = cross-sectional area of the solenoid
N= number of turns of the solenoid
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
From the question;
N = 183 turns
l = 2.09cm = 0.0209m
diameter, d = 9.49mm = 0.00949m
But;
A = π d² / 4 [Take π = 3.142 and substitute d = 0.00949m]
A = 3.142 x 0.00949² / 4
A = 7.1 x 10⁻⁵m²
Substitute these values into equation (i) as follows;
L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]
L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209
L = 143 x 10⁻⁶ H
L = 143 μH
Therefore the inductance in microhenrys of the Tarik's solenoid is 143
The inductance, in microhenrys, of Tarik's solenoid should be considered as the 143μH.
Calculation of the inductance:Since
The inductance (L) of a coil wire (e.g solenoid) should be provided by
L = μ₀N²A / l --------------(i)
here,
l = the length of the solenoid
A = cross-sectional area of the solenoid
N= number of turns of the solenoid
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
So,
N = 183 turns
l = 2.09cm = 0.0209m
diameter, d = 9.49mm = 0.00949m
So,
A = π d² / 4
= 3.142 x 0.00949² / 4
= 7.1 x 10⁻⁵m²
Now
L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]
L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209
L = 143 x 10⁻⁶ H
L = 143 μH
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If the momentum of a 1000 kg car travelling at 10 m/s was transferred completely to a 20.0 kg traffic barrier, what would the final speed of the barrier be
Answers
Answer:
500 m/s
Explanation:
Momentum, p is a product of mass and velocity. From law of conservation of momentum, the initial momentum equals final momentum
[tex]m_1v_1=m_2v_2\\1000*10=20v_2\\v_2=10000/20=500 m/s[/tex]
Here m and v represent mass and velocity respectively and subscripts 1 and 2 represent car and barrier respectively
Therefore, the velocity of barrier will be 500 m/s
There is a spot of paint on the front wheel of the bicycle. Take the position of the spot at time t=0 to be at angle θ=0 radians with respect to an axis parallel to the ground (and perpendicular to the axis of rotation of the tire) and measure positive angles in the direction of the wheel's rotation. What angular displacement θ has the spot of paint undergone between time 0 and 2 seconds? Express your answer in radians using three significant figures.
Answers
Answer:
[tex]\int\limits^2_0 {w} \, dt[/tex]
Explanation:
If Angular velocity w (omega ) is given, which is defined as, w = (change in angle)/(change in unit time).
then simply taking integral from 0 to 2 gives us the answer.
Answer:
Change in theta = 0.793rad
Explanation:
Please see attachment below.