Answer:
A gamma ray, light from a lamp, and a microwave are examples of electromagnetic radiation.
Explanation:
Gamma radiation, visible light (light from a lamp), and microwaves are all apart of the electromagnetic spectrum.
A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta
1 answer:
Answer:
<em>The depth will be equal to</em> <em>6141.96 m</em>
<em></em>
Explanation:
pressure on the submarine = 62 MPa = 62 x 10^6 Pa
we also know that = ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
= 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine = 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure , we have
62 x 10^6 = 10094.49h
depth h = <em>6141.96 m</em>
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Answer:
The separation distance between the slits is 16710.32 nm.
Explanation:
Given that,
Wavelength = 641 nm
Angle =4.4°
(a). We need to calculate the separation distance between the slits
Using formula of young's double slit
Where, d = the separation distance between the slits
m = number of order
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Put the value into the formula
Hence, The separation distance between the slits is 16710.32 nm.
Answer:
9.67 A
Explanation:
The weight of a student with a mass of m = 75 kg is:
where g=9.8 m/s^2 is the acceleration due to gravity.
We want the magnetic force on the wire to be equal to this weight. The magnetic force on the wire is
where
I is the current in the wire
L = 2.0 m is the length of the wire
B = 38 T is the magnetic field
is the angle between the direction of B and L
Since we want W=F, we can write
And we can solve it to find the current I:
Answer:
v = 19.33 m / s South
Explanation:
To solve this exercise we must use the conservation of momentum, for which we must define a system formed by the two cars, therefore the forces during the collision are internal and therefore the moment is conserved.
Since it is a vector quantity, we are going to work on each axis, the x axis is in the East-West direction
initial instant. Before the crash
p₀ = m 0 + M v₂ₓ
final instant. Right after the crash
p_f = (m + M) vₓ
p₀ = 0_pf
M v₂ₓ = (m + M) vₓ
In this case m is the mass of the car and M the mass of the SUV
vₓ = v₂ₓ (1)
in the Y axis (North - South direction)
initial instant
p₀ = m v_{1y} + M 0
final moment
p_f = (m + M) v_y
p₀ = p_f
m v_{1y} + M 0 = (m + M) v_y
v_y = (2)
With these speeds we can use the relationship between work and the variation of kinetic energy, in this part the two cars are already united.
W = ΔK
friction force work is
W = - fr d
the friction force is described by the equation
fr = μ N
Newton's second law
N-W = 0
N = W
we substitute
W = - μ (m + M) g d
as the car stops the final kinetic energy is zero and
the initial kinetic energy is
K₀ = ½ (m + M) v²
we substitute
- μ (m + M) g d = 0 - ½ (m + M) v²
μ g d = ½ v²
v² = 2 μ g d
the distance traveled can be found with the Pythagorean theorem
d =
d =
d = 8.40 m
let's calculate the speed
v² = 2 0.75 9.8 8.40
v = √123.48
v = 11.11 m / s
this velocity is in the direction of motion so we can use trigonometry to find the angles
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ (-5.48 / -6.37)
θ = 40.7º
Since the two magnitudes are negative, this angle is in the third quadrant, measured from the positive side of the x-axis in a counterclockwise direction.
θ'= 180 + 40.7
θ’= 220.7º
In the exercise they indicate the the sedan moves in the y-axis, therefore
sin θ'= v_y / v
v_y = v sin 220.7
v_y = 11.11 sin 220.7
v_y = -7.25 m / s
the negative sign indicates that it is moving south
To find the speed we substitute in equation 2
v_y =
v_{1y} = v_ y
let's calculate
v_{1y} = -7.25
v_{1y} = - 19.33 m/s
therefore the speed of the sedan is v = 19.33 m / s with a direction towards the South