A baseball is accelerated downward at 9.8 m/s^2 when the only vertical force acting upon the baseball is gravity.
Answer:
So the sound intensity level they would experience without the earplugs is 110.32dB.
Explanation:
Given data
Sound intensity by factor =215
Sound intensity level =87 dB
To find
Sound intensity level they would experience without the earplugs
Solution
First we need to find the new sound intensity level
So
The dB can be calculated as:
Substitute the given values
So the sound intensity level they would experience without the earplugs is 110.32dB.
They both are mechanical waves.
Explanation:
Echolocation and ultrasounds are both mechanical waves that require a medium to trace through. They both process and transfer information through waves; the difference is that we humans cannot hear ultrasound waves.
Answer:
Explanation:
The force exerted in a magnetic field is given as
F = q (v × B)
Where
F is the force entered
q is the charge
v is the velocity
B is the magnetic field
Given that,
The magnetic field is
B = 2•i + 4•j. T
The velocity of the electron is
v = 2•i + 6•j + 8•k. m/s
Also, the charge of an electron is
q = -1.602 × 10^-19 C.
Then note that,
V×B is the cross product of the speed and the magnetic field
Then,
F = q (V×B)
F = -1.602 × 10^-19( 2•i + 4•j +8•k × 2•i + 4•j)
Note
i×i=j×j×k×k=0
i×j=k. j×i=-k
j×k=i. k×j=-i
k×i=j. i×k=-j
F = -1.602 × 10^-19[(2•i + 4•j +8•k) × (2•i + 4•j)]
F = -1.602 × 10^-19 [2×2•(i×i) + 2×4•(i×j) + 4×2•(j×i) + 4×4•(j×j) + 8×2•(k×i) + 8×4•(k×j)]
F = -1.602 × 10^-19[4•0 + 8•k + 8•-k + 16•0 + 16•j + 32•-i]
F = -1.602 × 10^-19(0 + 8•k - 8•k + 0 + 16•j - 32•i)
F = -1.602 × 10^-19(16•j - 32•i)
F = -1.602 × 10^-19 × ( -32•i + 16•j)
F = 5.126 × 10^-18 •i - 2.563 × 10^-18 •j
Then, the x component of the force is
Fx = 5.126 × 10^-18 N
Also, the y component of the force is
Fy = -2.563 × 10^-18 N
Answer:
Pressure applied to the needle is 7528 Pa
Explanation:
As we know by poiseuille's law of flow of liquid through a cylindrical pipe
the rate of flow through the pipe is given as
now we know that
radius = 0.2 mm
Length = 6.32 cm
now we have
now we have
