Correct answer choice is:
D. A continuous transmission of energy from one location to the next.
Explanation:
Waves include the carrier of energy without the carrier of matter. In outcome, a wave can be characterized as a change that progresses into a medium, carrying energy from one spot (its source) to different spot without carrying matter.
Answer:
B)
The magnitude of induced emf in the conducting loop is 0.99 mV.
Explanation:
Rate of increase in magnetic field per unit time = 0.090 T/s
Area of the conducting loop = 110 cm^2 = 0.0110 m^2
Electromagnetic induction is the production of an emf or voltage in a coil of wire due to a changing magnetic field through the coil.
Induced e.m.f is given as:
EMF = (-N*change in magnetic field/time)*Area
EMF = rate of change of magnetic field per unit time * Area
EMF = 0.090 * 0.0110
EMF = 0.00099 V
EMF = 0.99 mV
Answer:
lymph nodes
tonsils and adenoids
thymus
Explanation:
-Arteries are the blood vessels that take the blood that contains oxygen from the heart to the tissues and are part of the circulatory system.
-Lymph nodes are glands that take care of filtering the fluid that goes through the lympathic system and are also important for the functioning of the immune system.
-Capillaries are blood vessels that connect the veins and arteries and are part of the circulatory system.
-Tonsils and adenoids are located in the throat and they help protect the body from diseases and they are part of immune system and the lympathic system.
-Veins are the vessels that take the blood to the heart and they are part of the circulatory system.
-Thymus is an organ in which the T cells develop and they help protect the body against virus and bacteria and it is part of the immune and lympathic systems.
According to this, cells or organs that are considered to be part of both the immune and lymphatic systems are:
lymph nodes
tonsils and adenoids
thymus
To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.
PART A) We will begin by finding the two net distances.
And the distance 'd' is
Through the free-body diagram the tension components are given by
Here we can watch that,
Dividing both expression we have that,
Replacing the values,
PART B) Using the vertical component we can find the tension,
