Answer:
Bob provided the greatest impulse.
Explanation:
Given:
Arnold exerts force = 300 N for 3 seconds
Bob exerts force = 500 N for 2 seconds
Cecil exerts force = 200 N for 4 seconds.
Find:
Highest impulse
Computation:
Impulse = Force × TIme
Arnold provides impulse = 300 N × 3 seconds = 900 N/s
Bob provides impulse = 500 N × 2 seconds = 1,000 N/s
Cecil provides impulse = 200 N × 4 seconds = 800 N/s
Therefore, Bob provided the greatest impulse.
A charged particle moves from point A to point B in an external electric field, and in the process its kinetic energy decreases fro, 87.6 J at A to 57.3 J at B. The electric potential at A is -48.0 V, and the electric potential at B is 18.0 V. What is the charge of the particle, including sign
Answer:
-0.46 C
Explanation:
The relationship between total kinetic energy, KE, and total electric potential, V is:
ΔKE = ΔV * q
where ΔKE = change in kinetic
ΔV = change in voltage
q = charge
The kinetic energy changes from 87.6 J to 57.3 J while the electric potential changes from -48.0 V to 18.0 V.
Therefore:
57.3 - 87.6 = (18.0 - (-48.0)) * q
-30.3 = 66q
=> q = -30.3 / 66
q = -0.46 C
The charge of the particle is -0.46 C.
Culture influence learning styles?
Answer: Cultures do have distinctive learning style patterns, but the great variation among individuals within groups means that educators must use diverse teaching strategies with all students. like, Chinese culture and Chinese learners, Adult learning stages and learning styles
Explanation: hope this helps☜(ﾟヮﾟ☜)
The Sun's energy comes from which nuclear reaction?
A. Nuclear fission
B. Gamma decay
C. Positron emission
D. Nuclear fusion
SUBMIT
Sun's energy comes from the nuclear fusion taking place inside. In nuclear fusion two light nuclei fuses together to form a heavy nuclei with the release of greater amount of energy.
What is nuclear fusion :Nuclear fusion is the process of combining two light nuclei to form a heavy nuclei. In this nuclear process, tremendous energy is released. This is the source of heat and light in stars.
On the other hand, nuclear fission is the process of breaking of a heavy nuclei into two lighter nuclei. Fission also produces massive energy. But in comparison, more energy is produced by nuclear fusion.
Nuclear fission is used in nuclear power generators. The light energy and heat energy comes form the nuclear fusion of hydrogens to form helium nuclei. Hence, option D is correct.
Find more on nuclear fusion:
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A 120-V rms voltage at 60.0 Hz is applied across an inductor, a capacitor, and a resistor in series. If the peak current in this circuit is 0.8484 A, what is the impedance of this circuit
Answer:
200 [tex]\Omega[/tex]
Explanation:
The computation of the impedance of the circuit is shown below:
Provided that
RMS voltage = 120 v
Frequency = 60.0 Hz
RMS current = 0.600 A
Based on the above information, the formula to compute the impedance is
[tex]Z=\dfrac{V_{max}}{I_{peak}}[/tex]
where,
[tex]V_{max} = \sqrt{2} \times V_{rms}[/tex]
[tex]= \sqrt{2} \times 120[/tex]
[tex]= 169.7 V[/tex]
And, [tex]I_Peak = 0.8484[/tex]
Now placing these above values to the formula
So, the impedance of the circuit is
[tex]= \frac{169.7}{0.8484}[/tex]
= 200 [tex]\Omega[/tex]
Mr Smith is working in a muddy garden. When he picks up a paving stone his feet sink deeper into the mud. Explain why his feet sink into the mud more when he picks up the paving stone. Write about pressure in your answer
because he is carrying more mass and as the ground is muddy his feet goes in due to the pull of gravity
A car travels 2500 m in 8 minutes. Calculate the speed at which the car travelled
Answer:
5.95m/s to 2 decimal places
Explanation:
In physics speed is measured in metres per second so convert 8mins to seconds
8x60=420 seconds
The formula needed:
Speed (m/s)= Distance (m)/Time (s)
2500/420=5.95m/s
An adiabatic nozzle is used to accelerate 6000 kg/hour of CO2 to 450 m/s. CO2 enters the nozzle at 1000 kPa and 500 C. The inlet area of the nozzle is 40 cm2. Find the inlet velocity and the exit temperature of CO2.
Answer:
a. inlet velocity = 60.8m/s
b. exit temperature = 686k
Explanation:
You throw a small ball upward and notice the time it takes to come back. If you then throw the same ball so that it takes twice as much time to come back, what is true about the motion of the ball the second time?(a) Its initial speed was twice the speed in the first experiment.(b) It traveled an upward distance that is twice the distance of the original toss.(c) It had twice as much acceleration on the way up as it did the first time.(d) The ball stopped at the highest point and had zero acceleration at that point.
Answer:
(a) Its initial speed was twice the speed in the first experiment.
Explanation:
In order to determine what means that in the second launching the flight time of the same ball is twice respect to the first launching, you use the following formula:
[tex]t=2\sqrt{\frac{2h}{g}}[/tex] (1)
h: height reached by the ball
g: gravitational constant
In the equation you take into account that t is the time that ball takes to go upward and go downward, that is the reason of the factor 2 before the square root.
Furthermore you use the fact that the maximum height reached by the ball is given by:
[tex]h=\frac{v_o^2}{2g}[/tex] (2)
Next, you replace the equation (2) into the equation (1):
[tex]t=2\sqrt{\frac{2v_o^2}{2g^2}}=2\frac{v_o}{g}[/tex]
Then, you can notice that if the initial velocity is twice the flight time of the ball is also twice.
Hence, the anwser is:
(a) Its initial speed was twice the speed in the first experiment.
A rock falls from a vertical cliff that is 4.0 m tall and experiences no significant air resistance as it falls. At what speed will its gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy
Answer:
About 6.26m/s
Explanation:
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
Divide both sides by mass:
[tex]gh=\dfrac{1}{2}v^2[/tex]
Since the point of equality of kinetic and potential energy will be halfway down the cliff, height will be 4/2=2 meters.
[tex](9.8)(2)=\dfrac{1}{2}v^2 \\\\v^2=39.4 \\\\v\approx 6.26m/s[/tex]
Hope this helps!
The gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.
Given data:
The height of vertical cliff is, h = 4.0 m.
Since, we are asked for speed by giving the condition for gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy. Then we can apply the conservation of energy as,
Kinetic energy = Gravitational potential energy
[tex]\dfrac{1}{2}mv^{2}=mgh[/tex]
Here,
m is the mass of rock.
v is the speed of rock.
g is the gravitational acceleration.
Solving as,
[tex]v=\sqrt{2gh}\\\\v=\sqrt{2 \times 9.8 \times 4.0}\\\\v =8.85 \;\rm m/s[/tex]
Thus, we can conclude that the gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.
Learn more about the conservation of energy here:
https://brainly.com/question/15707891
Two light pulses are emitted simultaneously from a source. The pulses take parallel paths to a detector 6.75 m away, but one moves through air and the other through a block of ice. Determine the difference in the pulses' times of arrival at the detector.
Answer:
Δt=7ns = 7x10⁻⁹
Explanation:
You have a tennis ball that you throw directly upward with initial speed v. The ball rises to height h, then falls back down. At time t, you catch the ball when it is moving downward with the same speed v. Determine the average velocity of the tennis ball during this time interval.
Answer:
v_average =0
Explanation:
In kinematics the average speed defined with the distance interval traveled in a given time interval
V = Δx /Δt
in this case the tennis ball travels a distance h to the highest point and a distance h from the highest point to the floor, so the distance you travel zero
Dy = h -h = 0
therefore the average speed should be zero
v_average =0
Tether ball is a game children play in which a ball hangs from a rope attached to the top of a tall pole. The children hit the ball, causing it to swing around the pole. What is the total initial acceleration of a tether ball on a 2.0 m rope whose angular velocity changes from 13 rad/s to 7.0 rad/s in 15 s
Answer:
a_total = 14.022 m/s²
Explanation:
The total acceleration of a uniform circular motion is given by the following formula:
[tex]a=\sqrt{a_c^2+a_T^2}[/tex] (1)
ac: centripetal acceleration
aT: tangential acceleration
Then, you first calculate the centripetal acceleration by using the following formula:
[tex]a_c=r\omega^2[/tex]
r: radius of the circular trajectory = 2.0m
w: final angular velocity of the ball = 7.0 rad/s
[tex]a_c=(2.0m)(7.0rad/s)^2=14.0\frac{m}{s^2}[/tex]
Next, you calculate the tangential acceleration. aT is calculate by using:
[tex]a_T=r\alpha[/tex] (2)
α: angular acceleration
The angular acceleration is:
[tex]\alpha=\frac{\omega_o-\omega}{t}[/tex]
wo: initial angular velocity = 13 rad/s
t: time = 15 s
Then, you use the expression for the angular acceleration in the equation (1) and solve for aT:
[tex]a_T=r(\frac{\omega_o-\omega}{t})=(2.0m)(\frac{7.0rad/s-13.0rad/s}{15s})=-0.8\frac{m}{s^2}[/tex]
Finally, you replace the values of aT and ac in the equation (1), in order to calculate the total acceleration:
[tex]a=\sqrt{(14.0m/s^2)^2+(-0.8m/^2)^2}=14.022\frac{m}{s^2}[/tex]
The total acceleration of the ball is 14.022 m/s²
A train slows down as it rounds a sharp horizontal turn, going from 86.0 km/h to 56.0 km/h in the 18.0 s that it takes to round the bend. The radius of the curve is 160 m. Compute the acceleration at the moment the train speed reaches 56.0 km/h. Assume the train continues to slow down at this time at the same rate.
Answer:
The acceleration of the train is 1.581 m/s² inward.
Explanation:
Given;
initial velocity of the train, u = 86.0 km/h = 23.889 m/s
final velocity of the train, v = 56.0 km/h = 15.556 m/s
change in time, Δt = 18 s
The total acceleration of particles moving along a curved path is given as vector sum of the tangential acceleration and radial acceleration
[tex]a = \sqrt{a_t^2 + a_r^2}[/tex]
where;
[tex]a_t[/tex] is the tangential acceleration
[tex]a_r[/tex] is radial acceleration
[tex]a_t = \frac{v-u}{t} \\a_t = \frac{15.556-23.889}{18} \\\\a_t = -0.463 \ m/s^2 \\\\a_t = 0.463 \ m/s^2 \ \ (inward)[/tex]
[tex]a_r = \frac{v^2}{r} \\\\a_r = \frac{15.556^2}{160} \\\\a_r = 1.512 \ m/s^2[/tex]
[tex]a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{(-0.463)^2+(1.512)^2} \\\\a = \sqrt{2.5005} \\\\a = 1.581 \ m/s^2[/tex]
Therefore, the acceleration at the moment the train speed reaches 56.0 km/h is 1.581 m/s² inward.
An ideal, or Carnot, heat pump is used to heat a house to a temperature of 294 K (21 oC). How much work must the pump do to deliver 3000 J of heat into the house (a) on a day when the outdoor temperature is 273 K (0 oC) and (b) on another day when the outdoor temperature is 252 K (-21 oC)
Answer:
a) [tex]W_{in} = 214.286\,J[/tex], b) [tex]W_{in} = 428.571\,J[/tex]
Explanation:
a) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.
[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.
Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 273\,K[/tex]. The Coefficient of Performance is:
[tex]COP_{HP} = \frac{294\,K}{294\,K-273\,K}[/tex]
[tex]COP_{HP} = 14[/tex]
Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.
[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]
The input work to deliver a determined amount of heat to the house:
[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]
If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 14[/tex], the input work that is needed is:
[tex]W_{in} = \frac{3000\,J}{14}[/tex]
[tex]W_{in} = 214.286\,J[/tex]
b) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.
[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.
Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 252\,K[/tex]. The Coefficient of Performance is:
[tex]COP_{HP} = \frac{294\,K}{294\,K-252\,K}[/tex]
[tex]COP_{HP} = 7[/tex]
Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.
[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]
The input work to deliver a determined amount of heat to the house:
[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]
If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 7[/tex], the input work that is needed is:
[tex]W_{in} = \frac{3000\,J}{7}[/tex]
[tex]W_{in} = 428.571\,J[/tex]
You throw a ball straight up into the air from the top of a building. The building has a height of 15.0 m. The ball reaches a height (measured from the ground) of 25.0 m and then it starts to fall back down.
a) Determine the initial velocity of the ball.
b) What is the velocity of the ball when it comes back down and is at the same height from which it was thrown?
c) How long will it take the ball to come back down to this height from the time at which it was first thrown?
d) Let’s say that you missed catching the ball on the way back down and it fell to the ground. How long did it take to hit the ground from the moment you threw it up?
e) What was the ball’s final velocity the moment before it hit the ground?
Answer:
a) vo = 14m/s
b) v = 14m/s
c) t = 2.85s
d) t = 0.829s
e) v = 22.12 m/s
Explanation:
a) To find the initial velocity of the ball yo use the following formula:
[tex]h_{max}=\frac{v_o^2}{2g}[/tex] (1)
hmax: maximum height reached by the ball but measured from the point at which the ball is thrown = 25.0m - 15.0m = 10.0m
vo: initial velocity of the ball = ?
g: gravitational acceleration = 9.8m/s^2
You solve the equation (1) for vo and replace the values of the other parameters:
[tex]v_o=\sqrt{2gh_{max}}}=\sqrt{2(9.8m/s^2)(10.0m)}=14\frac{m}{s}[/tex]
The initial velocity of the ball is 14m/s
b) To find the velocity of the ball when it is at the same position as the initial point where it was thrown, you can use the following formula:
[tex]v^2=2gh_{max}\\\\v=\sqrt{2gh_{max}}[/tex]
as you can notice, v = vo = 14m/s
The velocity of the ball is 14 m/s
c) The flight time of the ball is given by twice the time the ball takes to reach the maximum height. You use the following formula:
[tex]t=2\frac{v_o}{g}=2\frac{14m/s}{9.8m/s^2}=2.85s[/tex] (3)
The time is 2.85s
d) To find the time the ball takes to arrive to the ground after the ball passes the same height at which is was thrown, you can use the following formula:
[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex] (4)
y: 0 m (ball just after it impact the ground)
yo: initial position = 15.0 m
vo: in)itial velocity of the ball = 14m/s
t: time
You replace the values of the parameters in the equation (4) and obtain a quadratic formula:
[tex]0=15.0-14t-\frac{1}{2}(9.8)t^2\\\\[/tex]
You use the quadratic formula to find the roots t:
[tex]t_{1,2}=\frac{-(-14)\pm\sqrt{(-14)^2-4(4.9)(15)}}{2(-4.9)}\\\\t_{1,2}=\frac{14\pm22.13}{-9.8}\\\\t_1=0.829s\\\\t_2=-2.19s[/tex]
you choose the positive values because is has physical meaning
The time the ball takes to arrive to the ground is 0.829s
e) The final velocity is:
[tex]v=v_o+gt[/tex]
[tex]v=14m/s+(9.8m/s^2)(0.829s)=22.12\frac{m}{s}[/tex]
The final velocity is 22.14 m/s
An airplane is flying on a bearing of N 400 W at 500 mph. A strong jet-stream speed wind of 100 mph is blowing at S 500 W.
Required:
a. Find the vector representation of the plane and of the wind.
b. Find the resultant vector that represents the actual course of the plane.
c. Give the resulting speed and bearing of the plane.
Answer:
A. a (-321.393, 383.022) b (-76.40, -64.278)
B. (-397.991, 318.744)
C. a. resulting speed 509.9mph b. bearing of the plane = 51.6°
Explanation:
A cat of mass m climbs over an inclined plane of height h and then climbs down along the vertical side of the same inclined plane until it reaches the base of the said inclined plane. Calculate the work made by the gravitational force applied onto the cat by the Earth a) while climbing; b) while climbing down; c) total work.
Answer:
19.62mh or 2mgh
Explanation:
the mass of the cat = m
height of the inclined plane = h
NB: The work made by the gravitational force applied onto the cat by the Earth will be equal to the work done by the cat to overcome this force.
a) For climbing (directly) up the inclined plane, the work done by the cat against gravity will be
work = weight x distance
weight = mg = mass x acceleration due to gravity
where g acceleration due to gravity = 9.98 m/s^2
work = -mgh = -9.81mh
the negative sign indicates that the work is done against gravity
b) while climbing down the vertical side of the inclined plane, work done by the cat is the same amount of work the cat does in climbing (directly) up the inclined plane, if we disregard losses due to friction. The only advantage of the inclined plane is that the inclined plane allows the same work to be done with a smaller force exerted over a greater distance; by spreading the distance (now the length of the inclined plane), and reducing the force (weight of the cat) by resolving it into a smaller force that is normal to the inclined plane. All in all the work done is equal and is in accordance with the conservation of energy.
the work done by gravity on the cat on climbing down is
work = mgh = 9.81mh
c) total work done = final work - initial work = 9.81mh - (-9.81mh) = 19.62mh
circular saw blade of diameter 0.190 m starts from rest. In a time interval of 5.80 s it accelerates with constant angular acceleration to an angular velocity of 145 rad/s. Part A Find the angular acceleration. Express your answer in radians per second squared. α = nothing rad/s2 Request Answer Part B Find the angle through which the blade has turned. Express your answer in radians. θ = nothing rad Request Answer Provide Feedback
Answer:
Explanation:
We shall apply newton's law equation valid for rotational motion .
ωt = ω₀ + α t where ωt is angular velocity after time t , ω₀ is angular velocity at t= 0 and α is angular acceleration
145 = 0 + α x 5.8
α = 25 rad / s²
B )
θ = ω₀ t + 1/2 α t²
= 0 + .5 x 25 x 5.8 x 5.8
= 420.5 rad .
Electric charge can accumulate on an airplane in flight. You mayhave observed needle-shaped metal extensions on the wing tips andtail of an airplane. Their purpose is to allow charge to leak offbefore much of it accumulates. The electric field around the needleis much larger than the field around the body of the airplane and,can become large enough to produce dielectric breakdown of the air,discharging the airplane. To model this process, assume that twocharged spherical conductors are connected by a long conductingwire and a charge of 22.0 µC isplaced on the combination. One sphere, representing the body of theairplane, has a radius of 6.00 cm, and the other, representing thetip of the needle, has a radius of 2.00 cm.(a) What is the electric potential of eachsphere?(b) What is the electric field at the surface of each sphere?
Answer: (a) [tex]V_1 = V_2[/tex] = 24.75x[tex]10^{5}[/tex] V
(b) [tex]E_1 = 1.48.10^{11}[/tex] V/m
[tex]E_2 = 4.5.10^{11}[/tex] V/m
Explanation: Eletric Potential (V) is the amount of energy necessary to move a charged particle inside an electric field. It is calculated as:
[tex]V= \frac{k.q}{r}[/tex]
where:
k is coulomb's constant: k = 9.[tex]10^{9}[/tex] N.m²/C²
q is the charge of the object
r is the distance
Electric Field (E) is what surrounds an electric particle in a way that every particle inside the field is influenciated by it, through force of attraction or of repulsion. When related to electric potential, can be calculated as: E = [tex]\frac{V}{q}[/tex]
a) The ratio of the two charges is proportional to the ratio of the two radii:
[tex]\frac{q_1}{q_2} = \frac{r_1}{r_2}[/tex]
where 1 represents the sphere of the body of the airplane and 2 is the tip of the needle.
[tex]q_{1} = q_2.\frac{r_1}{r_2}[/tex]
[tex]q_1 = q_2.\frac{6}{2}[/tex]
[tex]q_1 = 3q_2[/tex] (1)
The combine charges of spheres results in a charge of 22.0µC, which means:
[tex]q_1 + q_2 =[/tex] 22.[tex]10^{-6}[/tex]
Substitute and resolve:
[tex]3q_2 + q_2[/tex] = 22.[tex]10^{-6}[/tex]
[tex]q_2 = \frac{22.10^{-6} }{4}[/tex]
[tex]q_2 = 5.5.10^{-6}[/tex] C
Using (1) to find the other charge:
[tex]q_1 = 3.5.5.10^{-6}[/tex]
[tex]q_1[/tex] = 16.5.[tex]10^{-6}[/tex] C
Now, to determine electric potential for each sphere:
Electric Potential for Sphere 1:
[tex]V_1 = \frac{9.10^{9}.16.5.10^{-6} }{6.10^{-2} }[/tex]
[tex]V_1 =[/tex] 24.75.[tex]10^{5}[/tex] V
Electric Potential for Sphere 2:
[tex]V_2 = \frac{9.10^{9}.5.5.10^{-6}}{2.10^{-2} }[/tex]
[tex]V_2 =[/tex] 24.75.[tex]10^{5}[/tex] V
The electric potential of each sphere is the same and has magnitude 24.75.[tex]10^{5}[/tex] V.
b) Electric Field for Sphere 1:
[tex]E_1 = \frac{24.75.10^{5} }{16.5.10^{-6} }[/tex]
[tex]E_1 = 1.48.10^{11}[/tex] V/m
Electric field for Sphere 2:
[tex]E_2 = \frac{24.75.10^{5} }{5.5.10^{-6} }[/tex]
[tex]E_2[/tex] = [tex]4.5.10^{11}[/tex] V/m
How far away is a train if you see the steam from it’s nozzle 5sec before you hear it’s sound(take the speed of sound is 331m/s)
Answer:
so the train is 1655m away
Explanation:
speed of sound=331m/s
time taken=5s
distance=?
as we know that
speed=distance/time
here we have to find the distance
speed×time=distance
distance=331m/s×5s
distance=1655m
Answer:
1655m is the answer
Explanation:
hope i am rightwhich of the following best describes a stable atom?
Learning Goal: To practice Problem-Solving Strategy: Simple Harmonic Motion II: Energy. A child's toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of the baby's crib so that when the baby pulls on the toy and lets go, the object oscillates horizontally with a simple harmonic motion. The amplitude of the oscillation is 6 cm and the maximum velocity achieved by the toy is 3.2 m/s . What is the kinetic energy K of the toy when the spring is compressed 4.1 cm from its equilibrium position
Answer:
K = 0.1357 J
Explanation:
This is a problem of the simple harmonium movement, where the system this former for ball and spring. This is represented by relate
x = A cos (wt + Ф)
They indicate the maximum speed of the system, let's find the speed
v = dx / dt
v = - A w sin (wt + Ф)
for maximum speed the sine is ±1
v = A w
w = v / A
let's reduce to the SI system
W = 6cm (1m / 100cm) = 0.06 m
w = 3.2 / 0.06
w = 53.33 rad / s
let's substitute the values in the equation
x = 0.06 cos (53.34 t)
phase angle is zero since the system is released from maxillary elongation
let's find the time for ax = 4.1 cm = 0.041 m
cos (53.34t) = x / 0.06
t = 1 / 53.34 cos -1 (x / 0.06)
t = 1 / 53.35 cos-1 (0.04 / 0.06)
t = 0.0187 0.818
t = 0,0153
v = Aw sin wt
v = 0.06 53.34 sin (53.34 0.0153)
v = 3.2 0.728
v = 2.33 m / s
now we can search for kinetic energy
K = ½ m v²
K = ½ 0.050 2.33²
K = 0.1357 J
If the number of loops in a coil around a moving magnet doubles, the emf created:_________
a. Doubles
b. Halves
c. Remains the same
Answer is a. Doubles
when the loops are increased in the coil then the magnetic field created doubles
A 5.31 kg object is swung in a vertical circular path on a string 2.99 m long. The acceleration of gravity is 9.8 m/s 2 . If the speed at the top of the circle is 6.2 m/s, what is the tension in the string when the object is at the top of the circle
Answer:
T = 120.3 N
Explanation:
Since, the tension in the rope is acting against both the centripetal force and the weight of the stone. As both act downward towards center of the circle and tension acts towards point of support that is upward. So, tension will be equal to the sum of centripetal force and weight of the stone:
Tension = Centripetal Force + Weight of Stone
T = mv²/r + mg
where,
m = mass of stone = 5.31 kg
r = radius of circle = length of string = 2.99 m
g = 9.8 m/s²
Therefore,
T = (5.31 kg)(6.2 m/s)²/(2.99 m) + (5.31 kg)(9.8 m/s²)
T = 68.27 N + 52.03 N
T = 120.3 N
The speed v of an object falling with a constant acceleration g can be expressed in terms of g and the distance traveled from the point of release, h, as v = kgp hq, where k, p, and q, are dimensionless constants. What must be the values of p and q?
Answer:
So the values are [tex]p= \frac{1}{2}[/tex] , [tex]q= \frac{1}{2}[/tex]
Explanation:
From the question we are told that
The equation is [tex]v = k [g^p][h^q][/tex]
Now dimension of v (speed ) is
[tex]v = m/s = LT^{-1}[/tex]
Now dimension of g (acceleration ) is
[tex]g= m/s^2 = LT^{-2}[/tex]
Now dimension of h (vertical distance ) is
[tex]h= m = L[/tex]
So
[tex]LT^{-1} = [ [LT^{-2}]^p][[ L]^q][/tex]
[tex]LT^{-1} = [ [T^{-2p}][[ L]^{p +q}][/tex]
Equating powers
[tex]1 =p+q[/tex]
[tex]-1 = -2p[/tex]
=> [tex]p= \frac{1}{2}[/tex]
and
[tex]q= 1 -\frac{1}{2} = \frac{1}{2}[/tex]
A transformer supplies 60 watts of power to a device that is rated at 20 volts (rms). The primary coil is connected to a 120-volt (rms) ac source. What is the current I1 in the primary coil?
Answer:
I = 0.5A
Explanation:
Hello,
Assuming the transformer is an ideal transformer, we can calculate the value of the current using the formula.
Data;
Power = 60 watt
Primary voltage = 120 volts (rms)
Power (P) = current (I) × voltage (V)
P = IV
This formula is the relationship between power, current and voltage.
Current = power / voltage
I = P / V
I = 60 / 120
I = 0.5A
The current in the primary coil is 0.5A
Location C is 0.021 m from a small sphere that has a charge of 5 nC uniformly distributed on its surface. Location D is 0.055 m from the sphere. What is the change in potential along a path from C to D?
Answer:
ΔV = -1321.73V
Explanation:
The change in potential along the path from C to D is given by the following expression:
[tex]\Delta V=-\int_a^bE dr[/tex] (1)
E: electric field produced by a charge at a distance of r
a: distance to the sphere at position C = 0.021m
b: distance to the sphere at position D = 0.055m
The electric field is given by:
[tex]E=k\frac{Q}{r^2}[/tex] (2)
Q: charge of the sphere = 5nC = 5*10^-9C
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
You replace the expression (2) into the equation (1) and solve the integral:
[tex]\Delta V=-kQ\int_a^b \frac{dr}{r^2}=-kQ[-\frac{1}{r}]_a^b[/tex] (3)
You replace the values of a and b:
[tex]\Delta V=(8.98*10^9Nm^2/C^2)(5*10^{-9}C)[\frac{1}{0.055m}-\frac{1}{0.021m}]\\\\\Delta V=-1321.73V[/tex]
The change in the potential along the path C-D is -1321.73V
A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pulley, which is a disk of radius 9.00 cm , has friction in its axle.What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium? (Answer should be in N m)
Answer:
The frictional torque is [tex]\tau = 0.2505 \ N \cdot m[/tex]
Explanation:
From the question we are told that
The mass attached to one end the string is [tex]m_1 = 0.341 \ kg[/tex]
The mass attached to the other end of the string is [tex]m_2 = 0.625 \ kg[/tex]
The radius of the disk is [tex]r = 9.00 \ cm = 0.09 \ m[/tex]
At equilibrium the tension on the string due to the first mass is mathematically represented as
[tex]T_1 = m_1 * g[/tex]
substituting values
[tex]T_1 = 0.341 * 9.8[/tex]
[tex]T_1 = 3.342 \ N[/tex]
At equilibrium the tension on the string due to the mass is mathematically represented as
[tex]T_2 = m_2 * g[/tex]
[tex]T_2 = 0.625 * 9.8[/tex]
[tex]T_2 = 6.125 \ N[/tex]
The frictional torque that must be exerted is mathematically represented as
[tex]\tau = (T_2 * r ) - (T_1 * r )[/tex]
substituting values
[tex]\tau = ( 6.125 * 0.09 ) - (3.342 * 0.09 )[/tex]
[tex]\tau = 0.2505 \ N \cdot m[/tex]
Answer:here to earn points
Explanation:
Dolphins of the open ocean are classified as Type II Odontocetes (toothed whales). These animals use ultrasonic "clicks" with a frequency of about 55 kHz to navigate and find prey. You may want to review (Pages 465 - 467) . Part A Suppose a dolphin sends out a series of clicks that are reflected back from the bottom of the ocean 75 m below. How much time elapses before the dolphin hears the echoes of the clicks
Answer:
0.1 sec
Explanation:
frequency of the clicks produced = 55 kHz = 55000 Hz
depth of the bottom of ocean from the dolphin = 75 m
we know that the speed of sound in water is generally accepted to be ≅ 1480 m/s.
the total distance traveled by the sound from the dolphin, to the bottom of the ocean, and then back to the dolphin = 2 x 75 = 150 m
time elapsed will then be
time = distance traveled ÷ speed of sound
time = 150/1480 ≅0.1 sec
A 28.5-g object moving to the right at 18.5 cm/s overtakes and collides elastically with a 12.5-g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision. (Take the positive direction to be to the right. Indicate the direction with the sign of your answer.)
Answer:
Velocity of each object is 17.43 cm/s towards the rightExplanation:
Momentum = Mass *velocity
Before Collision
Momentum of 28.5-g object moving to the right at 18.5 cm/s = 28.5 * 18.5
= 527.25 gcm/s
Momentum of 12.5-g object moving to the right at 15.0 cm/s = 12.5 * 15.0
= 187.5 gcm/s
Sum of their momentum before collision = 527.25 + 187.5 = 714.75 gcm/s
After collision
Momentum of the bodies = (28.5+12.5)v = 41v
v = common velocity of both objects
According to law of conservation of momentum, the sum of momentum of the bodies before collision is equal to their sum after collision.
41v = 714.75
v = 714.75/41
v = 17.43 cm/s (since their common velocity is positive, the direction will be to the right)
Since both objects move with the same velocity after collision, each object will have the same velocity which is 17.43 cm/s after the collision
