The question is incomplete, the concentration of qam and humulin is not given unless R is used concentration
Complete question:
A physician orders Humulin 50/50 44 units and Humulin N 40 units qam and Humulin R 35 units ac evening meal subcutaneously. How many total units of insulin are administered each morning?
Answer:
the total units of insulin admistered each morning
= 22 units of qam and humulin
Explanation:
given
44 units and Humnlin N
with concentration 50/100 = 1/2 = 0.5
∴ 44 × 0.5 ≈ 22 units in the morning
regular insulin administered each day
(22 + 35)units of qam and humulin
= 57units
The cost of energy delivered to residences by electrical transmission varies from $0.070/kWh to $0.258/kWh throughout the United States; $0.110/kWh is the average value.
Required:
At this average price, calculate the cost of:
a. leaving a 40-W porch light on for two weeks while you are on vacation?
b. making a piece of dark toast in 3.00 min with a 970-W toaster
c. drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer.
Answer:
Cost = $ 1.48
Cost = $ 0.005
Cost = $ 0.38
Explanation:
given data
electrical transmission varies = $0.070/kWh to $0.258/kWh
average value = $0.110/kWh
solution
when leaving a 40-W porch light on for two weeks while you are on vacation so cost will be
first we get here energy consumed that is express as
E = Pt .................1
here E is Energy Consumed and Power Delivered is P and t is time
so power is here 0.04 KW and t = 2 week = 336 hour
so
put value in 1 we get
E = 0.04 × 336
E = 13.44 KWh
so cost will be as
Cost = E × Unit Price .............2
put here value and we get
Cost = 13.44 × 0.11
Cost = $ 1.48
and
when you making a piece of dark toast in 3.00 min with a 970-W toaster
so energy consumed will be by equation 1 we get
E = Pt
power is = 0.97 KW and time = 3 min = 0.05 hour
put value in equation 1 for energy consume
E = 0.97 × 0.05 h
E = 0.0485 KWh
and we get cost by w\put value in equation 2 that will be
cost = E × Unit Price
cost = 0.0485 × 0.11
Cost = $ 0.005
and
when drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer
from equation 1 we get energy consume
E = Pt
Power Delivered = 5.203 KW and time = 40 min = 0.67 hour
E = 5.203 × 0.67
E = 3.47 KWh
and
cost will by put value in equation 2
Cost = E × Unit Price
Cost = 3.47 × 0.11
Cost = $ 0.38
There was a major collision of an asteroid with the Moon in medieval times. It was described by monks at Canterbury Cathedral in England as a red glow on and around the Moon. How long (in s) after the asteroid hit the Moon, which is 3.77 ✕ 105 km away, would the light first arrive on Earth?
Answer:
Explanation:
speed of light = 3 x 10⁸ m /s .
distance between moon and the earth = 3.77 x 10⁵ x 10³m .
Time taken by light to cover the distance
= distance / speed
= 3.77 x 10⁸ / 3 x 10⁸
= 1.256 s
An electric heater is constructed by applying a potential different of 120V across a nichrome wire that has a total resistant of 8 ohm .the current by the wire is
Answer:
15amps
Explanation:
V=IR
I=V/R
I = 120/8
I = 15 amps
A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC charge is released from rest at a point x = 40 cm, what is its kinetic energy the instant it passes the point x = 70 cm?
Answer:
Ek = 8,79 [J]
Explanation:
We are going to solve this problem, using the energy conservation principle
State 1 or initial state (charges at rest t=0)
E₁ = Ek + U₁
As charge are at rest Ek = 0
And U₁ has two components
U₁₂ = K * Q₁*Q₂ / 0,4 and U₃₂ = K*Q₃*Q₂ / 0,6
U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,4 ⇒ U₁₂ = 9*60*10*10⁻³/0,4
U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,6 ⇒ U₃₂ = - 9*20*10*10⁻³/0,6
U₁₂ = 540*10⁻2/0,4 [J] ⇒13,5 [J]
U₃₂ = - 180*10⁻² /0,6 [J] ⇒ - 3 [J]
Then E₁ = E₁₂ + E₃₂
E₁ = 10,5 [J]
At the moment of Q₂ passing x = 40 cm or 0,4 m
E₂ = Ek + U₂
We can calculate the components of U₂ in this new configuration
U₂ = U₁₂ + U₃₂
U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,7 ⇒ U₁₂ = 9*60*10*10⁻³/0,7
U₁₂ = 540*10⁻²/0,7 U₁₂ = 7,71 [J]
U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,3 ⇒ U₃₂ = - 9*20*10*10⁻³/0,3
U₃₂ = - 9*20*10⁻²/0,3
U₃₂ = - 6
U₂ = 7,71 -6
U₂ = 1,71 [J]
Then as
E₂ = Ek + U₂ and E₂ = E₁
Then
Ek + U₂ = E₁
Ek = 10,5 - U₂
Ek = 10,5 - 1,71
Ek = 8,79 [J]
A small glass bead charged to 8.0 nC is in the plane that bisects a thin, uniformly charged, 10-cm long glass rod and is 4.0 cm from the rod's center. The bead is repelled from the rod with a force of 940 μN.
Required:
What is the total charge on the rod?
Answer:
71nC is the total charge of the rod
Explanation:
See attached file
The total charge on the rod is equal to 3.3 × 10⁻⁸ C.
What is the force on a charge in an electric field?The force on the charge in a uniform electric field E is given by:
F = qE where q is charge in coulombs
The electric field due to the charge associated with the rod is given by:
[tex]E =\frac{kQ}{r\sqrt{r+\frac{L^2}{4} } }[/tex]
Where r is the distance between the bead and the rod, L is the length of the glass rod and Q is the charge on the rod.
The force experienced by the bead charged is,
[tex]F =\frac{kqQ}{r\sqrt{r+\frac{L^2}{4} } }[/tex]
From the above equation, we can find the value of Q as:
[tex]Q =\frac{Fr\sqrt{r+\frac{L^2}{4} } }{kq}[/tex]
Given, the value of force, F = 940μN = 940 ×10⁻⁹ C
The length of glass rod, L = 10cm = 0.1 m and r = 4cm = 0.04 m
[tex]Q =\frac{(940\times 10^{-6}N)(0.04m)\sqrt{(0.04m)+\frac{(0.1m)^2}{4} } }{(8.99\times 10^9 N.m^2/C^2)(8\times 10^{-9}C)}[/tex]
[tex]Q= 0.5228\times 10^{-6}\times\sqrt{0.0041}[/tex]
[tex]Q = 3.3\times 10^{-8} C[/tex]
Therefore, the total charge on the rod is 3.3 × 10⁻⁸ C.
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A uniform 2.0-kg rod that is 0.92 m long is suspended at rest from the ceiling by two springs, one at each end. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 29 N/m and 66 N/m. Find the angle that the rod makes with the horizontal.
Answer:
11.7°
Explanation:
See attached file
A box experiencing a gravitational force of 600 N. is being pulled to the right with a force of 250 NA 25 N. frictional force acts on the box as it moves to the right what is the net force in the Y direction
Answer:
32
Explanation:
Answer:
0
Explanation:.
A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wall at a point having coordinates (1.2, 1.9), where the units are meters, what is the distance of the fly from the corner of the room?
Answer:
[tex]D=2.25 m[/tex]
Explanation:
We can use the equation of a distance between to points:
[tex]D=\sqrt{(x_{1}-x_{0})^{2}+(y_{1}-y_{0})^{2}}[/tex]
Now, we know that the origin point (x₀,y₀) = (0,0) and the point (x₁,y₁) = (1.2,1.9)
[tex]D=\sqrt{(1.2-0)^{2}+(1.9-0)^{2}}[/tex]
[tex]D=2.25 m[/tex]
Therefore the distance of the fly will be D = 2.25 m from the corner.
I hope it helps you!
Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 350 newtons stretches a spring 30 centimeters. How much work is done in stretching the spring from 50 centimeters to 80 centimeters
Answer:
Explanation:
350 N force stretches the spring by 30 cm
spring constant K = 350 / 0.30 = (350 / 0.3) N / m
To calculate work done by a spring force we proceed as follows
spring force when the spring is stretched by x = Kx
This force is variable so work done by it can be calculated by integration
Work done by it in stretching from x₁ to x₂
W = ∫ F dx
= ∫ Kx dx with limit from x₁ to x ₂
= 1/2 K ( x₂² - x₁² )
Putting the given values of x₁ = 0.50 m , x₂ = 0.8 m
Work done
= 1/2 x (350 / 0.3)x ( 0.80² - 0.50² )
= 227.50 J
A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to
Complete Question
A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)
Answer:
The velocity is [tex]v = 3.79 *10^{5} \ m/s[/tex]
Explanation:
From the question we are told that
The magnitude of the electric field is [tex]E = 144 \ kV /m = 144*10^{3} \ V/m[/tex]
The magnetic field is [tex]B = 0.38 \ T[/tex]
The force due to the electric field is mathematically represented as
[tex]F_e = E * q[/tex]
and
The force due to the magnetic field is mathematically represented as
[tex]F_b = q * v * B * sin(\theta )[/tex]
Now given that it is perpendicular , [tex]\theta = 90[/tex]
=> [tex]F_b = q * v * B * sin(90)[/tex]
=> [tex]F_b = q * v * B[/tex]
Now given that it is not deflected it means that
[tex]F_ e = F_b[/tex]
=> [tex]q * E = q * v * B[/tex]
=> [tex]v = \frac{E}{B }[/tex]
substituting values
[tex]v = \frac{ 144 *10^{3}}{0.38 }[/tex]
[tex]v = 3.79 *10^{5} \ m/s[/tex]
An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 100mm long. If modulus of elasticity is for the aluminum is 85GN/m^2, calculate the total contraction on the bar due to compressive load of 180kN?
Answer:
1.228 x [tex]10^{-6}[/tex] mm
Explanation:
diameter of aluminium bar D = 40 mm
diameter of hole d = 30 mm
compressive Load F = 180 kN = 180 x [tex]10^{3}[/tex] N
modulus of elasticity E = 85 GN/m^2 = 85 x [tex]10^{9}[/tex] Pa
length of bar L = 600 mm
length of hole = 100 mm
true length of bar = 600 - 100 = 500 mm
area of the bar A = [tex]\frac{\pi D^{2} }{4}[/tex] = [tex]\frac{3.142* 40^{2} }{4}[/tex] = 1256.8 mm^2
area of hole a = [tex]\frac{\pi(D^{2} - d^{2}) }{4}[/tex] = [tex]\frac{3.142*(40^{2} - 30^{2})}{4}[/tex] = 549.85 mm^2
Total contraction of the bar = [tex]\frac{F*L}{AE} + \frac{Fl}{aE}[/tex]
total contraction = [tex]\frac{F}{E} * (\frac{L}{A} +\frac{l}{a})[/tex]
==> [tex]\frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85})[/tex] = 1.228 x [tex]10^{-6}[/tex] mm
18. Two spaceships are traveling in the same direction. With respect to an inertial frame of reference, spaceship A has a speed of 0.900c. With respect to the same inertial frame, spaceship B has a speed of 0.500c. Find the speed vAB of spaceship A relative to spaceship B.
Answer:
[tex]v_{A} -v_{B}=v_{AB}\\v_{AB} =0.9c-0.5c\\v_{AB}=0.4c[/tex]
Explanation:
Because the spaceships are traveling in the same direction, you should subtract the speed of spaceship A from the speed of spaceship B in order to calculate [tex]v_{AB}[/tex].
Calculate the flow rate of blood (of density 0.846 g/cm3 ) in an aorta with a crosssectional area of 1.36 cm2 if the flow speed is 48.5 cm/s. Answer in units of g/s.
Answer:
55.80 g/s
Explanation:
From the question,
Flow rate = density×Area×velocity.
φ = ρ×A×V................... Equation 1
Where φ = flow rate of blood, ρ = density of blood, A = cross sectional area of blood, V = velocity of blood.
Given: ρ = 0.846 g/cm³, A = 1.36 cm², V = 48.5 cm/s.
Substitute these values into equation 1
φ = 0.846×1.36×48.5
φ = 55.80 g/s
Hence, the flow rate of the blood = 55.80 g/s
A certain dam generates 120 MJ of mechanical (hydroelectric) energy each minute. If the conversion from mechanical to electrical energy is then 15% efficient, what is the dam's electrical power output in W?
Answer:
electric energy ( power ) = 300000 W
Explanation:
given data
mechanical (hydroelectric) energy = 120 MJ/min = 2000000 J/s
efficiency = 15 % = 0.15
solution
we know that Efficiency of electric engine is expression as
Efficiency = Mechanical energy ÷ electric energy ......................1
and here dam electrical power output is
put here value in equation 1
electric energy ( power ) = Efficiency × Mechanical energy ( power )
electric energy ( power ) = 0.15 × 2000000 J/s
electric energy ( power ) = 300000 W
Two radio antennas A and B radiate in phase. Antenna B is 120 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 40 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied.
(a) What is the longest wavelength for which there will be destructive interference at point Q?
(b) What is the longest wavelength for which there will be constructive interference at point Q?
Answer:
a. for destructive interference
λmax= 240m
b. for constructive interference
λmax = 120m
Explanation:
An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.10 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 37.5 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target
Answer:
The speed of the arrow after passing through the target is 30.1 meters per second.
Explanation:
The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:
[tex]m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}[/tex]
Where:
[tex]m_{a}[/tex], [tex]m_{t}[/tex] - Masses of arrow and target, measured in kilograms.
[tex]v_{a,o}[/tex], [tex]v_{a,f}[/tex] - Initial and final speeds of the arrow, measured in meters per second.
[tex]v_{t,o}[/tex], [tex]v_{t,f}[/tex] - Initial and final speeds of the target, measured in meters per second.
The final speed of the arrow is now cleared:
[tex]m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})[/tex]
[tex]v_{a,f} = v_{a,o} + \frac{m_{t}}{m_{a}} \cdot (v_{t,o}-v_{t,f})[/tex]
If [tex]v_{a,o} = 2.1\,\frac{m}{s}[/tex], [tex]m_{t} = 0.3\,kg[/tex], [tex]m_{a} = 0.0225\,kg[/tex], [tex]v_{t,o} = 2.10\,\frac{m}{s}[/tex] and [tex]v_{t,f} = 0\,\frac{m}{s}[/tex], the speed of the arrow after passing through the target is:
[tex]v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )[/tex]
[tex]v_{a,f} = 30.1\,\frac{m}{s}[/tex]
The speed of the arrow after passing through the target is 30.1 meters per second.
Indiana Jones is in a temple searching for artifacts. He finds a gold sphere with a radius of 2 cm sitting on a pressure sensitive plate. To avoid triggering the pressure plate, he must replace the gold with something of equal mass. The density of gold is 19.3.103 kg/m3, and the volume of a sphere is V = 4/3 Ar3. Indy has a bag of sand with a density of 1,602 kg/m3.
(A) What volume of sand must he replace the gold sphere with? If the sand was a sphere, what radius would it have?
Answer:
Volume of Sand = 0.4 m³
Radius of Sand Sphere = 0.46 m
Explanation:
First we need to find the volume of gold sphere:
Vg = (4/3)πr³
where,
Vg = Volume of gold sphere = ?
r = radius of gold sphere = 2 cm = 0.02 m
Therefore,
Vg = (4/3)π(0.2 m)³
Vg = 0.0335 m³
Now, we find mass of the gold:
ρg = mg/Vg
where,
ρg = density of gold = 19300 kg/m³
mg = mass of gold = ?
Vg = Volume of gold sphere = 0.0335 m³
Therefore,
mg = (19300 kg/m³)(0.0335 m³)
mg = 646.75 kg
Now, the volume of sand required for equivalent mass of gold, will be given by:
ρs = mg/Vs
where,
ρs = density of sand = 1602 kg/m³
mg = mass of gold = 646.75 kg
Vs = Volume of sand = ?
Therefore,
1602 kg/m³ = 646.75 kg/Vs
Vs = (646.75 kg)/(1602 kg/m³)
Vs = 0.4 m³
Now, for the radius of sand sphere to give a volume of 0.4 m³, can be determined from the formula:
Vs = (4/3)πr³
0.4 m³ = (4/3)πr³
r³ = 3(0.4 m³)/4π
r³ = 0.095 m³
r = ∛(0.095 m³)
r = 0.46 m
When you are told that the wind has a "Small Coriolis force" associated with it, what is that "small force" exactly
Answer:
Coriolis force is a type of force of inertia that acts on objects that is in motion within a frame of reference that rotates with respect to an inertial frame. Due to the rotation of the earth, circulating air is deflected result of the Coriolis force, instead of the air circulating between the earth poles and the equator in a straight manner. Because of the effect of the Coriolis force, air movement deflects toward the right in the Northern Hemisphere and toward the left in the Southern Hemisphere, eventually taking a curved path of travel.
A truck running at 115 km / h slows down at a rate of 4 m / s every 2 seconds. How far will it travel to a stop? choose the correct option 1) 335.04m 2) 205.04m 3) 255.04m
Answer:
3) 255.04 m
Explanation:
Given:
v₀ = 115 km/h = 31.944 m/s
v = 0 m/s
a = -(4 m/s) / 2s = -2 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (31.944 m/s)² + 2 (-2 m/s²) Δx
Δx = 255.11 m
Closest answer is option 3.
When we describe electric flux, we say that a surface is oriented in a certain direction with respect to an electric field. When we try to calculate how much electric field passes through the surface, we make use of the:_________.
1. Wedge Product
2. Dot Product
3. Cross Product
Answer:
2. Dot Product
Explanation:
The calculation of the electric flux gives an scalar result.
When we tray to calculate how much electric field passes trough a surface, we are calculating a scalar value. Furthermore, the concept of flux requires the calculation of a scalar value.
Also it is necessary to take into account that the magnitude of the flux trough a surface depends of the inclination of the surface respect to the direction of the electric field. This is taken into account sufficiently by a dot product.
Then, the answer is:
2. Dot Product
A very long, straight horizontal wire carries a current such that 8.25×1018 electrons per second pass any given point going from west to east.
What is the magnitude of the magnetic field this wire produces at a point 5.00 cm directly above it?
Answer: The magnitude of the magnetic field this wire produces is 5.56 × 10^-6 T
Explanation: Please see the attachments below
This question involves the concepts of Biot-Savart Law and magnetic field. It can be solved by the application of Biot-Savart Law on a current-carrying wire.
The magnitude of the magnetic field is "5.28 x 10⁻⁶ T".
Using the Biot-Savart Law to find out the magnitude of the magnetic field produced by the wire at a point 5 cm above it, directly:
[tex]B = \frac{\mu_o I}{2\pi r}\\\\[/tex]
where,
B = magnetic field = ?
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
I = current = [tex]\frac{ne}{t} = (8.25\ x\ 10^{18}\ elctrons/sec)(1.6\ x\ 10^{-19}\ C /electron)[/tex] = 1.32 A
r = radius = distance above wire = 5 cm = 0.05 m
Therefore,
[tex]B = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(1.32\ A)}{2\pi(0.05\ m)}[/tex]
B = 5.28 x 10⁻⁶ T
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The attached picture shows the illustration of magnetic field due to a wire.
The buoyant force on an object placed in a liquid is (a) always equal to the volume of the liquid displaced. (b) always equal to the weight of the object. (c) always equal to the weight of the liquid displaced. (d) always less than the volume of the liquid displaced.
Answer:
(c) always equal to the weight of the liquid displaced.
Explanation:
Archimedes principle (also called physical law of buoyancy) states that when an object is completely or partially immersed in a fluid (liquid, e.t.c), it experiences an upthrust (or buoyant force) whose magnitude is equal to the weight of the fluid displaced by that object.
Therefore, from this principle the best option is C - always equal to the weight of the liquid displaced.
A heavier car is always safer in a crash than a lighter car.
Answer:
not true because the mass from the heavy car will cause it to damage more
Explanation:
Answer: answer on edmentum is false your welcome
Explanation:
It is false because it's more heavy so more damage l.
A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a stationary particle with a charge of –5 × 10–6 C. The radius of the orbit is:
Answer:
r = 0.22m
Explanation:
To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.
Then, you have:
[tex]F_c=F_e=ma_c[/tex] (1)
m: mass of the particle = 20g = 20*10-3 kg
ac: centripetal acceleration = ?
q: charge of the particle = 5*10^-6C
Fe: electric force between the charges
The electric force is given by:
[tex]F_e=k\frac{qq'}{r^2}[/tex] (2)
r: radius of the orbit
q': charge of the particle at the center of the orbit = -5*10^-6C
Furthermore, the centripetal acceleration is:
[tex]a_c=\frac{v^2}{r}[/tex] (3)
v: speed of the particle = 7m/s
You replace the expressions (2) and (3) in the equation (1) and solve for r:
[tex]k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}[/tex]
Finally, you replace the values of all parameters in the previous expression:
[tex]r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m[/tex]
The radius of the circular trajectory is 0.22m
If you could see stars during the day, this is what the sky would look like at noon on a given day. The Sun is near the stars of the constellation Gemini. Near which constellation would you expect the Sun to be located at sunset?
Answer:
The sun will be located near the Gemini constellation at sunset
The friends spend some time thinking about a beam of light traveling from one medium to another medium with higher index of refraction, which strikes the boundary obliquely. Which of Tristan's statements is correct
Answer:
"When light moves from a material in which its speed is high to a material in which its speed is lower, the angle of refraction θ2is less than the angle of incidence θ1and the ray is bent toward the normal."
Explanation:
Refraction is a phenomenon that occurs when light rays change direction after passing through a surface or medium. This is also known as 'bending'. Snell's law provides the relationship between the angle of incidence and refraction in the equation below:
n₁sinФ₁ = n₂sinФ₂
where n1 and n2 represent the two media and theta refers to the angles formed. When light hits a medium with a high refractive index, the speed of light becomes slower.
So, Tristan is right when he says that, "When light moves from a material in which its speed is high to a material in which its speed is lower, the angle of refraction θ2is less than the angle of incidence θ1and the ray is bent toward the normal."
help me
Describe the different types of non contact forces.
Answer:
the correct answer is
All four known fundamental interactions are non-contact forces: Gravity, the force of attraction that exists among all bodies that have mass. ... Examples of this force include: electricity, magnetism, radio waves, microwaves, infrared, visible light, X-rays and gamma rays.
Explanation:
hope this helps you!!!!
Calculate the moment of inertia of a skater given the following information.
(a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius.
(b) The skater with arms extended is approximately a cylinder that is 74.0 kg, has a 0.150 m radius, and has two 0.750 m long arms which are 3.00 kg each and extend straight out from the cylinder like rods rotated about their ends.
Answer:
(a) I = 0.363 kgm^2
(b) I = 1.95 kgm^2
Explanation:
(a) If you consider the shape of the skater as approximately a cylinder, you use the following formula to calculate the moment of inertia of the skater:
[tex]I_s=\frac{1}{2}MR^2[/tex] (1)
M: mass of the skater = 60.0 kg
R: radius of the cylinder = 0.110m
[tex]I_s=\frac{1}{2}(60.0kg)(0.110m)^2=0.363kg.m^2[/tex]
The moment of inertia of the skater is 0.363 kgm^2
(b) In the case of the skater with his arms extended, you calculate the moment of inertia of a combine object, given by cylinder and a rod (the arms) that cross the cylinder. You use the following formula for the total moment of inertia:
[tex]I=I_c+I_r\\\\I=\frac{1}{2}M_1R^2+\frac{1}{12}M_2L^2[/tex] (2)
M1: mass of the cylinder = 74.0 kg
M2: mass of the rod = 3.00kg +3.00kg = 6.00kg
L: length of the rod = 0.750m + 0.750m = 1.50m
R: radius of the cylinder = 0.150
[tex]I=\frac{1}{2}(74.0kg)(0.150m)^2+\frac{1}{12}(6.00kg)(1.50m)^2\\\\I=1.95kg.m^2[/tex]
The moment of inertia of the skater with his arms extended is 1.95 kg.m^2
Find the length (in m) of an organ pipe closed at one end that produces a fundamental frequency of 175 Hz when air temperature is 18.0°C
Answer:
Length = 0.4882 m
Explanation:
given data
fundamental frequency = 175 Hz
air temperature = 18.0°C
solution
we will apply here fundamental frequency formula that is
F = [tex]\frac{v}{4L}[/tex] ....................1
here v = [tex]331 \sqrt{1+\frac{T}{273}}[/tex]
here 331 m/s is speed of sound in air
so v = [tex]331 \sqrt{1+\frac{18}{273}}[/tex] = 341.74 m/s
now put value in equation 1 we get
F = [tex]\frac{v}{4L}[/tex]
[tex]175 = \frac{341.74}{4L}[/tex]
Length = 0.4882 m
If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 55.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Answer:
14.79 kgm/s
Explanation:
Data provided in the question
Let us assume the mass of baseball = m = 0.145 kg
The Initial velocity of pitched ball = [tex]v_i[/tex] = 47 m/s
Final velocity of batted ball in the opposite direction = [tex]v_f[/tex]= -55m/s
Based on the above information, the change in momentum is
[tex]\Delta P = m(v_f -v_i)[/tex]
[tex]= 0.145 kg(-55m/s - 47m/s)[/tex]
= 14.79 kgm/s
Hence, the magnitude of the change in momentum of the ball is 14.79 kg m/s