Answer:
8.37 Joules
Explanation:
The amount of energy that a substance, such as animal tissue or rubber can take and yet return to its original condition is known as resilience. When we stress such a substance and then let it restore to its previous state, we are referring to the substance to maintain its elastic area of the stress-strain curve.
From the given information:
the resilence is 0.93
The amount of work done during stretching of the tendon = 9.00 J
Thus,
the work done when relaxing = 0.93 × 9.00 J
= 8.37 Joules
In young Goodman’s Brown hawthornes reveals his feelings about his Puritan ancestors when
Answer:
In "Young Goodman Brown," Hawthorne reveals his feelings about his Puritan ancestors when the dark man reveals that he helped Brown's forebears persecute others.
Explanation:
hey mate i know it is right so don't worry this is the correct
Answer:
In "Young Goodman Brown," Hawthorne reveals his feelings about his Puritan ancestors when the dark man reveals that he helped Brown's forebears persecute others. so that means the answer is D.
Explanation:
A. brown strives to resist his dark mission
B. faith expresses her anxieties about young brown's departure
C. brown discovers his catechism teacher is on speaking terms with the devil
D. the dark man reveals that he helped brown's forebears persecute others(correct answer)
How far did you travel in 10 hours if you drove at a constant speed of 5km/hr? *
Answer:
you drove 50km
Explanation:
10×5 hope this helps
Answer:
50 Km
Explanation:
This is how far you have got on your journey if traveling like this.
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Hope this Helps
According to the model, when was the universe at its most dense?
A) During the Dark Ages where matter increased in mass.
B) Just before the Big Bang where all matter existed in a singularity.
C) During the nuclear fusion events, as the atoms become more massive.
D) Current day, as the number of galaxies, solar systems, and planets have increased.
Answer:
The Answer is D
Explanation:
Hope this helps!!!!
a. A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation.
1. the force of the horse pulling on the cart
2. the force of the cart pulling on the horse
3. the force of the horse pushing on the road
4. the force of the road pushing on the horse
b. Suppose that the horse and cart have started from rest; and as time goes on, their speed increases in the same direction. Which one of the following conclusions is correct concerning the magnitudes of the forces mentioned above?
1. Force 1 exceeds Force 2.
2. Force 2 is less than Force 3.
3. Force 2 exceeds Force 4.
4. Force 3 exceeds Force 4.
5. Forces 1 and 2 cannot have equal magnitudes.
Answer:
a) F₁ = F₂, F₃ = F₄, b) the correct answer is 3
Explanation:
a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies
Forces 1 and 2 are action and reaction forces F₁ = F₂
Forces 3 and 4 are action and reaction forces F₃ = F₄
as it indicates that the
b) how the car increases if speed implies that force 1> force3
F₁ > F₃
therefore the correct answer is 3
A girl and her bicycle have a total mass of 40.0 kg. At the top of the hill her speed is 5.0 m/s, and her speed doubles as she rides down the hill. The hill is 10.0 m high and 100 m long. How much kinetic energy and potential energy is lost to friction
Answer:
The kinetic energy and potential energy lost to friction is 2,420 J.
Explanation:
Given;
total mass, m = 40 kg
initial velocity of the girl, Vi = 5 m/s
hight of the hill, h = 10 m
length of the hill, L = 100 m
initial kinetic energy of the girl at the top hill:
[tex]K.E_{i} = \frac{1}{2} mv_i^2 = \frac{1}{2} \times 40 \times (5)^2\\\\K.E_{i} = 500 \ J[/tex]
initial potential energy of the girl at the top hill:
[tex]P.E_{i} = mgh_i = 40 \times 9.8 \times 10\\\\P.E_{i}= 3920 \ J[/tex]
Total energy at the top of the hill:
E = 500 J + 3920 J
E = 4,420 J
At the bottom of the hill:
final velocity = double of the initial velocity = 2 x 5 m/s = 10 m/s
hight of the hill = 0
final kinetic energy of the girl at the bottom of the hill:
[tex]K.E_{f} = \frac{1}{2} mv_f^2 \\\\K.E_f = \frac{1}{2} \times 40 \times (10)^2 = 200 0 \ J[/tex]
final potential energy of the girl at the bottom of the hill:
[tex]P.E_f = mgh_f = 40 \times 9.8 \times 0 = 0[/tex]
Based on the principle of conservation of energy;
the sum of the energy at the top hill = sum of the energy at the bottom hill
The energy at the bottom hill is less due to energy lost to friction.
[tex]E_{friction} \ + E_{bottom}= E_{top}\\\\E_{friction} = E_{top} - E_{bottom}\\\\E_{friction} = 4,420 \ J - 2,000 \ J\\\\E_{friction} = 2,420 \ J[/tex]
Therefore, the kinetic energy and potential energy lost to friction is 2,420 J.
A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m/s in 4.24 s. What is the magnitude of the linear impulse experienced by a 67.0 kg passenger in the car during this time
Answer:
the impulse experienced by the passenger is 630.47 kg
Explanation:
Given;
initial velocity of the car, u = 0
final velocity of the car, v = 9.41 m/s
time of motion of the car, t = 4.24 s
mass of the passenger in the car, m = 67 kg
The impulse experienced by the passenger is calculated as;
J = ΔP = mv - mu = m(v - u)
= 67(9.41 - 0)
= 67 x 9.41
= 630.47 kg
Therefore, the impulse experienced by the passenger is 630.47 kg
A car moving in a straight line uniformly accelerated speed increased from 3 m / s to 9 m / s in 6 seconds. With what acceleration did the car move?
a.
2 m/s2
b.
1 m/s2
c.
0 m/s2
d.
3 m/s2
Answer:
b) 1 m/s
I am sure...........
g Calculate the final speed of a solid cylinder that rolls down a 5.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.
Answer:
[tex]V=8.08m/s[/tex]
Explanation:
From the question we are told that:
Height[tex]h=5.00m[/tex]
Mass [tex]m=0.750kg[/tex]
Radius [tex]r=4.00cm=>0.04m[/tex]
Generally the equation for Total energy is mathematically given by
[tex]mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2[/tex]
Therefore
[tex]V=\sqrt{\frac{4gh}{3}}[/tex]
[tex]V=\sqrt{\frac{4*9.8*5}{3}}[/tex]
[tex]V=8.08m/s[/tex]
why are cows is important?
Answer:
cause they give u milk
Explanation:
Answer:
Cows are important as they provide humans many things for survival. They provide milk, meat, and leather, all of these are important resources.
why the walls of tyres becomes warm as the car moves
Answer:
the particles vibrate inside the tyre
Explanation:
as the car moves kinetic energy is transfered in the tyres which causes the particles to vibrate inside the tyre so the kinetic store is. transferred into thermal
A lens with a focal length of 15 cm is placed 45 cm in front of a lens with a focal length of 5.0 cm .
Required:
How far from the second lens is the final image of an object infinitely far from the first lens?
Answer:
the required distance is 6 cm
Explanation:
Given the data in the question;
f₁ = 15 cm
f₂ = 5.0 cm
d = 45 cm
Now, for first lens object distance s = ∝
1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'
Now, image distance of first lens s' = 15cm
object distance of second lens s₂ will be;
s₂ = 45 - 15 = 30 cm
so
1/f₂ = 1/s₂ + 1/s'₂
1/5 = 1/30 + 1/s'₂
1/s'₂ = 1/5 - 1/30
1/s'₂ = 1 / 6
s'₂ = 6 cm
Hence, the required distance is 6 cm
The distance of the final image from the first lens will be is 6 cm.
What is mirror equation?The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).
The given data in the problem is;
f₁ is the focal length of lens 1= 15 cm
f₂ s the focal length of lens 2= 5.0 cm
d is the distance between the lenses = 45 cm
From the mirror equation;
[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]
If f₁ is the focal length of lens 1 is 15 cm then;
[tex]s'=15 cm[/tex]
f₂ s the focal length of lens 2= 5.0 cm
s₂ = 45 - 15 = 30 cm
From the mirror equation;
[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]
Hence the distance of the final image from the first lens will be is 6 cm.
To learn more about the mirror equation refer to the link;
https://brainly.com/question/3229491
If 1.02 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved through it?
Answer:
Explanation:
one electron has [tex]1.60217662*10^{-19}~coulombs~then\\\\1.02*10^{20}~electrons------->1.02*10^{20}*1.60217662*10^{-19}~coulombs= 16.3422~coulombs[/tex]
The image shows the right-hand rule being used for a current-carrying wire.
An illustration with a right hand with fingers curled and thumb pointed up.
Which statement describes what the hand shows?
When the current flows down the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
When the current flows up the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
When the current flows down the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
Answer:
The answer is (D): When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
Explanation:
1. Convert the following length into meters
a. 123.50mm
b. 560cm
c. 100dm
d. 125.89km
A boy throws a ball straight up with a speed of 21.5 m/s. The ball has a mass of 0.19 kg. How much gravitational potential energy will the ball have at the top of its flight? (Assume there is no air resistance.) A. 43.9 J B. 37.5 J C. 48.5 J D. 41.2 J
Answer:
Explanation:
The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation
[tex]v_f=v_0+at[/tex] where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...
0 = 21.5 + (-9.8)t and
-21.5 = -9.8t so
t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)
Now we will use that time to find out the max height of the object in the equation
Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:
Δx = [tex]21.5(2.19)+\frac{1}{2}(-9.8)(2.19)^2[/tex] which simplifies down a bit to
Δx = 47.1 - 23.5 so
Δx = 23.6 meters.
Now we can plug that in to the PE equation to find the PE of the object:
PE = (.19)(9.8)(23.6) so
PE = 43.9 J
If the loading is 0.4, the coinsurance rate is 0.2, the number of units of medical care is 100, and the number of units of medical care is 1. What is the premium of this insurance?
Answer:
72 is the premimum of the insurance.
Explanation:
Below is the given values:
The loading = 0.4
Coinsurance rate = 0.2
Number of units = 100
Total number of units = 100 * 0.4 = 40
Remaining units = 60 * 0.2 = 12
Add the 60 and 12 values = 60 + 12 = 72
Thus, 72 is the premimum of the insurance.
A scenario where reaction time is important is when driving on the highway. During the delay between seeing an obstacle and reacting to avoid it (or to slam on the brakes!) you are still moving at full highway speed. Calculate how much distance you cover in meters before you start to put your foot on the brakes if you are travelling 65 miles per hour.
Answer:
66.83 meters
Explanation:
After a quick online search, it seems that scientists calculate the average reaction time of individuals as 2.3 seconds between seeing an obstacle and putting their foot on the brakes. Now that we have this reaction time we need to turn the miles/hour into meters/second.
1 mile = 1609.34 meters (multiply these meters by 65)
65 miles = 104,607 meters
1 hour = 3600 seconds
Therefore the car was going 104,607 meters every 3600 seconds. Let's divide these to find the meters per second.
[tex]\frac{104,607}{3600} = \frac{29.0575 meters}{1 second}[/tex]
Now we simply multiply these meters by 2.3 seconds to find out the distance covered before the driver puts his/her foot on the brakes...
29.0575m * 2.3s = 66.83 meters
The efficiency of a machine can be increased by
Explanation:
the efficiency of a machine can be increased by reducing the friction
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What are the messing forces that would make the object be in equilibrium?
Answer:
A) 20 N, B) 20 N, & C) 8 N
Explanation:
For the object to be in equilibrium, the upward forces must be equal to the downward forces and the forward forces must be equal to the backward forces.
1. Determination of A and B.
Forward forces = Backward forces
A + 10 + B = 25 + 25
A + 10 + B = 50
Collect like terms
A + B = 50 – 10
A + B = 40
Assume A and B to be equal. Thus, A is 20 N and B is 20 N.
2. Determination of C
Upward forces = Downward forces
C + 112 = 20 + 100
C + 112 = 120
Collect like terms
C = 120 – 112
C = 8 N
Thus, for the object to be in equilibrium, A must be 20 N, B must be 20 N and C must be 8N.
the unit of area is called a derived unit.why?
Explanation:
the unit of area is called a derived unit because it is made of two fundamental unit metre and metre.
A hoop rolls with constant velocity and without sliding along level ground. Its rotational kinetic energy is:______a- half its translational kinetic energyb- the same as its translational kinetic energyc- twice its translational kinetic energyd- four times its translational kinetic energy
Answer:
The same as its translational KE.
The easy way to do this is to make up numbers and use them.
So, I'll say m=2 and r=3. I will also say v=3 .
Rot. Inertia of a hoop is mr^2. So the rot KE is: 1/2 (mr^2)(w^2)
note: (1/2*I*w^2)
Translational kinetic energy is basically normal KE, so 1/2(m)(v^2)
Now, lets plug our made up values in:
Rot Ke : 1/2 (9*2)(3/3) *note w = v/r
Tran Ke: 1/2(2)(9)
Rot Ke: 9
Tran Ke: 9
9=9, same.
Which of the following elements has the largest atomic radius?
Silicon
Aluminum
Sulfur
Phosphorous
Answer:
francium
Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.
A digital signal differs from an analog signal because it a.consists of a current that changes smoothly. b. consists of a current that changes in pulses. c.carries information. d. is used in electronic devices.
Answer:
d.it is used in electronic devices
Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m
Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.
Explanation:
Given: Mass = 5 kg
Spring constant = 6 N/m
Formula used to calculate period is as follows.
[tex]T = 2 \pi \sqrt\frac{m}{k}[/tex]
where,
T = period
m = mass
k = spring constant
Substitute the values into above formula as follows.
[tex]T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s[/tex]
Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.
what is the frequency of a wave related to
Answer:
Frequency is the number of complete oscillations or cycles or revolutions made in one second.
A ship is flying away from Earth at 0.9c (where c is the speed of light). A missile is fired that moves toward the Earth at a speed of 0.5c relative to the ship. How fast does the missile move relative to the Earth
Answer:
the required speed with which the missile move relative to the Earth is -0.727c
Explanation:
Given the data in the question;
relative velocity relation;
u' = u-v / 1 - [tex]\frac{uv}{c^2}[/tex]
so let V[tex]_B[/tex] represent the velocity as seen by an external reference frame; u=V[tex]_B[/tex]
and let V[tex]_A[/tex] represent the speed of the secondary reference frame; v=V[tex]_A[/tex]
hence, u' is the speed of B as seen by A
so
u' = V[tex]_B[/tex]-V[tex]_A[/tex] / 1 - [tex]\frac{V_BV_A}{c^2}[/tex]
now, given that; V[tex]_A[/tex] = 0.9c and V[tex]_B[/tex] = 0.5c
we substitute
u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{(0.5c)(0.9c)}{c^2}[/tex]
u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{c^2(0.5)(0.9)}{c^2}[/tex]
u' = ( 0.5c - 0.9c ) / 1 - (0.5 × 0.9)
u' = ( -0.4c ) / 1 - 0.45
u' = -0.4c / 0.55
u' = -0.727c
Therefore, the required speed with which the missile move relative to the Earth is -0.727c
A string that is under 50.0N of tension has linear density 5.0g/m. A sinusoidal wave with amplitude 3.0cm and wavelength 2.0m travels along the string. What is the maximum speed of a particle on the string
Answer:
9.42 m/s
Explanation:
Applying,
V' = Aω.............. Equation 1
Where V' = maximum speed of the string, A = Amplitude of the wave, ω = angular velocity.
But,
ω = 2πf................. Equation 2
Where f = frequency, π = pie
And,
f = v/λ................ Equation 3
Where, λ = wave length, v = velocity
Also,
v = √(T/μ)................. Equation 4
Where T = Tension, μ = linear density.
From the question,
Given: T = 50.0 N, μ = 5.0 g/m = 0.005 kg/m
Substitute into equation 4
v = √(50/0.005)
v = √(10000)
v = 100 m/s
Also Given: λ = 2.0 m
Substitute into equation 3
f = 100/2
f = 50 Hz.
Substitute the value of f into equation 2
Where π = constant = 3.14
ω = 2(3.14)(50)
ω = 314 rad/s
Finally,
Given: A = 3.0 cm = 0.03 m
Substitute into equation 1
V' = 0.03(314)
V' = 9.42 m/s
Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of the earth. The second satellite travels at aspeed that is less than v.At what distance from the center of the earth does the secondsatellite orbit?At a distance that is less than r.At a distance equal to r.At a distance greater than r.Now assume that a satellite of mass m is orbiting the earth at a distance r from the center of the earth with speed v_e. An identical satellite is orbiting the moon at thesame distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make onerevolution compare to the time T_e it takes the satellite orbiting the earth to make onerevolution?T_m is less than T_e.T_m is equal to T_e.T_m is greater than T_e.
Answer:
a. At a distance greater than r
b. T_m is greater than T_e.
Explanation:
a. Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed vat a distance r from the center of the earth. The second satellite travels at a speed that is less than v. At what distance from the center of the earth does the second satellite orbit?
Since the centripetal force on any satellite, F equals the gravitational force F' at r,
and F = mv²/r and F' = GMm/r² where m = mass of satellite, v = speed of satellite, G = universal gravitational constant, M = mass of earth and r = distance of satellite from center of earth.
Now, F = F'
mv²/r = GMm/r²
v² = GM/r
v = √GM/r
Since G and M are constant,
v ∝ 1/√r
So, if the speed decreases, the radius of the orbit increases.
Since the second satellite travels at a speed less than v, its radius, r increases since v ∝ 1/√r.
So, the distance the second satellite orbits is at a distance greater than r
b. An identical satellite is orbiting the moon at the same distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make one revolution compare to the time T_e it takes the satellite orbiting the earth to make one revolution?
Since the speed of the satellite, v = √GM/r where M = mass of planet
Since the satellite is orbiting at the same distance, r is constant
So, v ∝ √M
Since mass of earth M' is greater than mass of moon, M", the speed of satellite circling moon, v_m is less than v the speed of satellite circling earth at the same distance, r
Now, period T = 2πr/v where r = radius of orbit and v = speed of satellite
Since r is constant for both orbits, T ∝ 1/v
Now, since the speed of the speed of the satellite on earth orbit v is greater than the speed of the satellite orbiting the moon, v_m, and T ∝ 1/v, it implies that the period of the satellite orbiting the earth, T_e is less than the period of the satellite orbiting the moon, T_m since there is an inverse relationship between T and v. T_e is less T_m implies T_m is greater than T_e
So, T_m is greater than T_e.
What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?
Answer:
1.2 x 10^-12
Explanation:
3/2.5 x 10^-6/10^6
1.2 x 10^-6 x 10^-6
1.2 x 10^-12
The latent heat of vaporization of water is roughly 10 times the latent heat of fusion of water. The amount of heat required to boil away 1 kg of water is __________ the amount of heat required to melt 1 kg of ice.
Answer:
The amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice
Explanation:
let the latent heat of fusion of ice = L
then, the latent heat of vaporization of water = 10L
The heat of fusion of 1 kg of ice = 1 x L = L
The heat of vaporization 1 kg of water = 1 x 10L = 10L
Therefore, the amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice
