Answer:
A. Ein = 8.05*10^-4 V/m
B. Clockwise sense
Explanation:
A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:
[tex]\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}[/tex] (1)
Ein: induced electric field
ds: differential of a path of the ring
ФB: magnetic flux in the ring
The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):
[tex]\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}[/tex] (2)
dB/dt = -0.280T/s (it is decreasing)
A: area of the ring = π(r/2)^2= (π/4) r^2
r: radius of the ring = 4.60/2 = 2.30 cm
Then, you replace the values of all variables in the equation (2):
[tex]E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}[/tex]
hence, the induced electric field is 8.05*10^-4 V/m
B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.
The Gulf Stream off the east coast of the United States can flow at a rapid 3.8 m/s to the north. A ship in this current has a cruising speed of 8.0 m/s . The captain would like to reach land at a point due west from the current position.
At this heading, what is the ship's speed with respect to land?
Answer:
61.6° west of South
Explanation:
The ship goes to the south at an equal rate just like water flows to the north. Thus, the velocities would balance making the ship move towards the west.
Since we're dealing with water, the ship goes 3.8 m / s to the South, but a lot still remains to the west. Finding this would require us drawing a triangle. 3.8 m/s point down side and the hypotenuse is 8
cos(θ) = [adjacent/hypotenuse]
Cos θ = 3.8/8
Cos θ = 0.475
θ = cos^-1 (0.475)
θ = 61.6°
Therefore the angle is 61.6° west of South.
The average, year-after-year conditions of temperature, precipitation, winds, and cloud in an area are known as its
A.climate.
b.weather.
C. global warming
d. seasons
Answer:
a. global warming
Explanation:
that's the definitain of global warming
Answer:
A climate
Explanation:
A hornet circles around a pop can at increasing speed while flying in a path with a 12-cm diameter. We can conclude that the hornet's wings must push on the air with force components that are Group of answer choices down and backwards. down, backwards, and outwards. down and inwards. down and outwards. straight down.
Answer:
down, backwards, and outwards.
Explanation:
For a hornet that is accelerating in flight, this means that there is a net forward motion at a relatively constant vertical height above the ground.
For this flight, the wings beat downwards to counter the weight of the hornet due to gravity, keeping it at that height above the floor.
For the hornet to accelerate forward, there has to be a net backwards force by the wing on the air. This backwards force accelerates tr forward due to the absence of an equal opposing force in the opposite direction save for a little drag.
The wings also beat with forces directed outwards to provide centripetal force to keep the hornet stable. The absence of this would cause it to spiral out of control.
02
Blue light has a frequency of about 7.5 x 1014 Hz. Calculate the energy, in Joules, of a single photon associated with this frequency
Answer:
49.725× 10^-24J
Explanation:
The Energy associated with a Photon us defined as;
E = hf
Where h is Planck's constant = 6.63× 10^-34m2kg/s
f is the frequency= 7.5 x 10^14 Hz
Hence
E = 6.63× 10^-34 × 7.5 x 10^14 =49.725× 10^-24J
Having aced your Physics 2111 class, you get a sweet summer-job working in the International Space Station. Your room-mate, Cosmonaut Valdimir tosses a banana at you at a speed of 16 m/s. At exactly the same instant, you fling a scoop of ice cream at Valdimir along exactly the same path. The collision between banana and ice cream produces a banana split 8.2 m from your location 1.4 s after the banana and ice cream were launched.
1. How fast did you toss the ice cream?
2. How far were you from Valdimir when you tossed the ice cream?
Answer:
a
The speed is [tex]s = 5.857 m/s[/tex]
b
The distance is [tex]D = 22.4 \ m[/tex]
Explanation:
From the question we are told that
The speed of the banana is [tex]v = 16 \ m/s[/tex]
The distance from my location is [tex]d = 8.2 \ m[/tex]
The time taken is [tex]t = 1.4 \ s[/tex]
The speed of the ice cream is
[tex]s = \frac{d}{t}[/tex]
substituting values
[tex]s = \frac{8.4}{1.4}[/tex]
[tex]s = 5.857 m/s[/tex]
The distance of separation between i and Valdimir is the same as the distance covered by the banana
So
[tex]D = v * t[/tex]
substituting values
[tex]D = 16 * 1.4[/tex]
[tex]D = 22.4 \ m[/tex]
A man pushes a 25kg box up an incline 2.0m by applying a steady force of 95N parallel to the incline. The box moves up the incline at a steady speed. The incline makes an angle 15 degrees to the horizontal
a) What is the force of friction between the incline and the box
b)If the box is released at the top of the incline, what will its speed be at the bottom
Answer:
a) Ff = 19.29 N
b) v = 3.00 m/s
Explanation:
a) To calculate the friction force you use the second Newton Law in the incline plane, with an acceleration equal to zero, because the motion of the box has a constant velocity:
[tex]F-F_f-Wsin(\theta)=0\\\\[/tex] (1)
F: force applied by the man = 95N
Ff: friction force
W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N
θ: degree of the inclined plane = 15°
You solve the equation (1) for Ff and you replace the values of all variables in the equation (1):
[tex]F_f=-Wsin(\theta)+F\\\\F_f=-(245N)sin18\°+95N=19.29N[/tex]
b) To fins the velocity of the box at the bottom you use the following formula:
[tex]W_N=\Delta K[/tex] (2)
That is, the net work over the box is equal to the change in the kinetic energy of the box.
The net work is:
[tex]W_N=Mgsin(18\°)d-Ffd[/tex]
d: distance traveled by the box = 2.0m
[tex]W_N=245sin18\°(2.0m)N-19.29(2.0m)N=112.83J[/tex]
You use this value of the net work to find the final velocity of the box, by using the equation (2):
[tex]112.8J=\frac{1}{2}m[v^2-v_o^2]\\\\v_o=0m/s\\\\v=\sqrt{\frac{2(112.8J)}{m}}=\sqrt{\frac{225.67J}{25kg}}=3.00\frac{m}{s}[/tex]
The speed of the box, at the bottom of the incline plane is 3.00 m/s
Assume that the coefficient of static friction between the board and the box is not known at this point. What is the magnitude of the acceleration of the box in terms of the friction force f?
Answer:
Explanation:
From Newton's second Law of Motion,
F = ma
Ff + F = ma
Where F is the applied force, Ff is the frictional force, a is the acceleration and m is the mass of the object or box.
Magnitude of the acceleration:
a = Ff+F/m
This must act in the direction of F or the box would slide or accelerate off the negative side of the board (taking the direction of F to be positive
A 550 kg dragster accelerates from rest to a final speed of 110 m/s in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is its average power output in watts and horsepower if this takes 7.30 s
Answer:
[tex]52.25\times10^4W\\699.1 hp[/tex]
Explanation:
According to the energy conversation:
ΔK=[tex]-f_kd+W[/tex]
ΔK=[tex]K_f-K_i ; K=1/2 mv^2[/tex]
where,
[tex]k_i, k_f[/tex] are initial and final kinetic energy of the system.
[tex]v_i[/tex]= initial velocity of the system
[tex]v_f[/tex]=final velocity of the system
W= total work done on the system
[tex]f_k[/tex]= friction force
d= distance traveled
Given: [tex]v_f[/tex]=110m/s
d=400m
[tex]f_k[/tex]=1200N
[tex]v_i[/tex]=0m/s
t=7.3s
ΔK=[tex]-f_kd+W[/tex]
W= ΔK + [tex]f_kd[/tex]
=[tex]K_f-K_i+f_kd\\[/tex]
[tex]=1/2 mv_f^2-1/2 mv_i^2+f_kd\\=\frac{1}{2} \times 550\times110^2 - \frac{1}{2} \times 550\times0^2+ (1200\times400)\\=3807500[/tex]
[tex]P=\frac{W}{t} =\frac{3807500}{7.3} \\P=52.15 \times10^4w\\P=\frac{52.15 \times10^4}{746} \\P=699.1 hp[/tex]
What is the frequency if 140 waves pass in 2 minutes?
Answer:
1.16 Hz
Explanation:
frequency, basically, is the number of wave on 1 second
so, in math we write like this
f = n/t
n = number of waves
t = time to do that (in sec)
f = 140/120 = 7/6 Hz
f = 1.16 Hz
Block 1, of mass m1 = 2.50 kg , moves along a frictionless air track with speed v1 = 27.0 m/s. It collides with block 2, of mass m2 = 33.0 kg , which was initially at rest. The blocks stick together after the collision.A. Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.B. Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically.C. what is the change deltaK= Kfinal- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.
Answer:
a
The total initial momentum of the two-block system is [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
b
The magnitude of the final velocity of the two-block system [tex]v_f = 1.9014 \ m/s[/tex]
c
the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is
[tex]\Delta KE =- 847.08 \ J[/tex]
Explanation:
From the question we are told that
The mass of first block is [tex]m_1 = 2.50 \ kg[/tex]
The initial velocity of first block is [tex]u_1 = 27.0 \ m/s[/tex]
The mass of second block is [tex]m_2 = 33.0\ kg[/tex]
initial velocity of second block is [tex]u_2 = 0 \ m/s[/tex]
The magnitude of the of the total initial momentum of the two-block system is mathematically repented as
[tex]p_i = (m_1 * u_1 ) + (m_2 * u_2)[/tex]
substituting values
[tex]p_i = (2.50* 27 ) + (33 * 0)[/tex]
[tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
According to the law of linear momentum conservation
[tex]p_i = p_f[/tex]
Where [tex]p_f[/tex] is the total final momentum of the system which is mathematically represented as
[tex]p_f = (m_+m_2) * v_f[/tex]
Where [tex]v_f[/tex] is the final velocity of the system
[tex]p_i = (m_1 +m_2 ) v_f[/tex]
substituting values
[tex]67.5 = (2.50+33 ) v_f[/tex]
[tex]v_f = 1.9014 \ m/s[/tex]
The change in kinetic energy is mathematically represented as
[tex]\Delta KE = KE_f -KE_i[/tex]
Where [tex]KE_f[/tex] is the final kinetic energy of the two-body system which is mathematically represented as
[tex]KE_f = \frac{1}{2} (m_1 +m_2) * v_f^2[/tex]
substituting values
[tex]KE_f = \frac{1}{2} (2.50 +33) * (1.9014)^2[/tex]
[tex]KE_f =64.17 J[/tex]
While [tex]KE_i[/tex] is the initial kinetic energy of the two-body system
[tex]KE_i = \frac{1}{2} * m_1 * u_1^2[/tex]
substituting values
[tex]KE_i = \frac{1}{2} * 2.5 * 27^2[/tex]
[tex]KE_i = 911.25 \ J[/tex]
So
[tex]\Delta KE = 64.17 -911.25[/tex]
[tex]\Delta KE =- 847.08 \ J[/tex]
A tuba may be treated like a tube closed at one end. If a tuba has a fundamental frequency of 40.4 Hz, determine the first three overtones. Use 343 m/s as the speed of sound in air.
If the speed of sound is 337 m/s, determine the length of an open tube (open at both ends) that has a fundamental frequency of 233 Hz and a first overtone frequency of 466 Hz.
Answer:
Explanation:
fundamental frequency at closed pipe = 40.4 Hz
overtones are odd harmonics in closed pipe
first three overtones are
3 x 40.4 , 5 x 40.4 , 7 x 40.4 Hz
= 121.2 Hz , 202 Hz , 282.8 Hz .
speed of sound given is 337 , fundamental frequency is 233 Hz
wavelength = velocity of sound / frequency
= 337 / 233
= 1.446 m
for fundamental note in open pipe
wavelength /2 = length of tube
length of tube = 1.446 / 2
= .723 m
= 72.30 cm .
first overtone will be two times the fundamental ie 466. In open pipe all the harmonics are found , ie both odd and even .
N capacitors are connected in parallel to form a "capacitor circuit". The capacitance of first capacitor is C, second one is C/2 and third one is C/4, forth one is C/8 and so on. Namely, capacitance of a capacitor is one-half of the previous one. What is the equivalent capacitance of this parallel combination when N goes to inifinity?
Answer:
2C
Explanation:
The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.
So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.
Using the formula for the sum of the infinite terms of a geometric series, we have:
Sum = First term / (1 - rate)
Sum = C / (1 - 0.5)
Sum = C / 0.5 = 2C
So the equivalent capacitance of this parallel connection is 2C.
assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charged the battery
Answer:
The amount of water that will power a battery with that rating = 7.35 m³
Explanation:
The power rating for the battery is missing from the question.
Complete Question
Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charged the battery with power rating, 12 V, 50 Ampere-minutes
Solution
Potential energy possessed by water at that height = mgH
m = mass of the water = ρV
ρ = density of water = 1000 kg/m³
V = volume of water = ?
g = acceleration due to gravity = 9.8 m/s²
H = height of water = 50 cm = 0.5 m
Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J
Energy of the battery = qV
q = 50 A.h = 50 × 60 = 3,000 C
V = 12 V
qV = 3,000 × 12 = 36,000 J
Energy = 36,000 J
At a 100% conversion rate, the energy of the water totally powers the battery
(4900V) = (36,000)
4900V = 36,000
V = (36,000/4900)
V = 7.35 m³
Hope this Helps!!!
A sound level of 96 dB is how many times as intense as one of 90 dB?
Answer:
A sound level of 96 dB is 4 times as intense as one of 90 dB
Explanation:
The formula of the intensity level of sound in decibels is given as follows:
Intensity Level = 10 log₁₀(I/I₀)
where,
I = Intensity of Sound
I₀ = Reference Intensity Level = 10⁻¹² W/m²
Therefore, for 96 dB sound level:
96 = 10 log₁₀(I₁/10⁻¹²)
log₁₀(I₁/10⁻¹²) = 96/10
I₁/10⁻¹² = 10^9.6
I₁ = (10⁻¹²)(4 x 10⁹)
I₁ = 0.004 W/m²
For 90 dB sound level:
90 = 10 log₁₀(I₂/10⁻¹²)
log₁₀(I₂/10⁻¹²) = 90/10
I₂/10⁻¹² = 10^9
I₂ = (10⁻¹²)(10⁹)
I₂ = 0.001 W/m²
Therefore,
I₁/I₂ = 0.004/0.001
I₁ = 4 I₂
Hence, the sound level of 96 dB is 4 times as intense as one of 90 dB.
A car travels around an oval racetrack at constant speed. The car is accelerating:________.
A) at all points except B and D.
B) at all points except A, B, C, and D.
C) everywhere, including points A, B, C, and D.
D) nowhere, because it is traveling at constant speed.
2) A moving object on which no forces are acting will continue to move with constant:_________
A) Acceleration
B) speed
C) both of theseD) none of these
Answer:
1A,2D,3B
Explanation:
hope this helps
Q) A particle in simple harmonic motion starts its motion from its mean position. If T be the time period, calculate the ratio of kinetic energy and potential energy of the particle at the instant when t = T/12.
t\12 and the parties are spreading ever
Explanation:
my point is that you can get sick if
you sont wash your ha
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The froghopper, a tiny insect, is a remarkable jumper. Suppose a colony of the little critters is raised on Rhea, a moon of Saturn, where the acceleration due to gravity is only 0.264 m/s2 , whereas gravity on Earth is =9.81 m/s2 . If on Earth a froghopper's maximum jump height is ℎ and its maximum horizontal jump range is R, what would its maximum jump height and range be on Rhea in terms of ℎ and R? Assume the froghopper's takeoff velocity is the same on Rhea and Earth.
Answer:
Maximum height of jump on Rhea is 37.16 times of that on Earth, i.e 37.16h
Maximum range of jump on Rhea is 37.16 of times that on Earth, i.e 37.16R
Explanation:
The acceleration due to gravity on Rhea = 0.264 m/s^2
Acceleration due to gravity on earth here = 9.81 m/s^2
this means that the acceleration due to gravity g on earth is 9.81/0.264 = 37.16 times that on Rhea.
maximum height that can be achieved by the froghopper is given by the equation;
h = [tex]\frac{u^{2}sin^{2} \alpha}{2g}[/tex]
let us put all the numerator of the equation as k, since the velocity of take off is the same for Earth and Rhea. The equation is simplified to
h = [tex]\frac{k}{2g}[/tex]
for earth,
h = [tex]\frac{k}{2*9.81}[/tex] = [tex]\frac{k}{19.62}[/tex]
for Rhea,
h = [tex]\frac{k}{2*0.264}[/tex] = [tex]\frac{k}{0.528}[/tex]
therefore,
h on Rhea is [tex]\frac{k}{0.528}[/tex] ÷ [tex]\frac{k}{19.62}[/tex] = 37.16 times of that on Earth, i.e 37.16h
Equation for range R is given as
R = [tex]\frac{u^{2}sin 2\alpha}{g}[/tex]
following the same approach as before,
R on Rhea will be [tex]\frac{k}{0.264}[/tex] ÷ [tex]\frac{k}{9.81}[/tex] = 37.16 of times that on Earth, i.e 37.16R
Your new toaster has two separate toasting units, each of which consumes 600 watts of power when it is in use. When you operate one unit, a current of 5 amperes flowsthrough the wiring in your home and the wires waste about 1 watt of power handling that current. If you operate both toasting units at once, your toaster consumes 1200 watts and the current flowing through the wiring in your home doubles to 10 amperes. How much power will the wires in your home waste now
Answer:
1.92 Watt lost
Explanation:
Power rating of each toaster = 600 Watts
Current that flows = 5 Amperes
Wasted power = 1 Watt
Voltage of toaster can be gotten from P = [tex]I^{2}[/tex]R
where I = current
and R = Resistance
600 = [tex]5^{2}[/tex] x R
R = 600/25 = 24 Ohms.
According to joules loss due to heating of wire
Power loss P ∝ [tex]I^{2}[/tex]R
imputing values,
1 ∝ [tex]5^{2}[/tex] x 24
1 ∝ 600
to remove the proportionality sign, we introduce a constant k
1 = 600k
k = 1/600 = 0.00167
For the case where the current is doubled to 10 ampere, as the power doubles to 1200 W.
The resistance across the wire becomes
1200 = [tex]10^{2}[/tex]R
R = 1200/100 = 12 Ohms
power loss P = k x [tex]I^{2}[/tex]R
P = 0.0016 x [tex]10^{2}[/tex] x 12
P = 1.92 Watt lost
This question involves the concepts of power, current, and resistance.
The power wasted by the wires in the home for two units will be "4 watt".
POWER WASTAGEThe power wasted by the wires can be given in terms of current and resistance by the following formula:
[tex]P=I^2R\\\\\frac{P}{I^2}=R=Constant\\\\\frac{P_1}{I_1^2}=\frac{P_2}{I_2^2}[/tex]
where,
P₁ = Power wasted for one unit = 1 wattI₁ = current through wires for one unit = 5 AR = Resistance of wires = constantP₂ = Power wasted for two units = ?I₂ = Current through wires for two units = 10 ATherefore,
[tex]\frac{1\ watt}{(5\ A)^2}=\frac{P_2}{(10\ A)^2}\\\\P_2=\frac{(1\ watt)(100\ A^2)}{25\ A^2}[/tex]
P₂ = 4 watt
Learn more about power here:
https://brainly.com/question/7963770
Aparticlewhosemassis2.0kgmovesinthexyplanewithaconstantspeedof3.0m/s along the direction r = i + j . What is its angular momentum (in kg · m2/s) relative to the point (0, 5.0) meters?
Answer:
[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]
Explanation:
In order to calculate the angular momentum of the particle you use the following formula:
[tex]\vec{L}=\vec{r}\ X\ \vec{p}[/tex] (1)
r is the position vector respect to the point (0 , 5.0), that is:
r = 0m i + 5.0m j (2)
p is the linear momentum vector and it is given by:
[tex]\vec{p}=m\vec{v}=(2.0kg)(3.0m/s)(\hat{i+\hat{j}})=6\frac{kgm}{s}(\hat{i}+\hat{j})[/tex] (3)
the direction of p comes from the fat that the particle is moving along the i + j direction.
Then, you use the results of (2) and (3) in the equation (1) and solve for L:
[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]
The angular momentum is -30 kgm^2/s ^k
first law of equilibrium
Answer:
For an object to be an equilibrium it must be experiencing no acceleration.
Explanation:
Hope it helps.
A large box containing your new computer sits on the bed of your pickup truck. You are stopped at a red light. When the light turns green, you stomp on the gas and the truck accelerates. To your horror, the box starts to slide toward the back of the truck. Draw clearly labeled free-body diagrams for the truck and for the box. Indicate pairs of forces, if any, that are third-law action–reaction pairs. (The horizontal truck bed is not frictionless.)
Answer:
The description of that same situation has been listed throughout the explanation segment below.
Explanation:
When another huge box or container containing your new machine or device sits on someone's pick-up truck's bed, the third low portion of the operation response force. This same friction force of the box mostly on the truck bed as well as the friction force including its truck bed on either the box from either the immune response pair.So that the above seems to be the right answer.
World religions: Shinto
Most Shinto rituals are tied to
A) worshiping the kami.
B) the life-cycle of humans and the seasonal cycles of nature.
C) forgiveness of sins.
D) preparing for the afterlife.
Scenario 2: Use the following information to answer questions 3 and 4:
Your client, Jim, is interested in weight control. He weighs 75kg.
3. If Jim walks 3.3 mph (0% grade), how long must he walk to expend 300 kcal total?
A. 52 min
B. 42 min
C. 65 min
D. 99 min
4. If Jim exercises at an intensity of 6 kcal/min, what is the leg ergometer work rate?
A. 47 watts
B. 90 watts
C. 61 watts
D. 71 watts
Answer:
A. 52 min
.A. 47 watts
Explanation:
Given that;
jim weighs 75 kg
and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.
Using the following relation to determine the amount of calories burned per minute while walking; we have:
[tex]\dfrac{MET*weight (kg)*3.5}{200}[/tex]
here;
MET = energy cost of a physical activity for a period of time
Obtaining the data for walking with a speed of 3.3 mph From the standard chart for MET, At 3.3 mph; we have our desired value to be 4.3
However;
the calories burned in a minute = [tex]\dfrac{4.3*75 (kg)*3.5}{200}[/tex]
= 5.644
Therefore, for walking for 52 mins; Jim burns approximately 293.475 kcal which is nearest to 300 kcal.
4.
Given that:
mass m = 75 kg
intensity = 6 kcal/min
The eg ergometer work rate = ??
Applying the formula:
[tex]V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7[/tex]
where ;
[tex]V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}[/tex]
[tex]V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}[/tex]
[tex]V_O_2 ( intensity ) = 0.0012[/tex]
∴[tex]0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min[/tex]
Converting to watts;
Since; 6.118kg-m/min is = 1 watt
Then 291.66 kgm /min will be equal to 47.67 watts
≅ 47 watts
The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian Empire are marked with the Empire's symbol, an ellipse whose major axis is n times its minor axis (a=nb in the figure ).
How fast, relative to an observer, does an Empire ship have to travel for its markings to be confused with those of a Federation ship? Use c for the speed of light in a vacuum.
Express your answer in terms of n and c.
Complete question
The complete question is shown on the first uploaded image
Answer:
The velocity is [tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]
Explanation:
From the question we are told that
a = nb
The length of the minor axis of the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the Empire's symbol, (an ellipse)
Now this length seen by the observer can be mathematically represented as
[tex]h = t \sqrt{1 - \frac{v^2}{c^2} }[/tex]
Here t is the actual length of the major axis of of the Empire's symbol, (an ellipse)
So t = a = nb
and b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the Empire's symbol, (an ellipse)
i.e h = b
So
[tex]b = nb [\sqrt{1 - \frac{v^2}{c^2} } ][/tex]
[tex][\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}[/tex]
[tex]v^2 =c^2 [1- \frac{1}{n^2} ][/tex]
[tex]v^2 =c^2 [\frac{n^2 -1}{n^2} ][/tex]
[tex]v = c* \sqrt{1 - \frac{1}{n^2} }[/tex]
The velocity of an object is given by the expression v (t) = 3.00 m / s + (2.00 m / s ^ 3) t ^ 2. Determine the position of the object as a function of time if it is located at x = 1.00 m at time t = 0.00 s.
Answer: [tex]x=\frac{2}{3}t^3+3t+1[/tex]
Explanation:
Given
velocity of object is given by
[tex]v(t)=3+2t^2[/tex]
and we know change of position w.r.t time is velocity
[tex]\Rightarrow \dfrac{dx}{dt}=v[/tex]
[tex]\Rightarrow \dfrac{dx}{dt}=3+2t^2[/tex]
[tex]\Rightarrow dx=(3+2t^2)dt[/tex]
Integrating both sides we get
[tex]\Rightarrow \int_{1}^{x}dx=\int_{0}^{t}(3+2t^2)dt[/tex]
[tex]\Rightarrow x\mid _{1}^{x}=(3t+\frac{2}{3}t^3)\mid _{0}^{t}[/tex]
[tex]\Rightarrow x-1=3(t-0)+\frac{2}{3}(t^3-0)[/tex]
[tex]\Rightarrow x=\frac{2}{3}t^3+3t+1[/tex]
The first antiparticle, the positron or antielectron, was discovered in 1932. It had been predicted by Paul Dirac in 1928, though the nature of the prediction was not fully understood until the experimental discovery. Today, it is well accepted that all fundamental particles have antiparticles.
Suppose that an electron and a positron collide head-on. Both have kinetic energy of 3.58 MeV and rest energy of 0.511 MeV. They produce two photons, which by conservation of momentum must have equal energy and move in opposite directions. What is the energy Eloton of one of these photons?
Answer:
4.09 MeV
Explanation:
Find the given attachment
Refer to a situation where you exert a force F on a crate of mass M, moving it at a speed v a distance d across a floor in a time interval t. The quantity F d/t is?
a.) kinetic energy of the crate
b.) potential energy of the crate
c.) linear momentum of the crate
d.) work you do on the crate
e.) power you supply to the crate
Answer:
e.) power you supply to the crate
Explanation:
According to given data, we have:
F = Force exerted on the crate
M = Mass of the crate
v = Speed of motion of the crate
d = Distance traveled by the crate across the floor
t = Time interval passed
Now, we try to analyze the given quantity:
=> F d/t
=> (Force)(Displacement)/(Time)
but, (Force)(Displacement) = Work Done
Therefore,
=> Work Done/Time
but, Work Done/Time = Power
Therefore,
=> Power
Hence, the quantity F d/t is:
e.) power you supply to the crate
Representar con una escala de 1cm = 10N dos fuerzas que tengan igual dirección, distinto sentido y sus intensidades son de 40n y 60n, respectivamente.
Alguien que me lo hagaaaaaaa
Answer:
To solve this problem we just need to graph two forces with same direction, pointing to different sides with intensities of 40 N and 60 N.
The image attached shows these forces.
Notice that the vectors are parallel, that's because they have the same direction, but they point to different sides, and their magnitudes have a difference of 20 N.
man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a brick in each hand.The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 k g times m squared. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 k g times m squared, what is the resulting angular speed of the platform in rad/s? Express to 3 sig figs.
Answer:
w₂ = 22.6 rad/s
Explanation:
This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.
Let's write the moment two moments
initial instant. Before releasing bricks
L₀ = I₁ w₁
final moment. After releasing the bricks
[tex]L_{f}[/tex] = I₂W₂
L₀ = L_{f}
I₁ w₁ = I₂ w₂
w₂ = I₁ / I₂ w₁
let's reduce the data to the SI system
w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s
let's calculate
w₂ = 6.0/2.0 7.54
w₂ = 22.6 rad/s
Two students are pushing their stalled car down the street. If the net force exerted on
the car by the students is 1000 N at an angle of 20° below horizontal, the horizontal
component of the force is:
(a) greater than 1000 N.
(b) less than 1000 N.
(c) sum of the pushing force and the weight of the students.
(d) (a) and (b)
(e) (a) and (c)
Answer:
B
Explanation:
Because the force has 2 components (horizontal and vertical), the horizontal component must be smaller than the total force. The Pythagorean theorem only adds positive values (because they're squared) so it makes sense. Using trigonometry, 100*cos(-20) yields a horizontal force of around 939.7N, which is less than 1000N.
