Answer:
1) speed of a cyclist = 100 m/min.
2) speed of a car moving = 3 km/min.
Explanation:
1) speed of a cyclist = distance over time
= 2000 m / 20 min
= 100 m/min.
2) speed of car moving = distance over time
= 60 km / 20 min
= 3 km/min.
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
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J. Henry Alston was the first African American to publish his research findings on the perception of heat and cold in a major US psychology journal. Please select the best answer from the choices provided T F
Answer:
True
Explanation:
J. Henry Alston was known as a famous African American psychologist. He was known through his detailed study of the sensations of heat and cold which is a necessity for all humans.
He however became the first African American to publish his research findings on the perception of heat and cold in a major US psychology journal which gave him recognition in his field.
Answer:
True YOUR WELCOME
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 photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 1.5 m away. He then wants to take a picture of an object that is 30 cm away.
How far must the lens move to focus on this second object?
Answer:
The distance is [tex]z = 0.008 \ m[/tex]
Explanation:
From the question we are told that
The focal length is [tex]f = 50 \ mm = 50*10^{-3} \ m[/tex]
Generally the lens equation is mathematically represented as
[tex]\frac{1}{u} + \frac{1}{v} = \frac{1}{f}[/tex]
At image distance u = 1.5 m
[tex]\frac{1}{1.5} + \frac{1}{v} = \frac{1}{50 *10^{-3}}[/tex]
=> [tex]\frac{1}{50 *10^{-3}} - \frac{1}{1.5} = \frac{1}{v}[/tex]
=>[tex]v = 0.052 \ m[/tex]
At image distance [tex]u = 30\ cm = 0.30 \ m[/tex]
[tex]\frac{1}{0.3} + \frac{1}{v_1} = \frac{1}{50 *10^{-3}}[/tex]
=> [tex]\frac{1}{50 *10^{-3}} - \frac{1}{0.30 } = \frac{1}{v_1}[/tex]
=> [tex]v_1 = 0.06 \ m[/tex]
The distance the lens need to move is evaluate as
[tex]z = |v - v_1|[/tex]
[tex]z = |0.052 - 0.06|[/tex]
[tex]z = 0.009 \ m[/tex]
The coefficient of static friction between a 3.00 kg crate and the 35.0o incline is 0.300. What minimum force F must be applied perpendicularly to the incline to prevent the crate from sliding down
Answer:
So the minimum force is
32.2Newton
Explanation:
To solve for the minimum force, let us assume it to be F (N)
So
F=mgsinA
But
=>>>> coefficient of static friction x (F + mgcosA
=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)
So making F subject of formula
F + 24.0 = 56.2
F = 32.2N
It is found that an engine rejects 100.0 J while absorbing 125.0 J each cycle of operation.
(a) What is the efficiency of the engine?
(b) How much work does it perform per cycle?
A satellite travels around the earth at 37,000 km/hr. How far will it travel after 13 hours?
Explanation:
Distance = speed × time
d = (37,000 km/hr) (13 hr)
d = 481,000 km
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.
: 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:
The ability of sculptural material to resist forces of pressure, like gravity, is called its __________.
Remove the wood block from the water. Place the brick block into the water and let it sink. How much water does the brick block displace? Hint: This is the liquid level with the block minus the liquid level before the block is added.
Answer:
the volume desalinated that is placed is equal to the volume of the block.
Explanation:
For this exercise we use the Archimedean principle which states that the thrust is equal to the weight of the desalted liquid.
B = ρ g V
When we have the block of wood the volume of water desalinated in the volume of the block under water, this is because the wood floats in the water
When placing the block of clay (brick), it sinks, so the volume desalinated that is placed is equal to the volume of the block.
V_body = V_waer = l to h
One train is at the 20 Km position, another is at the 200Km position, both are approaching, the first with a speed of 30Kmh-1 and the second with a speed of 48 Kmh-1. If they leave at 7:12, what time do they cross? Answer: t = 2.30 H Time: 9:30 am.
Explanation:
The distance between them is 200 km − 20 km = 180 km.
The relative velocity is 30 km/h − (-48 km/h) = 78 km/h.
The time it takes is 180 km / (78 km/h) = 2.31 hours, or 2 hrs 18 min.
Therefore, the trains meet at 9:30 AM.
A source vibrating at constant frequency generates a sinusoidal wave on a string under constant tension. If the power delivered to the string is tripled, by what factor does the amplitude change
Answer:
n = 1,732 the amplitude must be increased by a factor of 1,732
Explanation:
The power delivered by a wave is given by
P = E / t
P = ½ μ w² v A²
let's apply this expression to our case the power tripled
3P₀ = ½ μ w² v A’²
let's write the amplitude function of a initial amplitude
A ’= n A₀
where n is a number
3 P₀ = (½ μy w² v A₀²) n²
3P₀ = P₀ n²
n = √ 3
n = 1,732
therefore the amplitude must be increased by a factor of 1,732
What is process in which long thin strips of flexible ribbon are produced by pouring onto a flat surface?
Answer:
Tape casting
Explanation:
Tape casting is also known as knife coating. This process is used specially in the production of thin surfaces of a ceramic work.
Tape casting is also the process which which involves long thin strips of flexible ribbon being produced by pouring onto a flat surface. It is then left to dry by applying a high temperature and then ready for use.
.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.
What is the wavelength of the electromagnetic radiation needed to eject electrons from a metal?
Answer:
λ = hc/(eV + h[tex]f_{0}[/tex])
Explanation:
Let the work function of the metal = ∅
the kinetic energy with which the electrons are ejected = E
the energy of the incident electromagnetic wave = hf
Then, we know that the kinetic energy of the emitted electron will be
E = hf - ∅
because the energy of the incident electromagnetic radiation must exceed the work function for electrons to be ejected.
This means that the energy of the incident e-m wave can be written as
hf = E + ∅
also, we know that the kinetic energy of the emitted electron E = eV
and the work function ∅ = h[tex]f_{0}[/tex]
we can they combine all equations to give
hf = eV + h[tex]f_{0}[/tex]
we know that f = c/λ
substituting, we have
hc/λ = eV + h[tex]f_{0}[/tex]
λ = hc/(eV + h[tex]f_{0}[/tex]) This is the wavelength of the e-m radiation needed to eject electrons from a metal.
where
λ is the wavelength of the e-m radiation
h is the Planck's constant = 6.63 x 10^-34 m^2 kg/s
c is the speed of e-m radiations in a vacuum = 3 x 10^8 m/s
e is the charge on an electron
V is the voltage potential on the electron
[tex]f_{0}[/tex] is the threshold frequency of the metal
What is the wavelength of the light entering an interferometer if 402 bright fringes are counted when the movable mirror moves 0.115 mm?
Answer:
The value is [tex]\lambda = 572 *10^{-9} \ m[/tex]
Explanation:
From the question we are told that
The number of bright fringes is n = 402
The displacement of the mirror is [tex]\delta l = 0.115 \ mm = 1.15 *10^{-4} \ m[/tex]
Generally the number of fringes is mathematically represented as
[tex]n = \frac{2 * \delta l }{\lambda }[/tex]
=> [tex]\lambda = \frac{2 * \delta l }{n }[/tex]
=> [tex]\lambda = \frac{2 * 1.15*10^{-4} }{402 }[/tex]
=> [tex]\lambda = 572 *10^{-9} \ m[/tex]
The measured value of the latent heat of vaporization of helium (at 1 atm) is
84 J/mol. Use this to estimate the binding energy of helium atoms in the liquid.
a. 2.6 x 10-6 eV
b. 3.6 x 10-4 eV
c. 5.1 x 10-4 eV
d. 8.6 x 10-3 eV
e. 3.6 x 10-2 eV
Answer:
d. 8.6 x 10-3 eV
Explanation:
Binding energy is energy that binds two atoms together . Energy is required to separate them . In the process of evaporation , energy is required to separate each molecules/ atoms of liquid so that they are separated and then get evaporated . That is why latent heat of evaporation is required .
1 mole of helium will contain 6.02 x 10²² atoms of it .
binding energy per atom = 84 / 6.02 x 10²² J
= 13.953 x 10⁻²² J
= 13.953 x 10⁻²² / 1.6 x 10⁻¹⁹ eV .
= 8.6 x 10⁻³ eV
Neutrons travelling at 0.400 m/s are directed through a pair of slits having 1.00 m separation. an array of detectors is placed 10.0 m from the slits.
a) What is the de Broglie wavelength of the neutrons?
b) How far off axis if the first zero-intensity point on the detector array?
c) When a neutron reaches a detector, can we say which slit the neutrons passed through? Explain.
Answer:
9.91*10^-7 m
4.955*10^-6 m
Explanation:
Given that
v = 0.4 m/s
d = 1 m
L = 10 m
h = 6.62*10^-34 Js
m(neutron) = 1.67*10^-27 kg
To find the debroglie wavelength of the neutron, we use the formula
λ = h/mv
Now, we plug in the values we have listed.
λ = 6.62*10^-34 / (1.67*10^-27 * 0.4)
λ = 6.62*10^-34 / 6.68*10^-28
λ = 9.91*10^-7 m
b)
y1 = L (m + ½) λ/d, where m = 0
y1 = L (0 + ½) λ/d
y1 = L (½) λ/d
y1 = L/2 * λ/d or
y1 = Lλ/2d
now, we substitute the values for each of them, we have
y1 = (10 * 9.91*10^-7) / (2 * 1)
y1 = 9.91*10^-6 / 2
y1 = 4.955*10^-6 m
c) no, we can not say the neuron passed through one slit
g stAn 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:
θ = 28.9
Explanation:
For this exercise let's use the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where we use index 1 for air and index 2 for water where the fish is
sin θ₂ = n₁ / n₂ sin θ₁
in this case the air repair index is 1 and the water 1.33
we substitute
sin θ₂ = 1 / 1.33 sin t 40
sin θ = 0.4833
θ = sin⁻¹ 0.4833
θ = 28.9
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.
Water flows through a 0.5 cm diameter pipe connected to a 1 cm diameter pipe. Compared to the speed of the water in the 0.5 cm pipe, the speed in the 1 cm pipe is
A
Answer:
Speed of water in the 0.5cm diameter will be faster because it has a smaller area
Since area x radius ² so if radius is reduced by 0.5 speed is increased by 4times in the 0.5 diameter pipe
Two particles, one with charge −7.97×10−6 C and the other with charge 6.91×10−6 C, are 0.0359 m apart. What is the magnitude of the force that one particle exerts on the other?
Answer:
-384.22N
Explanation:
From Coulomb's law;
F= Kq1q2/r^2
Where;
K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2
q1 and q2 = magnitudes of the both charges
r= distance of separation
F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2
F= -495.65 × 10^-3/ 1.29 × 10^-3
F= -384.22N
When stretched beyond it's elastic limit, a metal rod such as steel
Answer:
When an elastic material is stretched, depending on the stress, it might reach a point beyond which it will no longer return to its original size and shape. This point is called the elastic limit.
When the material (such as a metal rod (e.g steel, copper)) is stretched beyond its elastic limit, Hooke's law is no longer obeyed. i.e strain is no longer directly proportional to stress.
How does area affect the pressure?
The smaller the area the greater the pressure, while the bigger the smaller the pressure. So they are inversely proportional
light of wavelength 610 nm is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central meximum and the first minimum on the other side is 1.23°. What is the width of the slit?
Answer:
The width of the slit is [tex]d = 5.68 *10^{-5} \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 610 \ nm = 610 *10^{-9} \ m[/tex]
The angle is [tex]\theta = 1.23 ^o[/tex]
Generally the angle between the first minimum on one side and that the central maximum is evaluated as
[tex]\theta _1 = \frac{\theta}{2}[/tex]
=> [tex]\theta _1 = \frac{1.23}{2}[/tex]
=> [tex]\theta _1 = 0.615 ^o[/tex]
Generally the condition for minimum diffraction is mathematically represented as
[tex]d sin \theta_1 = n\lambda[/tex]
For first minimum n = 1
=> [tex]d = \frac{n \lambda }{ sin (\theta_1)}[/tex]
=> [tex]d = \frac{1 * 610 *10^{-9}}{ sin (0.615)}[/tex]
=> [tex]d = 5.68 *10^{-5} \ m[/tex]
A charged particle is projected with its initial velocity parallel to a uniform magnetic field. What is the resulting path
Answer:
The particle will continue in a straight line parallel to the field
Explanation:
If a charged particle moves parallel to a magnetic field, without cutting across the field, then there will be no force on the particle. Since there is no force on the charged particle from the magnetic field, its velocity will continue in its original path, which is parallel to the field.
A volleyball is released three times from three different heights: 5 m, 10 m, and 20 m. Which statement is correct about the volleyball?
The volleyball accelerates more quickly toward the ground when released from a higher height.
The volleyball accelerates the same amount toward the ground each time.
The volleyball accelerates more slowly toward the ground when released from a higher height.
The volleyball may accelerate more quickly or slowly depending on its mass.
A student is setting up an experiment with 10 different balls. The student wants to measure which ball hits the ground with the largest force when released from a box 5 feet above the ground. Which quantity should be measured to correctly calculate the force with which each ball hits the ground?
the initial velocity of each ball
the mass of each ball
the rate at which the velocity of each ball increases
the final velocity of each ball right before hitting the ground
Explanation:
1)
The answer is
The volleyball accelerates the same amount toward the ground each time.
Because gravitational acceleration is constant irrespective of altitude(height) or mass of the object.
2)
The answer is
the mass of each ball
because gravitational force is given by
F = mg
since g is a constant, he only needs to measure the mass if each ball
The concepts of free fall and Newton's second law allow to find the correct answers
1) Acceleration is constant
2) The mass of the body
The game of volleyball consists of throwing the balls from the serve to pass it to the other court, this is a two-dimensional movement type projectile launch, in this type of movement the acceleration is the constant on the y axis, it is called acceleration of the gravity and vale (g = 9.8 m / s²)
1) Let's review the different statements of the first part regarding acceleration.
a) The ball accelerates faster as it goes down
False. Acceleration is constant
b) The ball accelerates the same amount
True. Acceleration is constant and directed on the vertical axis
c) The ball accelerates slower or faster
False. Acceleration is constant
The vertical launch of kinematics, establishes that the ball as it descends it goes faster with an acceleration equal to the acceleration of gravity, in this part it is asked to look for the force of the ball when it reaches the ground, for this the second is used Newton law
F = m a
where in this case the acceleration is the acceleration of gravity
a = g
F = m g
Let's review the different claims
a) Initial velocity
False. To calculate the force you only need the mass since acceleration is constant
b) The mass
True. The product of the mass and the acceleration of gravity gives the force with which the ball hits the ground
c) Increasing speed
False. Increasing speed allows calculating acceleration not force
d) The final speed
False. Force does not depend on the speed of the body
In conclusion using the concepts of free fall and Newton's second law we can find the correct answers:
1) Acceleration is constant
2) Body mass
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Maggie completed a 10000-m race at an average speed of 160
m/min. If Tom took 12.5 fewer minutes to complete the race,
what was Tom's average speed?
Answer: 200m/min
Explanation:
Divide 10000m by 160m/min, you will get the answer 62.5. You then subtract 12.5 from 62.5 to understand what you will need your answer for the other person’s speed will be. 10000m divided by 50min is 200m/min.
An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the sign of the value in your answers.)
(a) What is the magnification?
(b) What is the image height?
m
Explanation:
Given that,
Size of object, h = 0.066 m
Object distance from the lens, u = 0.210 m (negative)
Focal length of the converging lens, f = 0.14 m
If v is the image distance from the lens, we can find it using lens formula as follows :
[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m[/tex]
(a) Magnification,
[tex]m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2[/tex]
(b) Magnification, [tex]m=\dfrac{h'}{h}[/tex]
h' is image height
[tex]-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m[/tex]
Hence, this is the required solution.