The question is incomplete. The complete question is :
A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magnetic field. A field of 1.2 T is directed along the positive z-direction, which is upward. (a)If the loop is removed from the field region in a time interval of 2.8 ms, find the average emf that will be induced in the wire loop during the extraction process.
Solution :
Let us consider a [tex]$\text{circular loo}p \text{ of wire}$[/tex] which has a [tex]\text{radius}[/tex] of r = [tex]15[/tex] cm.
It is oriented horizontally along the xy-plane and is located in the region of an [tex]$\text{uniform magnetic field}$[/tex], such that it points in the positive z direction and having a magnitude of B = 1.2 T.
Now if the loop [tex]$\text{is removed from the field region}$[/tex] in a time interval of Δt = 2.8 ms. Initially the magnetic field and the area points is in the same direction, so that the angle between them is Ф = 0°, thus the initial and the final fluxes are :
[tex]$\phi_{B,i}=BA \cos (\phi) = BA $[/tex] and [tex]$\phi_{B,f} = 0$[/tex]
Area A = [tex]$\pi r^2.$[/tex] The induced emf equals to the change in the flux, and is divided by the time that it takes to go from the initial flux, Δt and multiplied by the number of turns N = 1, i.e. ,
[tex]$\epsilon = -\frac{\Delta \phi_{B}}{\Delta t}$[/tex]
[tex]$=-\frac{0-(1.2 T)\pi(0.15^2)}{2.8 \times 10^{-3}}$[/tex]
= 30.27 V
Therefore, the emf generated is 30.27 V.
Two balls are thrown from a cliff. One is thrown directly up, the other directly down. Both balls have the same initial speed, and both hit the ground below the cliff but at different times. Which ball hits the ground at the greater speed: (a) the ball thrown upward, (b) the ball thrown downward, or (c) both the same
Answer:
both the same
Explanation:
When a ball is thrown vertically upwards, it experiences that same acceleration due to gravity as an object thrown directly downwards.
This means that if we ignore the effects of air resistance, and the two balls have the same initial speed, they are expected both expected to hit the ground at the same speed as a result of the principle of conservation of energy.
Answer:
Both balls will hit the ground at the same speed because air resistance is ignored.
Explanation:
a) Find the current in the 1 Ω resistor.
b ) Find the current in the 8 Ω resistor.
c ) Find the current in the 5 Ω resistor.
PLEASE HELP I NEED THIS TODAY
Answer:
a)I=V/R
39.5 amp
Explanation:
because the voltage in serious with 1ohm resistor
kung ako ang gagawa ng isang papel pananaliksik ang layunin kung ito ay
Explanation:
4. Alin sa mga sumusunod na awitin ang may tempong presto?
a. “Chua-ay”
c. “Akong Manok”
b. “Sitsiritsit
d. "Ili-ili Tulog Anay”
5. Pakinggan ang awiting “Sa Ugoy ng Duyan” Ano ang tempo nito?
a. mabagal
c. mabilis na mabilis
b. mabilis at mabagal d. katamtamang bilis
6. Alin sa sumusunod na elemento ng musika ang nakikilala sa pamamagitan ng pakikinig o pag-awit na may nipis at kapal na tunog?
a. descant
b. ostinato
c. tempo
d. texture
7. Alin sa 2-part vocal ang nasa ibabang bahagi ng musical score?
a. alto
b. forte
c. tempo
d. soprano
8. Alin sa 2-part vocal ang nasa itaas na bahagi ng musical score?
a. alto
b. forte
c. tempo
d. soprano
9. Ano ang tawag sa paulit-ulit na rhythmic pattern na ginagamit sa pansaliw ng awitin?
a. descant
b. ostinato
c. melodic ostinato
d. rhythmic ostinato
10. Ano ang tawag sa paulit-ulit na rhythmic pattern at may kasamang melody na ginagamit sa pansaliw ng awitin?
a. descant
b. ostinato
c. melodic ostinato
d. rhythmic ostinato
Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 ss to speed up from rest to its top speed of 1 rotation every 1.30 ss . The astronaut is strapped into a seat 5.90 mm from the axis. What is the astronaut's tangential acceleration during the first 40.0 s?
How many g's of acceleration does the astronaut experience when the device is rotating at top speed? Each 9.80 m/s^2 of acceleration is 1 g.
Answer:
speed = 0.9 mm/s
Explanation:
time, t = 40 s
initial angular speed, wo = 0 rad/s
final frequency, f = 1/1.03 rps = 0.97 rps
final angular speed, w = 2 x 3.14 x 0.97 = 6.1 rad/s
time, t = 40 s
distance, r = 5.9 mm
The angular acceleration is given y the first equation of motion.
[tex]w =wo + \alpha t\\6.1 = 0 +\alpha \times 40\\\alpha = 0.1525 rad/s^{2}[/tex]
The linear velocity is
[tex]v =5.9\times 10^{-3}\times 0.1525 = 9\times 10^{-4} m/s[/tex]
speed, v = 0.9 mm/s
What are the relationships between the temperature scales of Fahrenheit, Kelvin, Celsius, and Rankine
A disk of Radius R with a uniform distibution of mass"m" rotater about an axis perpendicular to its place at the rim with angular speed "w" the moment of Inertia of the disc about an axis through the contre MR² What is the KE of the disk?
Answer:
[tex]\frac{1}{2}mR^2\omega^2[/tex]
Explanation:
The rotational kinetic energy of an object is given by [tex]KE_r=\frac{1}{2}I\omega^2[/tex], where [tex]I[/tex] is the object's moment of inertia/rotational inertia and [tex]\omega[/tex] is the object's angular speed.
What we're given:
The object's moment of inertia: [tex]I=MR^2[/tex] The object's radius, mass, and angular speed: [tex]R, m, \omega[/tex], respectivelySince no numerical value is given for any of these, it is implied the desired answer will be an equation in terms of the variables given.
Substituting [tex]I=MR^2[/tex]:
[tex]KE_r=\boxed{\frac{1}{2}mR^2\omega^2}[/tex]
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the vertical speed be right before it hits the ground?
A. 0 m/s
B. 15 m/s
C. 40 m/s
D. 30 m/s
Answer:
Explanation:
The nice thing about parabolic motion is that the object launched from a certain height will have the same velocity coming down when it reaches that height again, just in the opposite direction. For us, that means if the velocity of the ball right off the ground is 30 m/s, then right before it hits the ground again it will be -30 m/s (the negative just means that the direction is the opposite). Your choice is D.
Tarzan, whose mass is 75 kg, is running from a cheetah. Tarzan is moving at 5 m/s when he grabs onto a hanging vine. How high off the ground does Tarzan swing ?
Answer:
Explanation:
His Kinetic energy = 1/2 m v^2
v = 5 m/s
m = 75 kg
Ke = 1/2 75 * 5^2
Ke = 937.5 Joules
This will be converted to PE when he reaches the maximum height he reaches. In other words KE = PE
PE = m * g * h
m = 75
g = 9.81
h = ?
PE = 937.5
937.5 = 75 * 9.81 * h
937.5 = 735.75 * h
937.5/735.75 = h
h= 1.27 meters
A weightlifter presses a 200 N weight 0.5 m over his head in 2 s. What is the power of the weightlifter
Answer:
50 watts
Explanation:
Applying,
Power (P) = Workdone (W)/Time(t)
But,
Work done (W) = Force (F)×distance(d)
Therefore,
P = Fd/t..................... Equation 1
Where P = power of the weightlifter, F = Force applied, d = distance, t = time.
From the question,
Given: F = 200 N, d = 0.5 m, t = 2 s
Substitute these values into equation 1
P = (200×0.5)/2
P = 100/2
P = 50 watts
A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary
Answer:
T = 98 N
Explanation:
The gravity of the earth is known to be 9.8 m/s²
Data:
m = 10 kgg = 9.8 m/s²T = ?Use formula:
[tex]\boxed{\bold{T=m*g}}[/tex]Replace and solve:
[tex]\boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}[/tex][tex]\boxed{\boxed{\bold{T=98\ N}}}[/tex]The tension in the rope is 98 Newtons.
Greetings.
А bus has started to move from
the rest with an acceleration of
0.25 m/s². find its final velocity
Suppose an astronomer observes a binary star system where the stars are separated by 2.0 AU , and they have an orbital period of 7.0 years . Using Newton's version of Kepler's Third Law, find the combined mass of the stars.
Answer:
4.408 [tex]\mathsf{M_{sun}}[/tex]
Explanation:
According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:
[tex](m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}[/tex]
combined mass (m₁+m₂) = 4.408 [tex]\mathsf{M_{sun}}[/tex]
One of the wavelengths of light emitted by hydrogen atoms under normal laboratory conditions is at ?0 = 656.3nm in the red portion of the electromagnetic spectrum. In the light emitted from a distant galaxy this same spectral line is observed to be Doppler-shifted to ? = 953.3nm , in the infrared portion of the spectrum.
How fast are the emitting atoms moving relative to the earth?
Answer:
1.07 × 10⁸ m/s
Explanation:
Using the relativistic Doppler shift formula which can be expressed as:
[tex]\lambda_o = \lambda_s \sqrt{\dfrac{c+v}{c-v}}[/tex]
here;
[tex]\lambda _o[/tex] = wavelength measured in relative motion with regard to the source at velocity v
[tex]\lambda_s =[/tex] observed wavelength from the source's frame.
Given that:
[tex]\lambda _o[/tex] = 656.3 nm
[tex]\lambda_s =[/tex] 953.3 nm
We will realize that [tex]\lambda _o[/tex] > [tex]\lambda_s[/tex]; thus, v < 0 for this to be true.
From the above equation, let's make (v/c) the subject of the formula: we have:
[tex]\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{c+v}{c-v}}[/tex]
[tex]\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2=\dfrac{c+v}{c-v}[/tex]
[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2-1}{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2+1}[/tex]
[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{656.3}{953.3} \Big)^2-1}{\Big(\dfrac{656.3}{953.3} \Big)^2+1}[/tex]
[tex]\dfrac{v}{c} =0.357[/tex]
v = 0.357 c
To m/s:
1c = 299792458 m/s
∴
0.357c = (299 792 458 × 0.357) m/s
= 107025907.5 m/s
= 1.07 × 10⁸ m/s
A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. If the volume of the balloon is 500.0 m^3 and the surrounding air is at 15.0°C. What must the temperature of the air in the balloon be for it to lift a total load of 290 kg (in addition to the mass of the hot air)? The density of air at 15.0°C and atmospheric pressure is 1.23kg/m^3.
Answer:
272° C
Explanation:
Given :
Volume of the balloon, V = 500 [tex]m^3[/tex]
The temperature of the surrounding air, [tex]T_{air} = 15^\circ C[/tex]
Total load, [tex]m_{T}[/tex] = 290 kg
Density of the air, [tex]$\rho_{air} = 1.23 \ kg/m^3$[/tex]
We known buoyant force,
[tex]$F_B = \rho_{air} V$[/tex]
For a 290 kg lift, [tex]$m_{hot} = \frac{F_B}{g} = 290 \ kg$[/tex]
[tex]$m=\rho V$[/tex]
∴ [tex]$m_{hot}=\rho_{hot} V ; \ \ \ \ \ \frac{F_B}{g}-m_{hot} = 290 \ kg$[/tex]
[tex]$(\rho_{air} - \rho_{hot}) V= 290 \ kg$[/tex]
[tex]$\rho_{hot} = \rho_{air}- \frac{290}{V} \ kg = 1.23 \ kg/m^3 - \frac{290 \ kg}{500 \cm^3}$[/tex]
[tex]$\rho_{hot}= 0.65 \ kg/m^3 =\frac{\rho M}{R T_{hot}}$[/tex]
∴ [tex]$\rho_{hot} T_{hot}= \rho_{air} T_{air}$[/tex]
[tex]$T_{hot}= T_{air}\left[\frac{\rho_{air}}{\rho_{hot}}\right]$[/tex]
[tex]$=288 \ K \times \frac{1.23 \ kg/m^3}{0.65 \ kg/m^3}$[/tex]
= 545 K
[tex]$=272^\circ C$[/tex]
Therefore, temperature of the air in the balloon is 272 degree Celsius.
To lift a load more than the weight of the balloon, the temperature of the air in the balloon has to be higher than the air in the surrounding.
The temperature of the air in the balloon to lift a total load of 290 kg is approximately 272.12°C.Reasons:
Given information are;
Volume of the balloon = 500.0 m³
Temperature of the surrounding air = 15.0°C
Density of air at 15.0°C = 1.23 kg/m³
Required:
The temperature required to lift 290kg.
Solution:
Let, [tex]\rho _{air , b}[/tex], represent the density of the air in the balloon, we have;
[tex]\rho _{air , b}[/tex] × 500.0 + 290 = 1.23 × 500
Therefore;
[tex]\displaystyle \rho _{air , b} = \frac{1.23 \times 500- 290}{500} = 0.65[/tex]
According to the Ideal Gas Law, we have;
ρ₁ × R × T₁ = ρ₂ × R × T₂
Therefore;
[tex]\displaystyle T_2 = \mathbf{\frac{\rho_1 \times T_1}{\rho_2}}[/tex]
Therefore;
[tex]\displaystyle T_2 = \frac{1.23\times288.15}{0.65} \approx 545.27[/tex]
The temperature of the balloon, T₂ ≈ 545.27 - 273.15 = 272.12
The temperature of the air in the balloon, T₂ ≈ 272.12 °C
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Charissa says that bread goes through a chemical change when it is toasted. Which of the following observations would least support her conclusion?
A. a change in the smell of the bread
B. a change in the color of the bread
C. a change in the texture of the bread
D. a change in the shape of the bread
Answer:
A. the change in smell
Explanation:
sry if its wrong it's what I know
The pair of forces described by Newton third law must be
Answer:
The answer is Newton's third law of motion states that every action has an equal and opposite reaction. This means that force always act in pairs
The pair of forces described by Newton third law must be in opposite direction.
What is Newton's third law of motion ?Every action have equal and opposite reaction. for example when we fire bullet from a gun, the gun will recoil back and bullet moves forward. In case of rocket, rocket is fired, thrust is reaction of force applied by the gas on the floor.
The motion of lift from an airfoil in which the air is diverted downward by the airfoil's action and the wing is pushed upward in response.
When a spinning ball moves, the air is deflected to one side, and the ball responds by travelling in the other direction.
A jet engine's motion generates thrust, and hot exhaust gases rush out the back of the engine, producing thrust in the opposite direction.
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A ballistic pendulum is a device for measuring the speed of a projectile. The projectile is launched horizontally and embeds in a stationary block on the end of a string. The block-projectile system swings upward after the collision, reaching a maximum height. Which of the following statements is correct about the collision between the projectile-block system?
a. Kinetic energy of the system is conserved.
b. Linear momentum of the system is conserved.
c. Linear momentum of the system is not conserved.
d. The total mechanical energy of the system is conserved
Use your understanding of heat loss to ESTIMATE the cost of the lost energy through one standard window during an average summer day in Maryland. Use $0.17 per kWh as your energy cost.
Answer:
The cost of energy is $ 0.34.
Explanation:
The energy is the capacity to do work.
The energy is a scalar quantity and its SI unit is Joule.
The commercial unit of energy is kWh.
Cost of 1 kWh energy = $ 0.17
energy loss by standard window is 2 kWh .
So, the cost of lost of energy is
Cost = $ 0.17 x 2 = $ 0.34
A 1.2 kg basketball is thrown upwards. What is the potential energy of the basketball at the top of its path if it reaches a height of 15.6 m?
Answer:
Answer is 183.6 J
Explanation:
Using the Physics reference sheet the formula for Potential energy is
(mass) x (gravity) x (height)
Mass= 1.2
Gravity I used is 9.81 (use 10 to get the answer most schools use)
Height= 15.6
A system has both potential energy (PE) and kinetic energy (KE). According to
the law of conservation of energy, what can happen to the total energy of the
system?
Answer:
A. It must stay the same, but kinetic energy (KE) can be transformed to PE and PE can be transformed to KE within the system.
Explanation:
Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;
a. Potential energy (PE): it is an energy possessed by an object or body due to its position above the earth.
Mathematically, potential energy is given by the formula;
[tex] P.E = mgh[/tex]
Where,
P.E represents potential energy measured in Joules.m represents the mass of an object. g represents acceleration due to gravity measured in meters per seconds square. h represents the height measured in meters.b. Kinetic energy (KE): it is an energy possessed by an object or body due to its motion.
Mathematically, kinetic energy is given by the formula;
[tex] K.E = \frac{1}{2}MV^{2}[/tex]
Where;
K.E represents kinetic energy measured in Joules. M represents mass measured in kilograms. V represents velocity measured in metres per seconds square.Furthermore, the total energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.
Mathematically, it is given by the formula;
Total energy = P.E + K.E
The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.
In this scenario, a system has both potential energy (PE) and kinetic energy (KE).
According to the law of conservation of energy, we can infer or deduce that the total energy of the system must stay the same because it cannot be destroyed, but kinetic energy (KE) can be transformed to potential energy (PE) and potential energy (PE) can be transformed to kinetic energy (KE) within the system.
By how many newtons does the weight of a 85.9-kg person lose when he goes from sea level to an altitude of 6.33 km if we neglect the earth's rotational effects
Answer:
[tex]Weight\ loss=1.6321N[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=85.9kg[/tex]
Altitude [tex]h= 6.33 km[/tex]
Let
Radius of Earth [tex]r=6380km[/tex]
Gravity [tex]g=9.8m/s^2[/tex]
Generally the equation for Gravity at altitude is mathematically given by
[tex]g_s=9.8(\frac{6380}{6380+6.33})^2[/tex]
[tex]g_s=9.781m/s^2[/tex]
Therefore
Weight at sea level
[tex]W_s=9.8*85.9[/tex]
[tex]W_s=841.82N[/tex]
Weight at 6.33 altitude
[tex]W_a=9.781*85.9[/tex]
[tex]W_a=840.2N[/tex]
Therefore
[tex]Weight loss=W_s-W_b[/tex]
[tex]Weight loss=841.82-840.2[/tex]
[tex]Weight loss=1.6321N[/tex]
Jason takes off from rest across level water on his jet-powered skis. The combined mass of Jason and his skis is 75 kg (the mass of the fuel is negligible). The skis have a thrust of 200 N and a coefficient of kinetic friction on water of 0.10. Unfortunately, the skis run out of fuel after only 48 s. What is Jason's top speed
Answer:
81.1 m/s
Explanation:
The net force of Jason is T - f = ma where T = thrust = 200 N f = frictional force = μN = μmg where μ = coefficient of kinetic friction of water = 0.10, m = mass of Jason plus skis = 75 kg, g = acceleration due to gravity = 9.8 m/s² and a = Jason's acceleration
So, T - f = ma
T - μmg = ma
a = T/m - μg
susbstμituting the values of the varμiables into the equation, we have
a = 200 N/75 kg - 0.1 × 9.8 m/s²
a = 200 N/75 kg - 0.1 × 9.8 m/s²
a = 2.67 m/s² - 0.98 m/s²
a = 1.69 m/s²
Using v = u + at, we find Jason's velocity v where u = initial velocity = 0 m/s (since he starts from rest), a = 1.69 m/s² and t = time = 48 s
So, v = u + at
v = 0 m/s + 1.69 m/s² × 48 s
v = 0 m/s + 81.12 m/s
v = 81.12 m/s
v ≅ 81.1 m/s
So, Jason's top speed is 81.1 m/s
If the child has a mass of 13.9 kg, calculate the magnitude of the force in newtons the mother exerts on the child under the following conditions. (b) The elevator accelerates upward at 0.898 m/s2. 148.702 N
The elevator accelerates upward at an acceleration, then the magnitude of the force is 148.84 N.
What is Force?The force is the action of push or pull which makes an object to move or stop.
Given the mass of child m =13.9 kg, acceleration a =0.898 m/s², then the force will be given by
F = m(g-a)
F = 13.9 x (9.81 - (-0.898))
F = 148.84 N
Thus, the magnitude of the force is 148.84 N.
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You’re working with a patient who suddenly falls. You should?
If you move 10 times farther away from a source of light, then how will the
apparent brightness of that source change?
it will become 10 times less bright
it will become 2 times less bright
its brightness will not change
O it will become 100 times less bright
is anyone online??just asking
Answer:
me...:(
Explanation:
Answer:
hello I'm online here thanks for the points (◔‿◔)
A FBD of a rocket launching into space should include:
Answer:
Explanation:
the force of the rocket engine pushing it up, the force of gravity pulling it down, maybe some force of air resistance as the rocket goes fast, hmmm Free Body Diagrams (FBD) should have any and all forces on the model, unless they are negligible . or so slight they really make little difference in the total outcome.
An astronaut hits a golf ball of mass m on the Moon, where there is no atmosphere and the acceleration due to gravity is g 6 , where g is the acceleration due to gravity on Earth. Assume that the golf club is in contact with the ball for a time t. Just after losing contact with the club, the ball has an initial speed v directed at an angle T above the horizontal. 6. What is the magnitude of the average force exerted by the club on the ball during contact
Answer:
The magnitude of the average force exerted by the club on the ball during contact = mv/t
Explanation:
Impulse exerted on the ball = Momentum of the ball = mass * velocity = m*v
As we know,
m*v = Integration of F.dt with limits 0 to T
Ft = mv
F = mv/t
The magnitude of the average force exerted by the club on the ball during contact = mv/t
A body of mass 5 kg is moved by a horizontal force of 0.5 N on a smooth frictionless table for 20 seconds. Calculate the change in kinetic energy.
A. 5 J
B. 20 J
C. 10 J
D. 30 J
Answer: 10 J
Explanation:
Trust ;)
At what speed was object A moving ?
Answer:
C
Explanation:
The answer is C because if you look at the 1 hour mark it shows 10km
Answer:It will be 10km/hour
Explanation: