Answer:
E) 80 N/m
Explanation:
Given;
mass of the block, m = 4.8 kg
displacement of the block, x = -0.5 m
velocity of the block, v = -0.8 m/s
acceleration of the block, a = 8.3 m/s²
From Newton's second law of motion;
F = ma
Also, from Hook's law;
F = -Kx
where;
k is the force constant
Thus, ma = -kx
k = -ma/x
k = -(4.8 x 8.3) / (-0.5)
k = 79.7 N/m
k ≅ 80 N/m
Therefore, the force constant of the spring is closest to 80 N/m
Can someone tell me a very very simple physics experiment topic that links to biology?
Explanation:
One idea would be to investigate the correlation between your pulse pressure and your pulse rate. To do this, you'll need a blood pressure monitor.
First, measure your resting pressure and rate. Then exercise for 30 seconds. Measure your new blood pressure and pulse rate. Wait for your pressure and rate to return to normal, then repeat the trial for 1 minute, 1.5 minutes, 2 minutes, etc.
List the results in a table. This should include the amount of exercise time, your pulse rate, your systolic pressure (the high number, which is your blood pressure during contraction of your heart muscle), and your diastolic pressure (the low number, which is your blood pressure between heartbeats). Calculate your pulse pressure (systolic minus diastolic) for each trial. Graph the pulse pressure on the x-axis, and your pulse rate (beats per minute) on the y-axis.
What do you hypothesize will be the shape of the graph? Consider Bernoulli's formula, which relates fluid pressure and flow. How close do the results match your hypothesis? What might explain any differences?
An ideal gas is at a temperature of 320 K. What is the average translational kinetic energy of one of its molecules
Answer:
6.624 x 10^-21 J
Explanation:
The temperature of the ideal gas = 320 K
The average translational energy of an ideal gas is gotten as
[tex]K_{ave}[/tex] = [tex]\frac{3}{2}K_{b}T[/tex]
where
[tex]K_{ave}[/tex] is the average translational energy of the molecules
[tex]K_{b}[/tex] = Boltzmann constant = 1.38 × 10^-23 m^2 kg s^-2 K^-1
T is the temperature of the gas = 320 K
substituting value, we have
[tex]K_{ave}[/tex] = [tex]\frac{3}{2} * 1.38*10^{-23} * 320[/tex] = 6.624 x 10^-21 J
un
An aircraft travelling at 600 km/h accelerates steadily at 10 km/h per second. Taking the
speed of sound as 1100 km/h at the aircraft's altitude, how long will it take to reach the
Sound barrier'?
Answer:
50 seconds
Explanation:
Acceleration = change in velocity / change in time
a = Δv / Δt
10 km/h/s = (1100 km/h − 600 km/h) / t
t = 50 s
How many electrons would have to be removed from a coin to leave it with a charge of +1.0 10-7 C?
Answer:
[tex]n=6.25\times 10^{11}[/tex]
Explanation:
We need to find the number of electrons that would have to be removed from a coin to leave it with a charge of [tex]+10^{-7}\ C[/tex]. Then the number of electrons be n. Using quantization of electric charge as :
q = ne
e is charge on an electron
[tex]n=\dfrac{q}{e}\\\\n=\dfrac{10^{-7}}{1.6\times 10^{-19}}\\\\n=6.25\times 10^{11}[/tex]
So, the number of electrons are [tex]6.25\times 10^{11}[/tex].
In a container of negligible mass, 020 kg of ice at an initial temperature of - 40.0 oC is mixed with a mass m of water that has an initial temperature of 80.0 oC. No heat is lost to the surroundings. If the final temperature of the system is 20.0 oC, what is the mass m of the water that was initially at 80.0 oC
Answer:
The mass is [tex]m_w = 0.599 \ kg[/tex]
Explanation:
From the question we are told that
The mass of ice is [tex]m_c = 0.20 \ kg[/tex]
The initial temperature of the ice is [tex]T_i = -40.0 ^oC[/tex]
The initial temperature of the water is [tex]T_{iw} = 80^o C[/tex]
The final temperature of the system is [tex]T_f = 20^oC[/tex]
Generally according to the law of energy conservation,
The total heat loss is = total heat gained
Now the total heat gain is mathematically represented as
[tex]H = H_1 + H_2 + H_3[/tex]
Here [tex]H_1[/tex] is the energy required to move the ice from [tex]-40^oC \to 0^oC[/tex]
And it mathematically evaluated as
[tex]H_1 = m_c * c_c * \Delta T[/tex]
Here the specific heat of ice is [tex]c_c = 2100 \ J \cdot kg^{-1} \cdot ^oC^{-1}[/tex]
So
[tex]H_1 = 0.20 * 2100 * (0-(-40))[/tex]
[tex]H_1 = 16800\ J[/tex]
[tex]H_2[/tex] is the energy to melt the ice
And it mathematically evaluated as
[tex]H_2 = m * H_L[/tex]
The latent heat of fusion of ice is [tex]H_L = 334 J/g = 334 *10^{3} J /kg[/tex]
So
[tex]H_2 = 0.20 * 334 *10^{3}[/tex]
[tex]H_2 = 66800 \ J[/tex]
[tex]H_3[/tex] is the energy to raise the melted ice to [tex]20^oC[/tex]
And it mathematically evaluated as
[tex]H_3 = m_c * c_w * \Delta T[/tex]
Here the specific heat of water is [tex]c_w= 4190\ J \cdot kg^{-1} \cdot ^oC^{-1}[/tex]
[tex]H_3 = 0.20 * 4190* (20-0))[/tex]
[tex]H_3 = 16744 \ J[/tex]
So
[tex]H = 16800 + 66800 + 16744[/tex]
[tex]H = 100344\ J[/tex]
The heat loss is mathematically evaluated as
[tex]H_d = m * c_h ( 80 - 20 )[/tex]
[tex]H_d = m_w * 4190 * ( 80 - 20 )[/tex]
[tex]H_d = 167600 m_w[/tex]
So
[tex]167600 m_w = 100344[/tex]
=> [tex]m_w = 0.599 \ kg[/tex]
Find the work done by the gas during the following stages. (a) A gas is expanded from a volume of 1.000 L to 4.000 L at a constant pressure of 2.000 atm. (b) The gas is then cooled at constant volume until the pressure falls to 1.500 atm
Answer:
a) 607.95 J
b) 0 J
Explanation:
a) Initial volume = 1 L = 0.001 m^3
final volume = 4 L = 0.004 m^3
pressure = 2 atm = 202650 Pa (1 atm = 101325 Pa)
work done by the gas on the environment = PΔV
P is the pressure = 101325 Pa
ΔV is the change in volume from the initial volume to the final volume
ΔV = 0.004 m^3 - 0.001 m^3 = 0.003 m^3
work done by the gas = 202650 x 0.003 = 607.95 J
b) If the gas is cooled at constant volume, then the gas does no work. For a gas to do work, there must be a change in its volume.
Therefore the work done in cooling at constant volume until pressure falls to 1.5 atm = 0 J
How are period and frequency related to each other?
A ranger in a national park is driving at 15.0 m/s when a deer jumps into the road 60 m ahead of the vehicle. After a reaction time, t, the ranger applies the brakes to produce an acceleration of a = -3.00 m/s2. What is the maximum reaction time allowed if she is to avoid hitting the deer?
Answer:
t = 5 s
Explanation:
In uniform rectilinear movement, the equation for final speed is:
vf = v₀ + a*t
In this case we need that the car stops just before 60 m after applied the brakes, then
vf = v₀ - a*t
vf = final speed = 0
v₀ = initial speed = 15 m/s
And negative acceleration is 3 m/s²
0 = 15 (m/s) - 3 ( m/s²)*t
t = 15 / 3 (m/s /m/s²)
t = 5 s
The point is that with that value ranger will hit the deer so in order to not to hit the deer that time should be smaller than 5 seconds
A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 540 nm, what is the thickness of the film?
Answer:
The thickness of the film is 4.32 μm.
Explanation:
Given;
index of refraction of the thin film on one beam, n₂ = 1.5
number of bright fringes shift in the pattern produced by light, ΔN = 8
wavelength of the Michelson interferometer, λ = 540 nm
The thickness of the film will be calculated as;
[tex]\delta N = \frac{2L}{\lambda} (n_2 -n_1)[/tex]
where;
n₁ and n₂ are the index of refraction on the beam
L is the thickness of the film
[tex]\delta N = \frac{2L}{\lambda} (n_2 -n_1)\\\\L = \frac{\lambda}{2} (\frac{N}{n_2-n_1} )\\\\L = \frac{540*10^{-9}}{2} (\frac{8}{1.5-1} )\\\\L = 4.32*10^{-6} \ m\\\\L = 4.32 \mu m[/tex]
Therefore, the thickness of the film is 4.32 μm.
If the person generates another pulse like the first but the rope is tightened, the pulse will move:_________.
A. at the same rate
B. faster
C. slower
Answer:
the correct answer is B
Explanation:
The speed of a pulse on a string is described by the expression
v =√T/μ
where v is the speed of the pulse, t is the tension of the string and μ the linear density of the string
When applying this equation to our case, if the string is taut, it implies that the tension has increased so that the pulse speed is FASTER
the correct answer is B
How do you solve this ?
Why is the ans C not B ?
Explanation:
Draw a free body diagram of the toolbox. There are two forces:
Weight force mg pulling down,
and applied force F pulling up.
Sum of forces in the y direction:
∑F = ma
F − mg = ma
45 N − 15 N = (3 kg) a
a = 10 m/s²
The answer should be B. It's possible the answer key has a mistake.
Read the passage about the pygmy shrew.
The pygmy shrew is the smallest mammal in North America. However, when comparing the amount of food eaten to its body weight, the pygmy shrew eats more food than any other mammal. It will consume two to three times its own body weight in food daily. One explanation is that the pygmy shrew uses energy at a high rate. In fact, its heart beats over one thousand times per minute.
What is the best explanation for what happens to the food's mass and energy when it is consumed by the pygmy shrew?
Answer:
A very high metabolism and a very small size.
Explanation:
The pygmy shrew is a very small mammal, that forages day and night. The metabolism of the Pygmy shrew is so high that it must eat at least every 30 minutes or it might die. The best explanation for what happens to the food's mass and energy is that most of the food mass is rapidly used fro building up of the shrew due to its very high metabolism, and a bigger portion of the food is lost from the surface of the body of the shrew, due to its very small size. The combination of these two factors; a very high metabolism (rapidly uses up food material, and generates a large amount of heat in a very short time) and the very small size (makes heat loss due to surface area to volume ratio high) explains what happens to the food mass and energy.
The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rev/min in 8.00s, starting from rest?
Answer:
Explanation:
From the question we are told that
The moment of inertia is [tex]I = 2.50 \ kg \cdot m^2[/tex]
The final angular speed is [tex]w_f = 400 rev/min = \frac{400 * 2\pi}{60} = 41.89 \ rad/s[/tex]
The time taken is [tex]t = 8.0 s[/tex]
The initial angular speed is [tex]w_i = 0\ rad/s[/tex]
Generally the average angular acceleration is mathematically represented as
[tex]\alpha = \frac{w_f - w_i }{t}[/tex]
=> [tex]\alpha = \frac{41.89}{8}[/tex]
=> [tex]\alpha = 5.24 \ rad/s^2[/tex]
Generally the torque is mathematically represented as
[tex]\tau = I * \alpha[/tex]
=> [tex]\tau = 5.24 * 2.50[/tex]
=> [tex]\tau = 13.09 \ N \cdot m[/tex]
Which of the following landforms would you NOT expect to see in the Inland South?
a) Swamplands
b) Flat, sandy plains
c) Rugged mountains
d) Rolling hills
e) Glaciers
Answer:
Option: e) Glaciers
Explanation:
Historically, the Inland South viewed as a backwater of the United States. The Inland South is known for its Appalachians mountains, swampland. Flatland, sandy soils, and meandering rivers are some of its features. The climate is hot and humid, which allow the area to grow as Pantanal floodplains. Glaciers will be one of the landforms that one will not expect to see in the Inland South.
On a day that the temperature is 10.0°C, a concrete walk is poured in such a way that the ends of the walk are unable to move. Take Young's modulus for concrete to be 7.00 109 N/m2 and the compressive strength to be 2.00 109 N/m2. (The coefficient of linear expansion of concrete is 1.2 10-5(°C−1).)
What is the stress in the cement on a hot day of 42.0°C? N/m2
Answer:
The stress is [tex]stress = 2688000 \ N[/tex]
Explanation:
From the question we are told that
The first temperature is [tex]T_1 = 10 ^o \ C[/tex]
The young modulus is [tex]Y = 7.00 *10^9\ N/m^2[/tex]
The compressive strength is [tex]\sigma = 2.00 *10^{9} \ N/m^2[/tex]
The coefficient of linear expansion is [tex]\alpha = 1.2 *10^{-5} \ ^o C ^{-1}[/tex]
The second temperature is [tex]T_2 = 42.0^o \ C[/tex]
Generally the change in length of the concrete is mathematically represented as
[tex]\Delta L = \alpha * L * [T_2 - T_1 ][/tex]
=> [tex]\frac{\Delta L}{L} = \alpha * [T_2 - T_1 ][/tex]
=> [tex]strain = \alpha * [T_2 - T_1 ][/tex]
Now the young modulus is mathematically represented as
[tex]Y = \frac{stress}{strain}[/tex]
=> [tex]7.00 *10^9 = \frac{stress}{\alpha(T_2 - T_1 ) }[/tex]
=> [tex]stress = \alpha (T_2 - T_1 ) * 7.00 *10^9[/tex]
=> [tex]stress = 1.2* 10^{-5} (42 - 10 ) * 7.00 *10^9[/tex]
=> [tex]stress = 2688000 \ N[/tex]
Marcel is helping his two children, Jacques and Gilles, to balance on a seesaw so that they will be able to make it tilt back and forth without the heavier child, Jacques, simply sinking to the ground. Given that Jacques, whose weight is WWW, is sitting at distance LLL to the left of the pivot, at what distance L1L1L_1 should Marcel place Gilles, whose weight is www, to the right of the pivot to balance the seesaw
Answer:
L1 = WL/w
Explanation:
Jacques's weight = W
Jacques's distance from the pivot = L
Gilles's weight = w
Gilles's distance from the pivot must be L1
For the two boy to balance each other, they must generate equal amount of torques around the pivot
Torque = distance x weight
For Jacques, the torque generated = WL
For Gilles, the torque generated = wL1
Balancing the two torques, we have
WL = wL1
the distance L1 must be equal to
L1 = WL/w
Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.
If she runs fhe same course again, what constant speed would let her finish in the same time as in the first race?
Answer:
The velocity is [tex]v = 4.76 \ m/s[/tex]
Explanation:
From the question we are told that
The first distance is [tex]d_1 = 4.0 \ km = 4000 \ m[/tex]
The first speed is [tex]v_1 = 5.0 \ m/s[/tex]
The second distance is [tex]d_2 = 1.0 \ km = 1000 \ m[/tex]
The second speed is [tex]v_2 = 4.0 \ m/s[/tex]
Generally the time taken for first distance is
[tex]t_1 = \frac{d_1 }{v_1 }[/tex]
[tex]t_1 = \frac{4000}{5}[/tex]
[tex]t_1 = 800 \ s[/tex]
The time taken for second distance is
[tex]t_1 = \frac{d_2 }{v_2 }[/tex]
[tex]t_1 = \frac{1000}{4}[/tex]
[tex]t_1 = 250 \ s[/tex]
The total time is mathematically represented as
[tex]t = t_1 + t_2[/tex]
=> [tex]t = 800 + 250[/tex]
=> [tex]t = 1050 \ s[/tex]
Generally the constant velocity that would let her finish at the same time is mathematically represented as
[tex]v = \frac{d_1 + d_2}{t }[/tex]
=> [tex]v = \frac{4000 + 1000}{1050 }[/tex]
=> [tex]v = 4.76 \ m/s[/tex]
The constant speed that will let her finish in the same time as in the first race is 4.76 m/s
Determination of the time taken for first 4 KmDistance = 4 Km = 4 × 1000 = 4000 mSpeed = 5 m/sTime 1 =?
Time 1 = distance / speed
Time 1 = 4000 / 5
Time 1 = 800 s
Determination of the time taken for the last 1 KmDistance = 1 Km = 1 × 1000 = 1000 mSpeed = 4 m/sTime 2 =?Time 2 = distance / speed
Time 2 = 1000 / 4
Time 2 = 250 s
Determination of the constant speedTotal distance = 4000 + 1000 = 5000 mTotal time = 800 + 250 = 1050 sConstant speed =?Constant speed = Total distance / total time
Constant speed = 5000 / 1050
Constant speed = 4.76 m/s
Learn more about average speed:
https://brainly.com/question/8819317
Find the position of the center of mass of two bodies points of masses m1 and m2 joined by a rod of mass negligible in length d. Find the acceleration of the center of mass for the case that m1 = 1 [kg] and m2 = 3 [kg] and the applied forces of the figure with d = 2 [m]
Explanation:
If m₁ is at the origin, then the center of mass is at:
x = (m₁ × 0 m + m₂ × d m) / (m₁ + m₂)
x = m₂ d / (m₁ + m₂)
If m₁ = 1 kg, m₂ = 3 kg, and d = 2 m:
x = (3 kg) (2 m) / (1 kg + 3 kg)
x = 1.5 m
Sum of forces in the x direction:
∑F = ma
16 N = (1 kg + 3 kg) aₓ
aₓ = 4 m/s²
Sum of forces in the y direction:
∑F = ma
20 N = (1 kg + 3 kg) aᵧ
aᵧ = 5 m/s²
You push a box across the floor with a force of 20 N. You push it 10 meters in 5 seconds. How much work did you do? How much power did you use? Enter your answer in the space provided.
Explanation:
ur answer is in attachment.
hope it helps u
mark as brainlist
follow for good ans
Explanation:
Force applied on the box = 20 N
Displacement, s = 10 m
Time taken = 5 sec
According to first condition of the question, we could find the value of work done
i.e
Work done = force × displacement
= 20 × 10
= 200 Joule
According to second condition of the question, we could find the value of power
i.e
Power = work done/Time taken
= 200/5
= 40 watt.
Hope it helpful!!!!!!!!!!!!
In a Young's double-slit experiment, light of wavelength 500 nm illuminates two slits which are separated by 1 mm. The separation between adjacent bright fringes on a screen 5 m from the slits is: CONVERT FIRST
Answer:
The value is [tex]y = 0.0025 \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 500 \ nm[/tex]
The distance of separation is [tex]d = 1\ mm = 0.001 \ m[/tex]
The distance from the screen is [tex]D = 5 \ m[/tex]
Generally the separation between the adjacent bright fringe is mathematically represented as
[tex]y = \frac{D * \lambda }{ d}[/tex]
=> [tex]y = \frac{5 * 500 *10^{-9}}{0.001}[/tex]
=> [tex]y = 0.0025 \ m[/tex]
Let a metallic rod 20 cm in length be heated to a uniform temperature of 100°C Suppose that at t=0 the ends of the bar are plunged into an ice bath at 0°C and thereafter maintained at this temperature, but that no heat is allowed to escape through the lateral surface. Find the time that will elapse before the center of the bar cools to a temperature of 5°C if the bar is made of (a) silver (b) aluminum (c) cast iron. The α2 values for silver, aluminum and cast iron are 1.71, 0.86 and 0.12 respectively.
(Round your answer to two decimal places. Use two-term approximation of the series.)
For silver, it takes _____ seconds to cool the bar to a temperature of 5°C
For aluminum, it takes _____ seconds to cool the bar to a temperature of 5°C
For cast iron, it takes _____ seconds to cool the bar to a temperature of 5°C
Answer:
a) t = 59 s , b) t = 107 s , c) t = 466 s
Explanation:
This is an exercise in thermal conductivity. The power dissipated or transferred is
P = Q / Δt = k A dT/dx
where Q is the thermal energy of the bar, k the constant of thermal conductivity.
If we assume that we are in a stable regime
dT / dx = (T₀ - [tex]T_{f}[/tex]) / L
the energy in the bar is
Q = m [tex]c_{e}[/tex] T₀
we substitute
m c_{e} T₀ / t = k A (T₀ -T_{f}) / L
t = c_{e} / k m L / A T₀ / (T₀ -T_{f})
let's use the concept of density
ρ = m / V
V = A L
m = ρ AL
t = c_{e} / k (ρ A L) L / A T₀ / (T₀ -T_{f})
t = [tex]c_{e}[/tex] /k ρ L² T₀/(T₀ -T_{f})
In this exercise, the initial temperature is T₀ = 100ºC, the final temperature is T_{f }= 5ºC and the length
L = L₀ / 2 = 20/2 = 10cm = 0.1m
a) case of silver
c_{e} = 234 J / kg ºC
k = 437 W / m ºC
ρ = 10,490 10³ kg / m³
let's calculate
t = 234/437 10.49 10³ 0.1² 100 / (100 -5)
t = 59 s
b) case materials aluminum
c_{e} = 900 J / kg ºC
k = 238 W / m ºC
ρ = 2.70 10³ kg / m³
t = 900/238 2.70 10³ 0.1² 100 / (100-5)
t = 107 s
c) iron material
c_{e} = 448 J / kg ºC
k = 79.5 W / m ºC
ρ = 7.86 10³ kg / m³
t = 448 / 79.5 7.86 10³ 0.1² 100 / (100-5)
t = 466 s
.If we had two unknown masses on opposite sides of the pivot, could we calculate both masses given just the information used in the experiment? Explain.
Answer:
No
Explanation:
No, it is impossible to calculate the two masses.
From the statement, there is one known mass on one of the side of the pivot and one "mystery" mass object on the other side of the pivot, so that we have to move that mystery mass or that known mass in a way that it balances both.
We need to know at least one mass. We cannot use any equilibrium condition involving torque with unknown masses.
Which of the following is not a Health-Related fitness part? Question 1 options: Body composition Power Flexibility Speed
Answer:
The correct answer is - speed.
Explanation:
Health related fitness have five major components that are muscular strength or power, endurance, body composition, cardiovascular endurance and flexibility.
All these five health related fitness helps an individual to loose weight, better sleep, enhance mood and prevents from various disease and many other beneficiary results.
Thus, the correct answer is : speed.
Bailey wants to find out which frozen solid melts the fastest: soda, ice, or orange juice. She pours each of the three liquids into the empty cubes of an ice tray, and then places the ice tray in the freezer overnight. The next day, she pulls the ice tray out and sets each cube on its own plate. She then waits and watches for them to melt. When the last part of the frozen liquid melts, she records the time.
Answer:
its 45 over 6
Explanation:the answer is in the question
Answer: Only the melted cube's shape changed.
Explanation:
Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron would the proton have to be? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2, e = 1.6 × 10-19 C, mproton = 1.67 × 10-27 kg, melectron = 9.11 × 10-31 kg)
Answer:
The value is [tex]r = 5.077 \ m[/tex]
Explanation:
From the question we are told that
The Coulomb constant is [tex]k = 9.0 *10^{9} \ N\cdot m^2 /C^2[/tex]
The charge on the electron/proton is [tex]e = 1.6*10^{-19} \ C[/tex]
The mass of proton [tex]m_{proton} = 1.67*10^{-27} \ kg[/tex]
The mass of electron is [tex]m_{electron } = 9.11 *10^{-31} \ kg[/tex]
Generally for the electron to be held up by the force gravity
Then
Electric force on the electron = The gravitational Force
i.e
[tex]m_{electron} * g = \frac{ k * e^2 }{r^2 }[/tex]
[tex]\frac{9*10^9 * (1.60 *10^{-19})^2 }{r^2 } = 9.11 *10^{-31 } * 9.81[/tex]
[tex]r = \sqrt{25.78}[/tex]
[tex]r = 5.077 \ m[/tex]
A person is lying on a diving board 3.00 m above the surface of the water in a swimming pool. She looks at a penny that is on the bottom of the pool directly below her. To her, the penny appears to be a distance of 8.00 m from her.
Required:
What is the depth of the water at this point?
Answer:
The depth of water at the point is [tex]d_A = 6.55 \ m[/tex]
Explanation:
From the question we are told that
The height of the person above water is [tex]d = 3.00 \ m[/tex]
The distance of the coin as seen by the person is [tex]d' = 8.00 \ m[/tex]
Generally the apparent depth is mathematically represented as
[tex]d_a = \frac{d_A}{n}[/tex]
Here [tex]d_A[/tex] is the actual depth of water while n is the refractive index of water with a constant value [tex]n = 1.33[/tex]
Now from the point the person is the apparent depth is evaluated as
[tex]d_a = d'-d[/tex]
=> [tex]d_a = 8 - 3[/tex]
=> [tex]d_a = 5 \ m[/tex]
So
[tex]5 = \frac{d_A}{1.33}[/tex]
=> [tex]d_A = 5 * 1.33[/tex]
=> [tex]d_A = 6.55 \ m[/tex]
: An experienced spear fisherman sees a small fish swimming in a tidal pool. If the fisherman sees the fish at approximately a 40o angle (measured from vertical), he knows that he must release his spear at what angle
Answer:
He should aim below 40°
Explanation:
Because
His eyes are being deceived about the real location of the fish, because the light coming from the fish is refracting away from the normal, if traced from the vertical as it passes into the air and to his eye . so, he will perceive the fish as being shallower than it really is. So if he throws the spear which will travel in a straight line directly at where he perceives the fish to be, the spear will pass above its head so he needs to need to aim lower than 40°
Explanation:
A relaxed biceps muscle requires a force of 25.0N for an elongation of 3.0 cm; under maximum tension, the same muscle requires a force 500N for the same elongation. Find the Young's modulus for the muscle tissue under each of these conditions if the muscle can be modeled as a uniform cylinder with an initial length of 0.200 m and a cross-sectional area of 50 cm^2.
3.3 x10^4N/m²
6.7 x105N/m²
Explanation:
Let the young modulus of the relaxed biceps be
Y= F¹Lo/ deta L1 x A
= 25 x0.2/ 0.03* 50cm²(1m²
0.0004cm^-²)
= 3.3x10^4N/m²
But young modules of muscle under maximum tension will be
Y= F"Lo/ deta L" x A
= 500x 0.2/ 0.03* 50cm²(1m²
0.0004cm^-²)
= 6.7 x10^5N/m²
The Young's Modulus of the relaxed muscle and the muscle under maximum tension is 3.3×10⁴ N/m² and 6.6×10⁵ N/m² respectively.
Young's Modulus:
Assuming the biceps muscle as a uniform cylinder with an initial length of L = 0.2 m and cross-sectional area of A = 50cm² = 0.05m²
(i) For the relaxed muscle:
Force required for elongation of ΔL = 0.03m is, F = 25 N
Young's Modulus (Y) = stress/strain
Now, stress = F/A,
and strain = ΔL/L
thus,
Y = (F/A) / (ΔL/L)
Y = FL/AΔL
Y = (25×0.2)/(0.05×0.03)
Y = 3.3×10⁴ N/m²
(ii)(i) For the muscle under maximum tension:
Force required for elongation of ΔL = 0.03m is, F = 500 N
Young's Modulus (Y) = stress/strain
Now, stress = F/A,
and strain = ΔL/L
thus,
Y = (F/A) / (ΔL/L)
Y = FL/AΔL
Y = (500×0.2)/(0.05×0.03)
Y = 6.6×10⁵ N/m²
Learn more about Young's Modulus :
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Light of wavelength 575 nm passes through a double-slit and the third order bright fringe is seen at an angle of 6.5^degree away from the central fringe. What is the separation between the double slits? a) 5.0 mu m b) 10 mu m c) 15 mu m d) 20 mu m e) 25 mu m
Answer:
The correct option is C
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 575 *10^{-9} \ m[/tex]
The angle is [tex]\theta = 6.5^o[/tex]
The order of maxima is n = 3
Generally for constructive interference
[tex]dsin \theta = n * \lambda[/tex]
=> [tex]d = \frac{n * \lambda }{ sin \theta }[/tex]
=> [tex]d = \frac{3 * 575 *10^{-9} }{ sin 6.5 }[/tex]
=> [tex]d = 15.24 *10^{-6} \ m[/tex]
=> [tex]d = 15 \mu m[/tex]
Calculate the radius of curvatuire of the concave lens based on the measured focal length.
Answer:
R₁ = (n -1) f
Explanation:
In geometric optics the focal length and the radius of curvature are related, for the case of a lens
1 / f = (n₂-n₀) (1 / R₁ - 1 / R₂)
where f is the focal length, n₂ is the refractive index of the material, n₀ is the refractive index of the medium surrounding the material, R₁ and R₂ are the radius of curvature of each of the material's
In our case, the most common is that the lens is in the air, so n1 = 1, in many cases one of the surfaces is flat, so its radius of curvature R₂ = ∞.
1 / f = (n-1) (1 / R₁)
we look for the radius of curvature R₁
1 / R₁ = 1 / f (n-1)
R₁ = (n -1) f
With this expression we can find the radius of curvature of a concave-plane lens