Answer:
Explanation:
Maximum emf produced = nωAB. where n is no of turns , A is area , B is magnetic field and ω is angular velocity .
n = 100 , A = .035 m²
B = .05 T
ω = 2π f , f is no of revolution per second by coil
= 2 x 3.14 x 6 x 3
= 113.04 rad /s
Maximum emf produced = 100 x 113.04 x .035 x .05
= 19.78 volt .
Q3, A ball of mass 5.0 kg moving with a Velocity of 10.0 ms collides
with a 15.0 kg ball moves with a Velocity of 4 ms! If both balls
Stick together after Collision, Calculate their Common Velocity after Impact if they initially moves in The Same direction, and Opposite direction.
Answer:
Their common velocity after the collision will be 5.5m/s
Explanation:
look at the attachment above ☝️
Which is an example of using a physical model to describe a leg bone?
Answer: C
Explanation:
3. Categorize each statement as true or false. True False as the fringe order increase for a diffraction grating the fringe brightness decreases
Answer:
True
Explanation:
According to the given situation, the statement is correct as it is secondary maxima of declining strength in the diffraction pattern.
Therefore, the fringe brightness decreases, the fringe order increase for a diffraction grating. So, the right answer is true.
So from the above explanation, the given statement is true
what is a light year
Answer:
A light-year is the distance light travels in one year.
Answer:
Explanation:
a unit of astronomical distance equivalent to the distance that light travels in one year, which is 9.4607 × 1012 km (nearly 6 million million miles).
A 125 kg wrecking ball on a 15.4 m
long cable is pulled back to an
angle of 33.5° and released. How
much KE does it have at the
bottom of its swing?
(Unit = J)
here is the answer hope it will help
Rock X is released from rest at the top of a cliff is on earth. A short time later, Rock y is released from rest from the same point as rock X. Both rocks fall for several seconds before landing on the ground directly below the cliff. Frictional forces are considered to be negligible.
Which of the following graph correctly shows the vertical velocity of rock X as a function of time? Take the positive direction to the upward.
The graph of the velocity and the time can be shown by option D.
What is the correct graph?We know that the movement of an object as it is falling under gravity would have a constant acceleration. The constant acceleration means that the velocity of the object is also held a constant.
We now have to look at the graphs as we have them here. The graph as it has been shown has the the velocity on the vertical axis and it has the time on the horizontal axis. The gradient of the slope is what we would refers to as the acceleration of the body.
We also need to recall that the acceleration has to be a constant since tje tow object would have to reach the ground at the same time if they are released from the same height at the same time.
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PLEASE HELP! I'LL GIVE BRAINLEST. PLEASE A RIGHT ANSWER
Answer:
The answer is a - is the amount of matter in this object
Explanation:
What type of force is jumping a trampoline?
Answer:
Tension
Explanation:
A 1400-kg sports car (including the driver) crosses the rounded top of a hill (radius = 87.0 m ) at 18.0 m/s .
a) Determine the normal force exerted by the road on the car.
b)Determine the normal force exerted by the car on the 65.0- kg driver.
c)Determine the car speed at which the normal force on the driver equals zero.
(a) The normal force exerted by the road on the car is 8,506.2 N.
(b) The normal force exerted by the car on the driver is 394.9 N.
(c) The speed of the car at which the normal force on driver is zero is 29.2 m/s.
Normal force exerted by the road on the carThe normal force exerted by the road on the car is calculated as follows;
Normal force = weight of the car - centripetal force of car
Weight of the car = (1400 x 9.8) = 13,720 N
Centripetal force of the car = (1400 x 18²)/87 = 5,213.8 N
Normal force = 13,720 N - 5,213.8 N
Normal force = 8,506.2 N
Normal force exerted on the driverNormal force = weight of driver - centripetal force of driver
Weight of driver = (65 x 9.8) = 637 N
Centripetal force of driver = (65² x 18²)/(87) = 242.07 N
Normal force = 637 N - 242.07 N = 394.9 N
Speed at which normal force on driver is zeroN = mg - mv²/r
0 = mg - mv²/r
mv²/r = mg
v²/r = g
v² = rg
v = √rg
v = √(87 x 9.8)
v = 29.2 m/s
Thus, the normal force exerted by the road on the car is 8,506.2 N.
The normal force exerted by the car on the driver is 394.9 N.
The speed of the car at which the normal force on driver is zero is 29.2 m/s.
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. In a classic clip on America’s Funniest Home Videos, a sleeping cat rolls gently off the top of a warm TV set. Ignoring
air resistance, calculate (a) the position and (b) the velocity of the cat after 0.100 s, 0.200 s, and 0.300 s.
After 0.100 s, the cat would be 0.049 m below its initial position, with a velocity of 0.98 m/s. After 0.200 s, the cat would be 0.196 m below its initial position, with a velocity of 1.96 m/s. Finally, after 0.300 s, the cat would be 0.441 m below its initial position, with a velocity of approximately 2.94 m/s.
To calculate the position and velocity of the sleeping cat after certain time intervals, we need to make some assumptions and use basic principles of physics.
Assuming the cat falls vertically downward due to gravity and ignoring air resistance, we can apply kinematic equations to solve the problem.
(a) Position:
Using the equation for the position of an object in free fall, h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is time, we can determine the position of the cat.
After 0.100 s: h = (1/2)(9.8 m/s^2)(0.100 s)^2 = 0.049 m (approximately)
After 0.200 s: h = (1/2)(9.8 m/s^2)(0.200 s)^2 = 0.196 m (approximately)
After 0.300 s: h = (1/2)(9.8 m/s^2)(0.300 s)^2 = 0.441 m (approximately)
(b) Velocity:
The velocity of the cat can be calculated using the equation v = gt, where v is the velocity and t is time.
After 0.100 s: v = (9.8 m/s^2)(0.100 s) = 0.98 m/s (approximately)
After 0.200 s: v = (9.8 m/s^2)(0.200 s) = 1.96 m/s (approximately)
After 0.300 s: v = (9.8 m/s^2)(0.300 s) = 2.94 m/s (approximately)
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The moon weighs 7x1022kg and we are about 380,000,000m away from the moon. If you weigh 50kg, how much gravitational force does the moon have on you? (G=6.7x10-¹¹)
The gravitational force that the moon has on a person with a mass of 50 kg is approximately 1.15 N.
The gravitational force between two objects depends on their masses and the distance between them. This force is given by the formula:
F = (G × m₁ × m₂) / r² where F is the gravitational force, m₁ and m₂ are the masses of the two objects, r is the distance between them, and G is the gravitational constant, which has a value of 6.7 × 10⁻¹¹ N m²/kg².
Using this formula, we can find the gravitational force that the moon has on a person with a mass of 50 kg.
The mass of the moon is 7 × 10²² kg, and the distance between the moon and the person is 380,000,000 m.
Therefore, we have:
m₁ = 50 kg
m₂ = 7 × 10²² kg
r = 380,000,000 m
G = 6.7 × 10⁻¹¹ N m²/kg²
Substituting these values into the formula, we get:
F = (G × m₁ × m₂) / r²
F = (6.7 × 10⁻¹¹ × 50 kg × 7 × 10²² kg) / (380,000,000 m)²
F = 1.15 N
Therefore, the gravitational force that the moon has on a person with a mass of 50 kg is approximately 1.15 N.
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What is the gravitational potential energy of a 4 kg bowling ball 1 m above the ground?
Answer:
40 J
Explanation:
Gpe= mgh
=4*10*1
=40
A car is moving at 5 m/s. 2 seconds later, it is traveling at 17 m/s. How far did the car go?
Equation 1: Vf = Vo+at
Equation 2: Vf2=V0+2ad
Equation 3: Xf=X0+vot+12at2
Equation 4: Xf=X0+(vf+vo2)t
Answer:
gh
Explanation:
a collision at 30 mph will take any loose object in your car and give it the same force as if it were:
A collision at 30 mph will take any loose object in your car and give it the same force as if it were: shot from a canon.
What is the force of impact at 30 mph?Your body weight times the speed of the moving object equals the crash force. If you are holding your child while driving, you run the risk of crushing them if there is an accident. A 100-pound adult becomes a 3,000-pound force on the youngster in an accident at 30 miles per hour.About 40% of those struck by a car traveling at 30 mph will die as a result of their injuries, according to the U.S. Department of Transportation (DOT). To give you an idea, consider that 5 percent of people would not survive being hit by a car doing 20 mph.
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Cheetahs can accelerate to a speed of 20.0 m/s in 2.50 s and can continue to accelerate to reach a top speed of 30.0 m/s. Assume the acceleration is constant until the top speed is reached and is zero thereafter. Let the +x-direction point in the direction the cheetah runs. Express the cheetah’s top speed (vtop) in miles per hour (mi/h).
Starting from a crouched position, how much time (t-accel) does it take a cheetah to reach its top speed, and what distance (d) does it travel in that time?
Answer:
The cheetah’s initial velocity is 0 m/s. Let’s use the following equation to determine the acceleration.
Explanation:
Acceleration = (v/t) = (20/2.5) = 8m/sec^2.
1) Time to 29m/sec. = (v/a) = (29/8) = 3.625 secs. (3.6)
Distance covered to top speed = (v^2/2a) = 52.5625 metres (52.6).
2) Distance to run at top speed = (120 - 52.5625) = 67.4375 metres.
Time = (d/v) = (67.4375/29) = 2.325 secs.
Total time to warthog = (2.325 + 3.625) = 5.95 secs.
Cheetah's top speed will be "67.12 mi/h". A complete solution is provided below.
Given that:
Speed,
20.0 m/sTime,
2.50 sAcceleration to top speed,
30.0 m/sNow,
→ Acceleration (a) of cheetah will be:
= \(\frac{20}{2.5}\)
= \(8 \ m/s^2\)
→ Top speed (\(V_{top}\)) of cheetah
= \(\frac{30}{1609}\times 3600\)
= \(67.12 \ mi/h\)
→ The time required (t) to reach the top speed will be:
= \(\frac{30}{8}\)
= \(3.75 \ sec\)
and,
→ The distance travelled (d) by cheetah will be:
= \(\frac{1}{2}\times 8\times (3.75)^2\)
= \(\frac{1}{2}\times 8\times 14.0625\)
= \(56.25\)
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Two long, charged, thin-walled, concentric cylindrical shells have radii of 3.0 cm and 6.0 cm . The charge per unit length is 5 . 0 × 10 - 6 on the inner shell and - 7 . 0 × 10 - 6 on the outer shell. What are the (a) magnitude E and (b) direction (radially inward or outward) of the electric field at radial distance r=4.0 cm ? What are (c) E and (d) the direction at r=8.0 cm ?
The electric field in the direction is 4.5×10^5 N/C.
We denote the inner and outer cylinders with subscripts i and o, respectively.
(a) Since r_i<r=4.0cm<r_o,
E(r)= λ_i / 2πϵ_0r = 5.0×10^−6C/m / 2π(8.85×10^−12C^2/N.m)(4.0×10^−2m) = 2.3×10^6 N/C
(b) The electric field E(r) points radially outward.
(c) Since r>r_o,
E(r=8.0cm)= λ_i+λ _o / 2πϵ_0r = 2π(8.85×10^−12 C/N.m 2)(8.0×10^2m)
5.0×10^−6C/m−7.0×10^−6C/m =−4.5×10^5 N/C
or ∣E(r=8.0cm)∣=4.5×10^5 N/C.
(d) The minus sign indicates that E(r) points radially inward.
An electric field is a physical phenomenon that arises from the presence of charged particles. When a charged particle, such as an electron or a proton, is placed in a region of space, it creates an electric field that permeates that space. The electric field is a vector field, meaning it has both magnitude and direction, and is represented by electric field lines. The strength of the electric field is proportional to the magnitude of the charge that created it and inversely proportional to the distance from the charge.
Electric fields play a crucial role in many aspects of physics and engineering, including electronics, electrostatics, and electromagnetism. They are responsible for many everyday phenomena, such as the attraction and repulsion of magnets, the operation of electric motors and generators, and the functioning of electronic devices like smartphones and computers.
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A wheelbarrow is an example of which simple machine?
O pulley
O lever
O screw
wheel and axel
What is the number of waves that pass a fixed ping in a given amount of time?
Long-term health effects are something to consider at nearly every age. What are some behaviors or choices a person could make during midlife that could be obstacles to later-life success? These may include habits, negative perceptions, unmanaged stress, or other health-related behaviors.
Certain behaviors and choices during this time can have significant impacts on later-life success are poor diet, lack of exercise, smoking and chronic stress.
Poor diet: Consuming a diet that is high in saturated fats, processed foods, and sugar can lead to weight gain, high blood pressure, and other chronic conditions that can have lasting effects on health.
Lack of exercise: Failing to exercise regularly can increase the risk of chronic diseases such as heart disease, diabetes, and certain cancers, which can impact the quality of life in later years.
Smoking: Smoking is a major risk factor for many chronic diseases, including heart disease, stroke, and lung cancer. Quitting smoking during midlife can significantly reduce the risk of these diseases.
Alcohol abuse: Drinking excessively can increase the risk of liver disease, heart disease, and other chronic health conditions that can negatively impact later-life success.
Chronic stress: Unmanaged stress can contribute to the development of mental health problems, such as anxiety and depression, as well as chronic physical health conditions, such as hypertension and heart disease.
By making healthy choices, managing stress, and maintaining positive perceptions and attitudes, individuals can increase their chances of achieving success in later life.
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Why could it be argued that the respiratory system is most critical to sustaining life?
Explanation:
Energy is the most important ingredient for life. Organisms use energy in diverse ways. Scientifically, energy is defined as the ability to do work. Without this ability, organisms would not exist.
So, the most important process is one that can furnish the body with energy.
The respiratory system happens to be the one that furnishes the body with energy. During respiration, the energy needs of the body is met by series of processes. Oxygen is taken in and use to liberate calories from chemical substances packed with energy. So, without respiration, the bodily energy demands will not be met.A force of 1000N is used to kick a football of mass 0.8kg find the velocity with which the ball moves if it takes 0.8 sec to be kicked.
The velocity of the ball is 100m/s
The first step is to write out the parameters;
The force used to kick the ball is 1000N
The mass of the ball is 0.8 kg
Time is 0.8 seconds
Therefore the velocity can be calculated as follows
F= Mv-mu/t
1000= 0.8(v) - 0.8(0)/0.8
1000= 0.8v- 0.8/0.8
Cross multiply both sides
1000(0.8) = 0.8v
800= 0.8v
divide both sides by the coefficient of v which is 8
800/0.8= 0.8v/0.8
v= 1000m/s
Hence the velocity is 1000m/s
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State Lami's theorem
Answer:
Lami's Theorem states:
“When three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces”.
Explanation:
The design of interior spaces is relatively unimportant to good
architecture?
A bubble of air has a diameter of 1 mm when it is 0.5 m under the surface of water ( coefficient of surface tension 0.073 N/m). Find the gauge pressure inside the bubble.
The gauge pressure inside the bubble. is -101310.4 Pa.
The negative sign shows that the pressure inside the bubble is lower than the atmospheric pressure. Hence, the bubble will rise to the surface of the water.
How do we calculate?We apply Laplace's law to find the absolute pressure inside the bubble:
ΔP = 2γ/r
where ΔP is the pressure difference across the curved surface of the bubble, γ is the coefficient of surface tension of water, and r is the radius of curvature of the bubble.
r = 0.5 mm = 0.0005 m
Substituting the given values, we have:
ΔP = 2 × 0.073 N/m ÷ 0.0005 m
ΔP = 14.6 × 10^(-3) Pa
The atmospheric pressure at sea level is approximately 101325 Pa. Therefore, the gauge pressure inside the bubble is:
P_gauge = ΔP - P_atm
P_gauge = 14.6 × 10^(-3) Pa - 101325 Pa
P_gauge = -101310.4 Pa
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4. Answer the following questions in terms of a wave's frequency, wavelength, amplitude, and
energy.
a. Describe the characteristics of a high pitch wave.
b. Describe the characteristics of a high-volume wave.
a. A high-pitch wave is characterized by a high frequency and a short wavelength. The frequency determines the pitch of the sound, with higher frequencies corresponding to higher pitches.
The wavelength is the distance between two consecutive peaks or troughs of the wave and is inversely proportional to the frequency. Therefore, a high-pitch wave has a shorter wavelength.
The amplitude of the wave, which is the height of the peak or the depth of the trough, is not directly related to the pitch of the sound, but it does determine the volume or intensity of the sound.
b. A high-volume wave is characterized by a high amplitude and a relatively long wavelength. The amplitude determines the volume or intensity of the sound, with higher amplitudes corresponding to louder sounds.
The wavelength of the wave does not directly affect the volume of the sound, but it can affect how the sound is perceived in different environments.
In general, longer wavelengths are more effective at traveling through obstacles such as walls and are better at penetrating long distances, whereas shorter wavelengths are more easily scattered and attenuated in the atmosphere.
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What is a quantity that has magnitude and direction?
O speed
O vector
O motion
O reference point
Answer:
vector
Explanation:
vector quantity have both magnitude and direction whereas all the other option are scalar quantity
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A 4.0-kg mass is moving to the right at 3.0 m/s. An 8.0 kg mass is moving to the left at 2.0 m/s. If after collision the two
masses join together, what is their velocity after collision?
O-0.33 m/s
O-0.20 m/s
O +1.4 m/s
O +2.3 m/s
Answer:
- 0.33 m/s
Explanation:
An illustration is shown above,
In this case, since the two objects move in opposite directions before collision, then move together, the formula to be used is,
m1u1 - m2u2 = (m1 + m2)v
Where,
m1 = mass of the first object
u1 = initial velocity of the first object
v1 = final velocity of the first object
m2 = mass of the second object
u2 = initial velocity of the second object
v2 = final velocity of the second object
Therefore,
(4.0 • 3.0) - (8.0 • 2.0) = (4.0 + 8.0)v
12 - 16 = 12v
-4 = 12v
Divide both sides by 12,
-4 / 12 = 12v / 12
-1 / 3 = v
v = -0.33 m/s
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A penny is dropped from the top of a tower. It hits the ground below after 2.5 s. How tall was the tower?
A. 24,5m
B. 3.92m
C. 30.6m
D. 60m
Answer:
C. 30.6m
Explanation:
To find the height of the tower, we are to use Newtons law of motion to solve this problem. Since the penny is falling from the top of the tower, it is acted by the acceleration due to gravity. The formula to be used is:
\(H=ut+\frac{1}{2}gt^2\)
Where H is the height of the tower, t is the time taken to hit the ground, u is the initial velocity and g is the acceleration due to gravity.
Given that, t = 2.5 s, g =9.8 m/s², u = 0 m/s (at the top of tower)
\(H=ut+\frac{1}{2}gt^2\\\\H=0(2.5)+ \frac{1}{2}(9.8)(2.5)^2\\\\H=30.6\ m\)
1. A pelican flying along a horizontal path drops a fish from a height of 5.4 m while traveling 5.0 m/s. How far does the fish travel horizontally before it hits the water below?
The horizontal distance the fish travels before hitting the water is independent of the height the fish is dropped from, and the speed at which the pelican is traveling. The horizontal distance is determined only by the time the fish is in the air, which can be found using the equation:
Distance = Speed * Time
To find the time, we use the equation:
Time = sqrt(2 * Distance / Acceleration due to gravity)
Where distance is the height the fish is dropped from (5.4m) and the acceleration due to gravity is 9.8 m/s^2
Time = sqrt(2 * 5.4 / 9.8) = sqrt(2 * 5.4 / 9.8) = sqrt(1.08) = 1.03s
So the fish is in the air for 1.03s and the horizontal distance it travels is given by:
Distance = Speed * Time = 5.0m/s * 1.03s = 5.15m
So the fish will travel horizontally 5.15m before hitting the water.
Many biological systems are well-described by the laws of statistical physics. A simple yet often powerful approach is to think of a system as having only two states. For example, an ion channel may be open or closed. In this problem, consider a simple model of membrane channels for ions: The system is described by a Boltzmann distribution with only two states, with energies ε1 (open) and ε2 (closed). Assume the "open" state is the state of higher energy, so that ε1 > ε2.
If the probability of finding an ion channel open is popen and the probability of finding the ion channel closed is pclosed, which of the expressions below best represents the relative probability of open to closed, R = popen/pclosed? Use the notation z1 = e-ε1/kBT and z2 = e-ε2/kBT
a. z1-z2
b. z2-z1
c. z1/z2
d. z2/z1
e. Something else
Answer:
z1/z2
Explanation:
we have no quantum effects therefore we can make use of Maxwell Boltzmann distribution in the description of this system.
using the boltzman distribution the probability of finding a particle in energy state
\(P_{ei} = \frac{gie^{-ei/kol} }{z}\)
we have
gi to be degeneration of the ith state
ei to be energy of ith state
\(z=e^{-ei/kbt}\) summation
\(P_{ope} = \frac{e^{-ei/kBt} }{z} = \frac{Z_{1} }{Z}\)
We have R to be equal to
\(\frac{P_{ope} }{P_{Close} } = \frac{Z1}{Z2}\)