Answer:
(a) 0.204 Weber
(b) 0.22 Volt
Explanation:
N = 100, radius, r = 10 cm = 0.1 m, B = 0.0650 T, angle is 90 degree with the plane of coil, so theta = 0 degree with the normal of coil.
(a) Magnetic flux, Ф = N x B x A
Ф = 100 x 0.0650 x 3.14 x 0.1 0.1
Ф = 0.204 Weber
(b) B1 = 0.0650 T, B2 = 0.1 T, dt = 0.5 s
dB / dt = (B2 - B1) / dt = (0.1 - 0.0650) / 0.5 = 0.07 T / s
induced emf, e = N dФ/dt
e = N x A x dB/dt
e = 100 x 3.14 x 0.1 x 0.1 x 0.07 = 0.22 V
The potential energy of the rock is 30,000 J
Explanation:
The mechanical energy of an object is equal to the sum of its gravitational potential energy (PE) and its kinetic energy (KE):
where
PE is the gravitational potential energy, which is the energy possessed by the object due to its position in the gravitational field
KE is the kinetic energy, which is the energy possessed by the object due to its motion
In this problem, the rock is at rest, so its kinetic energy is zero:
KE = 0
Therefore, the energy of the rock is just equal to its potential energy, which is:
where
m = 10.2 kg is the mass of the rock
is the acceleration of gravity
h = 300 m is the height of the rock above the ground
Substituting and solving, we find
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Answer:
100 J, 225 J
Explanation:
The kinetic energy of an object is given by:
where
m is the mass of the object
v is the velocity of the object
In this problem, the initial kinetic energy of the object is
K = 25 J
Then, the velocity is doubled, which means
v' = 2v
Therefore, the new kinetic energy will be
Therefore, the kinetic energy has quadrupled:
Later, the velocity is tripled, which means
v'' = 3v
Therefore, the new kinetic energy will be
Therefore, the kinetic energy has increased by a factor of 9:
Answer:
2.2nC
Explanation:
Call the amount by which the spring’s unstretched length L,
the amount it stretches while hanging x1
and the amount it stretches while on the table x2.
Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,
we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,
applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,
where ke is the Coulomb constant. Combining these,
we get q = √(mgx2(L+x2)²/x1ke =2.2nC
Answer:
Warming signs before fad diet includes:
1. When a special type of diet is recommended for weight loss.
2. When quick weight loss is promised instead of advice on the need to exercise, eat healthy foods including fruits and vegetables, and adopt a good feeding behavior.
3. Promises rapid weight loss in a short period of time.
After taking the fad diet the following might be experienced:
1. Constipation.
2. Weakness and fatigue.
3. More accumulation of fats in the body in the long run after a short term normal BMI (Body mass index).
4. Lower metabolic rates over time.
Explanation:
weight loss is a practice adopted by people for various reasons ranging from the desire to stay healthy to, ability to pursue a career that requires a good Body Mass Index. For instance, A BMI (Body mass index) which is equal to a person's weight (kg) / (height(m))^2 of 30 and higher is considered obese.
Fad diet which may give a temporary result of weight loss is not considered totally healthy as it doesn't capture the need to eat a healthy balanced diet instead focuses on more routine consumption of a particular kind of diet.
A healthy weight loss plan captures the need to exercise, reduce body fats while eating healthy. Eating fruits, vegetables are also encouraged.
