Answer:
Bone
Explanation:
Diagnostic radiology include the use of non-invasive imaging scans to diagnose a patient.
The voltages used in diagnostic tubes range from roughly 20 kV to 150 kV and thus the highest energies of the X-ray photons range from roughly 20 keV to 150 keV.
The tests and equipment used sometimes involves low doses of radiation to create highly detailed images of an area.
He needs to improve his endurance time.
By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s
Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:
mass m = 0.170 kg
initial speed u = 6 m/s.
Distance covered s = 61 m
To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V
To do this, let us first calculate the kinetic energy at which the ball move.
K.E = 1/2m
K.E = 1/2 x 0.17 x 
K.E = 3.06 J
The work done on the ball is equal to the kinetic energy. That is,
W = K.E
But work done = Force x distance
F x S = K.E
F x 61 = 3.06
F = 3.06/61
F = 0.05 N
From here, we can calculate the acceleration of the ball from Newton second law
F = ma
0.05 = 0.17a
a = 0.05/0.17
a = 0.3 m/
To calculate the final velocity, let us use third equation of motion.
=
+ 2as
=
+ 2 x 0.3 x 61
= 36 + 36
= 72
V = 
V = 8.485 m/s
Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.
Learn more about dynamics here: brainly.com/question/402617
Answer:
power requirement is 23.52 ×
W
Explanation:
given data
flow rate q = 2 m³/s
elevation h = 1200 m
density of the water ρ = 1000 kg/m³
to find out
power requirement
solution
we will get power by the power equation that is
power = ρ× Q× g× h ...................1
put here all value we get power
power = ρ× Q× g× h
power = 1000 × 2 × 9.8 × 1200
power = 23.52 ×
so power requirement is 23.52 ×
W
Answer:9.34 A/s
Explanation:
Given
radius of solenoid 
Emf induced 
no of turns per meter n=450
we know Induced EMF is given by

Magnetic Field is given by

thus 
Area of cross-section
where
solving integration we get

where r=distance from axis
R=radius of Solenoid


