At the end of the previous experiment, aclumsy scientist drops the coil, while still in the magnetic field, and still oriented with its plane perpendicular to the magnetic field, denting it and changing its shape to a semi-circle. The new shape has the same perimeter, but a different area, and it takes 0.036s to deform. What isthe average induced EMF during this mishap
Answer:
hello your question has some missing parts below are the missing parts
A Circular, 10-turn coil has a radius of 10.7 cm and is oriented with its plane perpendicular to a 0.2-T magnetic field.
answer : 1 volt
Explanation:
Determine the Average induced EMF during this mishap
A' = A/2 ( for a semi circle )
where A = [tex]\frac{\pi r^2}{2}[/tex]
To determine the Average induced EMF apply the relation below
| E | = η * [tex]\frac{\beta A}{T}[/tex] ----- ( 1 )
Replace A in equation 1 with A = [tex]\frac{\pi r^2}{2}[/tex]
hence equation becomes : | E | = η * βπr^2 / 2T'
where : T' = 0.0365 , β = 0.2 , η = 10 , r = 0.107
∴| E | = 0.999 ≈ 1volts
Match the descriptions with the graphs !
Answer:
Graph 1 matches with B, 2 with A, and 3 with C.
Explanation:
Graph 2 shows a car whose distance part of the graph is not going up or down, while the time going up. That means that the car is stopped. Graph 1 shows a straight line, meaning that the car is traveling at a constant speed. Graph 3 is a curved line, meaning the speed of the car is changing somehow, and since the line is becoming more horizontal, the car is getting slower.
BRAINLEST FOR CORRECT ANSWER PLEASE
Which has more momentum: a 3 kg sledgehammer swung at 1.5 m/s OR a 4 kg sledgehammer swung at 0.9 m/s? SHOW YOUR WORK
Answer:
Sledgehammer A has more momentum
Explanation:
Given:
Mass of Sledgehammer A = 3 Kg
Swing speed = 1.5 m/s
Mass of Sledgehammer B = 4 Kg
Swing speed = 0.9 m/s
Find:
More momentum
Computation:
Momentum = mv
Momentum sledgehammer A = 3 x 1.5
Momentum sledgehammer A = 4.5 kg⋅m/s
Momentum sledgehammer B = 4 x 0.9
Momentum sledgehammer B = 3.6 kg⋅m/s
Sledgehammer A has more momentum
A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a small pebble is caught in the tread of one of them. (A) What is the angular acceleration of the pebble during those two seconds
Answer:
[tex]a=2.5\ m/s^2[/tex]
Explanation:
Given that,
Initial speed, u = 5 m/s
Final speed, v = 10 m/s
Time, t = 2 s
The radius of the tire of the bike, r = 35 cm
We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.
[tex]a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2[/tex]
So, the required angular acceleration of the pebble is equal to [tex]2.5\ m/s^2[/tex].
We observe that a moving charged particle experiences no magnetic force. From this we can definitely conclude that: (A) No magnetic field exists in that region of space. (B) The particle must be moving parallel to the magnetic field. (C) Either no magnetic field exists or the particle is moving parallel to B. (D) Either no magnetic field exists or the particle is moving vertical (perpendicular) to B.
Answer:
(C) Either no magnetic field exists or the particle is moving parallel to B.
Explanation:
The moving charged particle tends to show no magnetic forces. Due to its moving in the same direction and will be parallel to the field. As when the magnetic fed and charged particle velocity is parallel then there will be no force. As the magnetic field are made by the generation of the current with the field line concentric to the current-carrying wire.How much work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s?
50 J,
12.5 J,
25 J,
49 J
Answer:
The correct option is (b).
Explanation:
We need to find the work done to increase the speed of a 1 kg toy car by 5 m/s.
We know that, the work done is equal to the kinetic energy of an object i.e.
[tex]W=\Delta K\\\\W=\dfrac{1}{2}mv^2\\\\W=\dfrac{1}{2}\times 1\times 5^2\\W=12.5\ J[/tex]
So, 12.5 J of work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s.
Which of the following would produce the most power?
b
ОООО
A mass of 10 kilograms lifted 10 meters in 10 seconds
A mass of 5 kilograms lifted 10 meters in 5 seconds
A mass of 10 kilograms lifted 10 meters in 5 seconds
A mass of 5 kilograms lifted 5 meters in 10 seconds
d
Answer:
A mass of 10 kilograms lifted 10 meters in 5 seconds.
Explanation:
Power can be defined as the energy required to do work per unit time.
Mathematically, it is given by the formula;
[tex] Power = \frac {Energy}{time} [/tex]
But Energy = mgh
Substituting into the equation, we have
[tex] Power = \frac {mgh}{time} [/tex]
Given the following data;
Mass = 10kg
Height = 10m
Time = 5 seconds
We know that acceleration due to gravity is equal to 9.8 m/s²
[tex] Power = \frac {10*9.8*10}{5} = 490 Watts [/tex]
Hence, a mass of 10 kilograms lifted 10 meters in 5 seconds would produce the most power.
A U-tube with a cross-sectional area of 1.00 cm2 is open to the atmosphere at both ends. Water is poured into the tube until the water rises part-way along the straight sides, and then 5.00 cm3 of oil is poured into one end. As a result, the top surface of the oil ends up 0.550 cm higher than the water surface on the other side of the U. What is the density of the oil
Answer:
0.89 g/cm^3 = 890 kg/m^3
Explanation:
Cross sectional area of U-tube ( A ) = 1.00 cm^2
volume of oil ( V ) = 5.00 cm^3
change between top surface = 0.550 cm
height of oil = 5 cm ( volume / area )
height of water = 5 - 0.550 = 4.45 cm
pressure at the oil-water junction = Pressure on the second side of the U-tube at same level
Po * g * Hoil = Pw * g * Hwater
Po * 5 = 1 * 4.45
∴ Density of oil ( Po ) = 4.45 / 5 g/cm^3 = 0.89 g/cm^3
A 744 N force is applied to an object to reach an acceleration of 24 m/s2. What is the objects mass?
31kg
Explanation:
F = ma
m = F/a
m = 744N/24m/s^2
m = 31kg
(*Newton's Second Law*)
On a spaceship designed to support a multiyear voyage to the outer planets of the solar system, plants will be grown to provide oxygen and to recycle carbon dioxide. Because the spaceship will be so far from the sun, an artificial light source will be needed to support photosynthesis. Which wavelengths of light should be used to maximize plant growth?
Answer:
A mixture of blue & red light.
Explanation:
During photosynthesis, the oxygen delivered emanates from water particles and if a weighty isotope of oxygen atom was noticed in delivered sub-atomic oxygen, the water atoms were marked with the hefty isotope.
In order to maximize the growth rate of the plant, the required wavelength of light to be used is a mixture of blue & red light. This is on the grounds that as the absorption optima of plant's photoreceptors are at wavelength frequency of red and blue light, subsequently the combination of red and blue light would be ideal for plant growth and development.
The productivity of red (650–665 nm) LEDs on plant development is straightforward on the grounds that these wavelength frequencies entirely fit with the retention pinnacle of chlorophylls and phytochrome, while the enhanced blue light presented the possibility that development under regular light could be mirrored utilizing blue and red LEDs with negligible use of energy.