Answer:
a= 0.063 m
b = 0.116 m
Explanation:
First of all, we need the spring constant in order to solve this problem. You are not giving that data, but I will tell you how to solve this assuming a value of k, In this case, let's assume the value of k is 1600 N/m. (I solved an exercise like this before, using this value).
Now, we need to use the expressions to calculate the distance of the spring.
The elastic potential energy (Uel) is given with the following formula:
Uel = 1/2 kx²
Solving for x:
x = √2*Uel/k
Replacing the data in the above formula (And using the value of k os 1,600):
x = √2 * 3.2 / 1600
x = 0.063 m
b) For this part, we need to apply the work energy theorem which is:
K1 + Ugrav1 + Uel1 + Uo = K2 + Ugrav2 + Uel2
Since in this part, the exercise states that the book is dropped, we can say that the innitial and the end is 0, therefore, K1 = K2 = 0.
The spring at first is not compressed, so Uel1 = 0, and Uo which is the potential energy of other factors, is also 0, because there are no other force or factor here. Therefore, our theorem is resumed like this:
Ugrav1 = Uel2
The potential energy from gravity is given by:
Ug = mgy
And as the spring is placed vertically, we know the height which the book is dropped, so the distance y is:
y = x + h
And this value of x, is the one we need to solve. Replacing this in the theorem we have:
mg(h+x) = 1/2kx²
g would be 9.8 m/s²
Now, replacing the data:
1.2*9.8(0.8 + x) = 1/2*1600x²
Rearranging and solving for x we have:
1.2*9.8*2(0.8 + x) = 1600x²
18.82 + 23.52x = 1600x²
1600x² - 23.52x - 18.82 = 0
Now we need to solve for x, using the general formula:
x = - (-23.52) ± √(-23.52)² - 4 * 1600 * (-18.82) / 2*1600
x = 23.52 ± √553.19 + 120,448 / 3200
x = 23.52 ± 347.85 / 3200
x1 = 23.52 + 347.85 / 3200 = 0.116 m
x2 = 23.52 - 347.85 / 3200 = -0.101 m
Using the positive value, we have that the distance is 0.116 m.
A child bounces in a harness suspended from a door frame by three elastic bands. If each elastic band stretches 0.300 m while supporting a 7.15-kg child at rest, what is the force constant for each elastic band? Assume that each spring supports 1/3 of the child's weight.
Answer:
The force constant for each elastic band is 77.93 N/m
Explanation:
Hooke's law of a spring or an elastic band gives the relation between elastic force (Fe) and stretching (x), the magnitude of that force is:
[tex] F_{e}= kx [/tex] (1)
With k, the elastic force constant. The three elastic bands support the child’s weight (W) and maintain him at rest, so by Newton’s second law for one of the elastic bands:
[tex] \sum F=0 [/tex] (2)
[tex]\frac{W}{3}-F_{e}=0\Rightarrow F_{e}=\frac{W}{3} [/tex] (3)
Using (1) on (3):
[tex] kx=\frac{W}{3}\Rightarrow k=\frac{mg}{3x} [/tex](4)
[tex] k=\frac{7.15\,kg*9.81\,\frac{m}{s^{2}}}{3*0.300\,m}\simeq\mathbf{77.93\frac{N}{m}} [/tex]
The force constant of each elastic band is 77.87 N/m.
To find the force constant for each elastic band, we use Hooke's law
Hooke's lawHooke's law states that the force applied to an elastic material is directly proportional to its extension provided the elastic limit is not exceeded
W = ke............ Equation 1Where:
W = Weight of the child on one elastic bandk = Force constante = extensionMake k the subject of the equation
k = W/e.............. Equation 2From the question,
Given:
W = (7.15×9.8)/3 = 23.36 Ne = 0.3 mSubstitute these values into equation 2
k = 23.36/0.3k = 77.87 N/mHence, The force constant of each elastic band is 77.87 N/m.
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An aluminum cup of 150 cm3 capacity is completely filled with glycerin at 23°C. How much glycerin will spill out of the cup if the temperature of both the cup and glycerin is increased to 41°C? (The linear expansion coefficient of aluminum is 23 × 10-6 1/C°. The coefficient of volume expansion of glycerin is 5.1 × 10-4 1/C°.
Answer:
1.19cm^3 of glycerine
Explanation:
Let Vo= 150cm^3 for both aluminum and glycerine, using expansion formula:
Volume of spill glycerine = change in volume of glycerine - change in volume of aluminum
Volume of glycerine = coefficient of volume expansion of glycerine * Vo* change in temperature - coefficient of volume expansion of Aluminum*Vo* change temperature
coefficient of volume expansion of aluminum = coefficient of linear expansion of aluminum*3 = 23*10^-6 * 3 = 0.69*10^-4 oC^-1
Change in temperature = 41-23 = 18oC
Volume of glycerine that spill = (5.1*10^-4) - (0.69*10^-4) (150*18) = 4.41*10^-4*2700 = 1.19cm3
We define the specific heat of a material as the energy that must be transferred to 1.0 kg of that material in order to cause it to warm 1.0∘C. Part A What happens to the specific heat if we transfer twice that much energy?
Answer:
No change in the specific heat.
Explanation:
Specific heat is an intrinsic property that it has not depend upon on the amount of substance or energy added, it depends upon the material.
So, when twice the amount of energy is transferred, specific heat of the material does not change rather the energy that is twice in the amount to 1 kg of that material cause it to warm 2.0° C.
A rifle, which has a mass of 5.50 kg., is used to fire a bullet, which has a massof m = 65.0 grams., at a "ballistics pendulum". The ballistics pendulum consistsof a block of wood, which has a mass of M = 5.00 kg., attached to two stringswhich are L = 125 cm long. When the block is struck by the bullet the blockswings backward until the angle between the ballistics pendulum and thevertical reaches a maximum angle ofa= 38.0o.a. What will be the maximum gravitational energy contained in the ballisticspendulum when it reaches the maximum angle?b. What was the velocity of the block of wood immediately after being struck bythe bullet?c. What was the velocity of the bullet immediately before it strikes the block ofwood?d. How much work was done by the bullet as it lodged in the block of wood?e. What will be the recoil velocity of the rifle?f. How much energy was released when the bullet was fired?
Answer:
Part a)
[tex]U = 13 J[/tex]
Part b)
[tex]v = 2.28 m/s[/tex]
Part c)
[tex]v = 177.66 m/s[/tex]
Part d)
[tex]W = 1012.7 J[/tex]
Part e)
[tex]v = 2.1 m/s[/tex]
Part f)
[tex]E = 1037.2 J[/tex]
Explanation:
Part a)
As we know that the maximum angle deflected by the pendulum is
[tex]\theta = 38^o[/tex]
so the maximum height reached by the pendulum is given as
[tex]h = L(1 - cos\theta)[/tex]
so we will have
[tex]h = L(1 - cos38)[/tex]
[tex]h = 1.25(1 - cos38)[/tex]
[tex]h = 0.265 m[/tex]
now gravitational potential energy of the pendulum is given as
[tex]U = mgh[/tex]
[tex]U = 5(9.81)(0.265)[/tex]
[tex]U = 13 J[/tex]
Part b)
As we know that there is no energy loss while moving upwards after being stuck
so here we can use mechanical energy conservation law
so we have
[tex]mgh = \frac{1}{2}mv^2[/tex]
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2(9.81)(0.265)}[/tex]
[tex]v = 2.28 m/s[/tex]
Part c)
now by momentum conservation we can say
[tex]mv = (M + m) v_f[/tex]
[tex]0.065 v = (5 + 0.065)2.28[/tex]
[tex]v = 177.66 m/s[/tex]
Part d)
Work done by the bullet is equal to the change in kinetic energy of the system
so we have
[tex]W = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)v_f^2[/tex]
[tex]W = \frac{1}{2}(0.065)(177.66)^2 - \frac{1}{2}(5 + 0.065)2.28^2[/tex]
[tex]W = 1012.7 J[/tex]
Part e)
recoil speed of the gun can be calculated by momentum conservation
so we will have
[tex]0 = mv_1 + Mv_2[/tex]
[tex]0 = 0.065(177.6) + 5.50 v[/tex]
[tex]v = 2.1 m/s[/tex]
Part f)
Total energy released in the process of shooting of gun
[tex]E = \frac{1}{2}Mv^2 + \frac{1}{2}mv_1^2[/tex]
[tex]E = \frac{1}{2}(5.50)(2.1^2) + \frac{1}{2}(0.065)(177.6^2)[/tex]
[tex]E = 1037.2 J[/tex]
An 18.8 kg block is dragged over a rough, horizontal surface by a constant force of 156 N acting at an angle of angle 31.9◦ above the horizontal. The block is displaced by 19.9 m, and the coefficient of kinetic friction is 0.209.
a. Find the work done by the 156 N force. The acceleration of gravity is 9.8 m/s^2
b. Find the magnitude of the work done by the orce of friction
c. What is the sign of the work done by the frictional force?
d. Find the work done by the normal force
Answer:
a) W = 2635.56 J
b) Wf = 423.27 J
c) c) The Sign of the work done by the frictional force (Wf) is negative (-)
d) W=0
Explanation:
Work (W) is defined as the Scalar product of the force (F) by the distance (d) that the body travels due to this force .
The formula for calculate the work is :
W = F*d*cosα
Where:
W : work in Joules (J)
F : force in Newtons (N)
d: displacement in meters (m)
α :angle that form the force (F) and displacement (d)
Known data
m = 18.8 kg : mass of the block
F= 156 N,acting at an angle θ = 31.9◦°: angle above the horizontal
μk= 0.209 : coefficient of kinetic friction between the cart and the surface
g = 9.8 m/s²: acceleration due to gravity
d = 19.9 m : displacement of the block
Forces acting on the block
We define the x-axis in the direction parallel to the movement of the cart on the floor and the y-axis in the direction perpendicular to it.
W: Weight of the cart : In vertical direction downaward
N : Normal force : In vertical direction the upaward
F : Force applied to the block
f : Friction force: In horizontal direction
Calculated of the weight of the block
W= m*g = ( 18.8 kg)*(9.8 m/s²)= 184.24 N
x-y components of the force F
Fx = Fcosθ = 156 N*cos(31.9)° = 132.44 N
Fy = Fsinθ = 156 N*sin(31.9)° = 82.44 n
Calculated of the Normal force
Newton's second law for the block in y direction :
∑Fy = m*ay ay = 0
N-W+Fy= 0
N-184.24+82,44= 0
N = 184.24-82,44
N = 101.8 N
Calculated of the kinetic friction force (fk):
fk = μk*N = (0.209)*( 101.8)
fk = 21.27 N
a) Work done by the F=156N.
W = (Fx) *d *cosα
W = (132.44 )*(19.9)(cos0°) (N*m)
W = 2635.56 J
b) Work done by the force of friction
Wf = (fk) *d *cos(180°)
Wf = (21.27 )*(19.9) (-1) (N*m)
Wf = - 423.27 J
Wf = 423.27 J :magnitude
c) The Sign of the work done by the frictional force is negative (-)
d) Work done by the Normal force
W = (N) *d *cos(90°)
W = (101.8 )*(19.9) (0) (N*m)
W = 0
The work done by the 156 N force is 2630.77 J. The magnitude of the work done by the friction force is 743.14 J. The work done by the normal force is zero.
Explanation:a. The work done by a force is calculated by multiplying the force applied by the displacement in the direction of the force. In this case, the force is applied at an angle of 31.9 degrees above the horizontal, so we need to find the component of the force in the horizontal direction.
Fx = F * cos(θ) = 156 N * cos(31.9°) = 132.3 N
The work done is given by W = Fx * d = 132.3 N * 19.9 m = 2630.77 J
b. The magnitude of the work done by the force of friction can be calculated using the formula:
Wfriction = friction force * displacement = μ * m * g * d, where μ is the coefficient of kinetic friction, m is the mass of the block, g is the acceleration due to gravity, and d is the displacement. Substituting the given values:
Wfriction = 0.209 * 18.8 kg * 9.8 m/s2 * 19.9 m = 743.14 J
c. The work done by the frictional force is negative, indicating that it acts against the motion of the block.
d. The work done by the normal force is zero because the displacement of the block is perpendicular to the direction of the normal force.
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Saturn has a radius of about 9.0 earth radii, and a mass 95 times the Earth’s mass. Estimate the gravitational field on the surface of Saturn compared to that on the Earth. Show your work.
Answer:The gravitational field on Saturn can be calculated by the following formula;
Explanation:
What would be the wavelength of the fundamental and first two overtones produced by an organ pipe of length L that is closed at one end and open at the other?
A) L, ½ L, ¼ L
B) ½ L, ¼ L, 1/6 L
C) 2L, L, ½ L
D) 4L, 2L, L
E) 4L, 4/3 L, 4/5 L
Answer:
option E
Explanation:
given,
organ pipe of length L
using formula ,
[tex]L = (2n - 1)\dfrac{\lambda}{4}[/tex]
n is the number of nodes
[tex]\lambda= \dfrac{4L}{(2n - 1)}[/tex]
now at n = 1
[tex]\lambda_1= \dfrac{4L}{(2(1) - 1)}[/tex]
[tex]\lambda_1= 4L[/tex]
now at n = 2
[tex]\lambda_2= \dfrac{4L}{(2(2) - 1)}[/tex]
[tex]\lambda_2= \dfrac{4}{3}L[/tex]
now at n = 2
[tex]\lambda_3= \dfrac{4L}{(2(3) - 1)}[/tex]
[tex]\lambda_3= \dfrac{4}{5}L[/tex]
The correct answer is option E
Answer:
The correct answer is option E
Explanation:
A ball (mass 0.40 kg) is initially moving to the left at 30 m/s. After hitting the wall, the ball is moving to the right at 20 m/s. What is the impulse of the net force on the ball during its collision with the wall?
Answer:
Impulse, J = 20 m/s
Explanation:
Given that,
Mass of the ball, m = 0.4 kg
Initial speed of the ball, u = -30 m/s (left)
Final speed of the ball after hitting, v = 20 m/s (right)
Let J is the impulse of the net force on the ball during its collision with the wall. The change in momentum of an object is equal to the impulse imparted to it. It is given by :
[tex]J=m(v-u)[/tex]
[tex]J=0.4\ kg(-30-20)\ m/s[/tex]
J = -20 m/s
So, the magnitude of impulse of the net force on the ball during its collision with the wall is 20 m/s.
The impulse on the ball is 20 Ns.
What is impulse?This can be defined as the product of force and time. The impulse of a force acting on a body is also equal to the change in momentum of the body
To calculate the impulse of the net force on the ball during the collision, we use the formula below.
Formula:
I = m(v-u)............ Equation 1Where:
I = Impulse on the ballv = Final velocity of the ballu = Initial velocity of the ballm = mass of the ball.From the question,
Given:
m = 0.40 kgv = 20 m/s u = -30 m/s (to the right)Substitute these values into equation 1
I = 0.40[20-(-30)]I = 0.40(20+30)I = 0.40(50)I = 20 Ns.Hence, the impulse on the ball is 20 Ns.
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HELP!! Two guitarists are tuning their instruments with each other. They both play a specific note at the same time. One guitarist is playing at a frequency 677 Hz. The guitarist's amplifier picks up a beat frequency of 35 Hz. What are the two possible frequencies the second guitarist could be playing in order to produce this beat frequency?
The two possible frequencies could be ; 642 Hz or 712 Hz
Explanation:
For beats; f₁-f₂=fb where f₁=frequency of the first guitar, f₂=frequency of the second guitar, fb =frequency of beat
In this case, the two possibilities are;
f₁-f₂=fb or f₂-f₁=fb
f₁=677 Hz and fb=35 Hz
then;
f₁-x=fb
677-x=35
x=677-35=642 Hz
or
x-f₁=fb
x-677=35
x=35+677 =712 Hz
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Keywords : Frequency, beats, guitarist's, Hz
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Nuclear fusion differs from nuclear fission because nuclear fusion reactions
Answer:
Nuclear fusion reactions produce more energy.( basic difference)
Explanation:
Nuclear fusion reactions are different from nuclear fusion reactions :- In nuclear fission one heavy and unstable nuclear breaks into two parts(smaller) and produce energy( very high energy than normal reaction). whereas in nuclear fusion two small and unstable nuclei fuse and produce very high amount of energy( even higher than nuclear fusion).
Answer:
they are different because nuclear fission is one heavy and unstable nuclear breaks into two parts, or half, and produce very high energy. nuclear fusion two small fuse and produce very high amount of energy
Explanation:
Two 2.0 kg bodies, A and B, collide. The velocities before the collision are
v→A=(15i^+30j^) m/s
and
v→B=(−10i^+5.0j^) m/s
. After the collision,
v→A=(−5.0i^+20j^) m/s
. What are (a) the final velocity of B and (b) the change in the total kinetic energy (including sign)?
Answer:
Explanation:
Given
mass of body A [tex]m_a=2 kg[/tex]
mass of body [tex]m_b=2 kg[/tex]
Velocity before Collision is [tex]u_a=15\hat{i}+30\hat{j}[/tex]
[tex]u_b=-10\hat{i}+5\hat{j} m/s[/tex]
after collision [tex]v_a=-5\hat{i}+20\hat{j} m/s[/tex]
let [tex]v_b_x[/tex] and [tex]v_b_y[/tex] velocity of B after collision in x and y direction
conserving momentum in x direction
[tex]m_a\times 15+m_b\times (-10)=m_a\times (-5)+m_b\times (v_b_x)[/tex]
as [tex]m_a=m_b[/tex] thus
[tex]15-10=-5+v_b_x[/tex]
[tex]v_b_x=10 m/s[/tex]
Conserving momentum in Y direction
[tex]m_a\times 30+m_b\times 5=m_a\times 20+m_b\times (v_b_y)[/tex]
[tex]30+5=20+v_b_y[/tex]
[tex]v_b_y=15 m/s[/tex]
thus velocity of B after collision is
[tex]v_b=10\hat{i}+15\hat{j}[/tex]
(b)Change in total Kinetic Energy
Initial Kinetic Energy of A And B
[tex]K.E._a=\frac{1}{2}\times 2(\sqrt{15^2+30^2})^2=1125 J[/tex]
[tex]K.E._b=\frac{1}{2}\times 2(\sqrt{10^2+5^2})^2=125 J[/tex]
Total initial Kinetic Energy =1250 J
Final Kinetic Energy of A And B
[tex]K.E._a=\frac{1}{2}\times 2(\sqrt{5^2+20^2})^2=425 J[/tex]
[tex]K.E._b=\frac{1}{2}\times 2(\sqrt{10^2+15^2})^2=325 J[/tex]
Final Kinetic Energy [tex]=425+325=750 J[/tex]
Change [tex]\Delta K.E.=1250-750=500 J[/tex]
Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawater where the speed of sound is 1522 m/s. When the dolphin is swimming directly away at 7.2 m/s, the marine biologist measures the number of clicks occurring per second to be at a frequency of 2674 Hz. What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?
Answer:
12.64968 Hz
Explanation:
v = Velocity of sound in seawater = 1522 m/s
u = Velocity of dolphin = 7.2 m/s
f' = Actual frequency = 2674 Hz
From Doppler effect we get the relation
[tex]f=f'\frac{v-u}{v}\\\Rightarrow f=2674\frac{1522-7.2}{1522}\\\Rightarrow f=2661.35032\ Hz[/tex]
The frequency that will be received is 2661.35032 Hz
The difference in the frequency will be
[tex]2674-2661.35032=12.64968\ Hz[/tex]
Final answer:
The difference between the observed frequency and the actual frequency of clicks emitted by the dolphin, calculated using the Doppler Effect, is approximately 12.7 Hz.
Explanation:
The question involves calculating the observed frequency of sound due to the Doppler Effect, which occurs when the source of the sound is moving relative to the observer. In this case, the dolphin is the source of sound waves (clicks), moving away from the marine biologist, who acts as the observer.
The formula for the Doppler Effect for a source moving away from the observer is given by:
f' = f * (v / (v + vs))
Where:
f' is the observed frequency
f is the actual frequency emitted by the source
v is the speed of sound in the medium (1522 m/s in seawater)
vs is the speed of the source relative to the medium (7.2 m/s)
The marine biologist measures the observed frequency (f') as 2674 Hz. We can rearrange the formula to solve for the actual frequency (f):
f = f' * (v + vs) / v
f = 2674 Hz * (1522 m/s + 7.2 m/s) / 1522 m/s
After performing the calculation:
f ≈ 2686.7 Hz
The difference in frequency is then:
Δf = f - f'
Δf ≈ 2686.7 Hz - 2674 Hz
Δf ≈ 12.7 Hz
Therefore, the difference between the observed frequency and the actual frequency of clicks emitted by the dolphin is approximately 12.7 Hz.
hat are the wavelengths of peak intensity and the corresponding spectral regions for radiating objects at (a) normal human body temperature of 37°C, (b) the temperature of the filament in an incandescent lamp, 1500°C, and (c) the temperature of the surface of the sun, 5800 K?
Answer:
(a)
[tex]\lambda _{m}=9.332 \times 10^{-6}m[/tex]
(b)
[tex]\lambda _{m}=1.632 \times 10^{-6}m[/tex]
(c) [tex]\lambda _{m}=4.988 \times 10^{-7}m[/tex]
Explanation:
According to the Wein's displacement law
[tex]\lambda _{m}\times T = b[/tex]
Where, T be the absolute temperature and b is the Wein's displacement constant.
b = 2.898 x 10^-3 m-K
(a) T = 37°C = 37 + 273 = 310 K
[tex]\lambda _{m}=\frac{b}{T}[/tex]
[tex]\lambda _{m}=\frac{2.893\times 10^{-3}}{310}[/tex]
[tex]\lambda _{m}=9.332 \times 10^{-6}m[/tex]
(b) T = 1500°C = 1500 + 273 = 1773 K
[tex]\lambda _{m}=\frac{b}{T}[/tex]
[tex]\lambda _{m}=\frac{2.893\times 10^{-3}}{1773}[/tex]
[tex]\lambda _{m}=1.632 \times 10^{-6}m[/tex]
(c) T = 5800 K
[tex]\lambda _{m}=\frac{b}{T}[/tex]
[tex]\lambda _{m}=\frac{2.893\times 10^{-3}}{5800}[/tex]
[tex]\lambda _{m}=4.988 \times 10^{-7}m[/tex]
A Loyola University student decided to depart from Earth after her graduation to find work on Mars. Before building a shuttle, she conducted careful calculations. A model for the velocity of the shuttle, from liftoff at t = 0 s until the solid rocket boosters were jettisoned at t = 74 s, is given by v(t)= 0.001490833t^3-.08389t^2+23.52t+7.3(in feet per second). Using this model, estimate the global maximum value and global minimum value of the ACCELERATION of the shuttle between liftoff and the jettisoning of the boosters.
Answer: Minimum global acc when you substitute t = 0 in equation 2
Acc = 23.52m/s^2.
Global max acc = 35.595684524m/s^2
Explanation:
(t)= 0.001490833t^3-.08389t^2+23.52t+7.3(in feet per second). ...equation 1.
Equation 1 is the model showing the velocity pathway..
To derive the equation of acceleration.
We differentiate equation 1
DV(t)/Dt = acc. = .004472499t^2 - 0.16778t + 23.52...equ 2
Minimum global acc when you substitute t = 0 in equation 2
Acc = 23.52m/s^2.
Maximum global acc when you substitute the = 74s
Acc = 24.491404524 -12.41572 +23.52
Global max acc = 35.595684524m/s^2
To find the global maximum and minimum values of the acceleration of a rocket, we compute the second derivative of the velocity function, which is the acceleration function. We then evaluate the acceleration function at the given time points and at any critical points within this period.
Explanation:To find the global maximum and minimum values of the acceleration of a rocket, we first need to determine the function for the acceleration, a(t), which is the second derivative of the given velocity function, v(t). The given velocity function v(t) = 0.001490833t^3 - 0.08389t^2 + 23.52t + 7.3. The first derivative, v'(t), gives us the acceleration function and can be computed as v'(t) = 0.004472499t^2 - 0.16778t + 23.52. Computing the second derivative, v''(t), we get a(t) = 0.008944998t - 0.16778.
To find the maximum and minimum values of this acceleration within the given time frame (0 to 74 sec), we evaluate the function at these time points and at any critical points within this interval. The critical points are obtained by solving the equation a'(t) = 0, which yield the value for which the acceleration is maximum or minimum.
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The specific heat capacity of methane gas is 2.20 j/g-k. How many joules of heat are needed to raise the temperature of 5.00 g of methane from 36.0°c to 75.0°c?
Answer:
429 J joules of heat are needed to raise the temperature of 5.00 g
Explanation:
given data
specific heat capacity = 2.20 j/g-k
methane = 5.00 g
temperature range = 36.0°c to 75.0°c
to find out
How many joules of heat are needed to raise the temperature of 5.00 g
solution
we will apply here formula for heat that needed to raise the temperature
Q = m × S × Δt ..............................1
m is mass of methane and S is specific heat capacity and Δt is change in temperature
put here value we get
Q = m × S × Δt
Q = 5 × 2.2 × ( 75 - 36 )
Q = 5 × 2.2 × 39
Q = 429 J
so 429 J joules of heat are needed to raise the temperature of 5.00 g
Blocks I and II, each with a mass of 1.0 kg, are hung from the ceiling of an elevator by ropes 1 and 2.What is the force exerted by rope 1 on block I when the elevator is traveling upward at a constant speed of 2.0 m/s?What is the force exerted by rope 1 on block II when the elevator is stationary?
When a body in the elevator is hung then-
If the elevator is at rest the only its weight is acting on it,
a.If the elevator is moving upward with constant velocity the wight of the body will not change.
b.If the elevator is moving upward with constant acceleration the weight of the body will increase.
c.If the elevator is moving downward with constant velocity the weight of the body will remain the same.
d.If the elevator is moving downward with constant acceleration the weight of the body will decrease.
Answer:
Part a)
[tex]T = 9.8 N[/tex]
Part b)
[tex]T = 9.8 N[/tex]
Part c)
If elevator is accelerating upwards then we have
So tension will be more than the weight
Now if the elevator is moving with uniform velocity
tension will be equal to the weight of the block
Explanation:
Part a)
As we know that rope 1 is connected to block 1 at its lower end
So here we have elevator is moving at constant velocity in upward direction
so we have
[tex]T - mg = 0[/tex]
[tex]T = mg[/tex]
[tex]T = 1\times 9.8[/tex]
[tex]T = 9.8 N[/tex]
Part b)
now for block II we have
[tex]T - mg = ma[/tex]
[tex]T - mg = 0[/tex]
[tex]T = 1 \times 9.8[/tex]
[tex]T = 9.8 N[/tex]
Part c)
If elevator is accelerating upwards then we have
[tex]T - mg = ma[/tex]
[tex]T = mg + ma[/tex]
So tension will be more than the weight
Now if the elevator is moving with uniform velocity
[tex]T - mg = 0[/tex]
[tex]T = mg[/tex]
so tension will be equal to the weight of the block
The gravitational constant G was first measured accurately by Henry Cavendish in 1798. He used an exquisitely sensitive balance to measure the force between two lead spheres whose centers were 0.19 m apart. One of the spheres had a mass of 188 kg, while the mass of the other sphere was 0.93 kg.What was the ratio of the gravitational force between these spheres to the weight of the lighter sphere?
The ratio is found by dividing the gravitational force, determined by the universal law of gravitation, by the weight of the lighter sphere, which is the product of its mass and the acceleration due to gravity.
Explanation:The ratio of the gravitational force between the spheres to the weight of the lighter sphere can be found by dividing the gravitational force by the weight of the lighter sphere. The gravitational force (F) between two objects can be found using the universal law of gravitation: F = G * (m1*m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them. Plugging in the given values, we can find the gravitational force. The weight of the lighter sphere can be found by multiplying its mass by the acceleration due to gravity (9.8 m/s^2). Divide the former by the latter to get the answer.
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The ratio of the gravitational force to weight can be calculated using Newton's law of gravitation and the definition of weight. The gravitational constant G used in the calculations was accurately measured by Henry Cavendish in 1798. This allows the measurement of minute gravitational attractions, pivotal in understanding the strength of gravitation.
Explanation:The ratio of the gravitational force between the two spheres to the weight of the lighter sphere is obtained through the gravitational law and the definition of weight. The universal gravitational constant G was accurately measured by English scientist Henry Cavendish in 1798 using an incredibly sensitive balance. The gravitational force (F) among the two spheres is computed through: F = G * m1 * m2 / r², where m1 and m2 are the masses of the spheres and r is the distance between the centers of the spheres, and G is the gravitational constant (6.67 × 10−11 N·m²/kg²).
Weight (W) is calculated as the mass (m) of an object multiplied by the acceleration due to gravity (g), which is approximately 9.8 m/s² on Earth. Thus, the weight of the lighter sphere is: W = m * g. Therefore, the ratio of the gravitational force to the weight of the lighter sphere is calculated by dividing the gravitational force by the weight of the lighter sphere. The tiny gravitational attraction between ordinary-sized masses measured by Cavendish is significant in the world of physics, as it determines the strength of one of the nature's fundamental forces.
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A cheerleader lifts his 37.4 kg partner straight up off the ground a distance of 0.817 m before releasing her. The acceleration of gravity is 9.8 m/s 2. If he does this 27 times, how much work has he done?Answer in units of J
Answer:
W = 8085.064 J
Explanation:
given,
mass of the cheerleaders partner = 37.4 Kg
distance above which she was lift = 0.817 m
acceleration due to gravity = 9.8 m/s²
number of time she was picked = 27 times
work he done = ?
now,
Work done will be equal to the potential energy into number times she was lifted.
Work done = N m g h
W = 27 x 37.4 x 9.8 x 0.817
W = 8085.064 J
work done by his partner is equal to W = 8085.064 J
A 1.2 kg block of mass is placed on an inclined plane that has a slope of 30° with respect to the horizontal, and the block of mass is released. A friction force Fk = 2.0 N acts on the block as it slides down the incline. Find the acceleration of the block down the incline.
Answer:
a≅3.33
Explanation:
a=(P×sin∝-Fk)÷1.2Answer:
The acceleration of the block down the incline is 3.23 m/s²
Explanation:
Fk = frictional force
m = mass
g = acceleration due to gravity
F1 = force in direction of motion
theeta = inclination angle
For Force on inclined plane
F1 = m*g*sin(theeta)
F1 = (1.2)*(9.8)*sin(30)
F1 = 5.88-N
now,
Fnet = F1 - Fk
Fnet = 5.88 - 2
Fnet = 3.88-N
now for acceleration,
Fnet = m*a
a = Fnet / m
a = 3.88 / 1.2
a = 3.23 m/s²
What is the mass of an object traveling at 20 m/s with a kinetic energy of 4000 J?
Answer:
20 kg
Explanation:
Kinetic energy=½*Mass * velocity²
4000= ½* m*20²
8000=400m
m=8000/400
m=20 kg
Find the network done by friction on a box that moves in a complete circle of radius 1.82 m on a uniform horizontal floor. The coefficient of kinetic friction between the floor and the box is 0.25, and the box weighs 65.0 N. A)-0 B)-370JC) -190 J D-1800 J
Answer:
the network done by friction on a box that moves in a complete circle is 185.7 joules
Explanation:
Step one
Given
Radius of circle =1.82m
Circumference of the circle =2*pi*r
=2*3.142*1.82=11.43
Hence distance =11.43m
Coefficient of friction u=0.25
Weight of box =65N
We know that work =force*distance
But the limiting force =u*weight
Hence the net work done by friction
Wd=0.25*65*11.43
Wd=185.7 joules
The work done by friction on a box moving in a complete circle is zero because the force of friction is always perpendicular to the direction of the box’s displacement.
Explanation:The subject of this question is the work done by friction on a box moving in a circle. When an object moves in a circle, it is the force of friction that prevents the object from sliding off the circular path and helps maintain a curved path. However, it is important to note that friction does no work when an object moves in a complete circle because the force of friction is at every point perpendicular to the direction of the box’s velocity.
To calculate work, we generally use the formula W = F * d * cos(θ), where F is the force (which is the kinetic friction force in this case), d is the displacement of the box (the path covered), and θ is the angle between the force and the direction of displacement. As the box moves in a complete circle, the angle is always 90 degrees. The cosine of 90 degrees is 0, hence the work done by kinetic friction in this case would be zero (0).
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A 24 kg child slides down a 3.3-m-high playground slide. She starts from rest, and her speed at the bottom is 3.0 m/s.a. What energy transfers and transformation occurs during the slide?b.What is the total change in the thermal energy of the slide and the seat of her pants?
Answer:
(a) Potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.
(b) [tex]U=668.16\ J[/tex]
Explanation:
Given:
mass of the child, [tex]m=24\ kg[/tex]height of the slide, [tex]h=3.3\ m[/tex]initial velocity of the child at the slide, [tex]v_i=0 m.s^{-1}[/tex]final velocity of the child at the bottom of slide, [tex]v_f=3\ m.s^{-1}[/tex](a)
∴The initial potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.
Initial potential energy:
[tex]PE=m.g.h[/tex]
[tex]PE=24\times 9.8\times 3.3[/tex]
[tex]PE=776.16\ J[/tex]
Kinetic energy at the bottom of the slide:
[tex]KE=\frac{1}{2} m.v^2[/tex]
[tex]KE= 0.5\times 24\times 3^2[/tex]
[tex]KE= 108\ J[/tex]
(b)
Now, the difference in the potential and kinetic energy is the total change in the thermal energy of the slide and the seat of her pants.
This can be given as:
[tex]U=PE-KE[/tex]
[tex]U=776.16-108[/tex]
[tex]U=668.16\ J[/tex]
Methane gas absorbs red light, and methane clouds reflect blue light, giving Uranus and Neptune their distinctive blue colors. Why do Uranus and Neptune have methane clouds, but Jupiter and Saturn do not
Answer:
Jupiter and Saturn have high temperatures and they have very large gravitational pull compared to Uranus and Neptune.
Explanation:
Jupiter and Saturn have high densities meaning any methane on them is dragged down to their hot surface which prevents methane from ever forming clouds.
Answer:
Jupiter and Saturn have high temperatures and they have very large gravitational pull compared to Uranus and Neptune.
Explanation:
The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly the windows transfer heat by conduction, calculate the rate of conduction in watts through a 3.00-m^2 window that is 0.635 cm thick (1/4 in) if the temperatures of the inner and outer surfaces are 5.00ºC and −10.0ºC, respectively. This rapid rate will not be maintained—the inner surface will cool, and even result in frost formation. (answer in ×10^{3} W)
Answer:
[tex]5.9527559\times 10^3\ W[/tex]
Explanation:
Q = Heat
t = Thickness = d = 0.635 cm
[tex]k_g[/tex] = Heat conduction coefficient of glass = 0.84 W/m °C (general value)
[tex]\Delta T[/tex] = Change in temperature
A = Area = 3 m²
Power is given by
[tex]P=\frac{dQ}{dt}=\frac{kA\Delta T}{d}\\\Rightarrow P=\frac{k_gA\Delta T}{d}\\\Rightarrow P=\frac{0.84\times 3(5-(-10))}{0.00635}\\\Rightarrow P=5.9527559\times 10^3\ W[/tex]
The rate of conduction in watts through the window is [tex]5.9527559\times 10^3\ W[/tex]
The rate of heat conduction out of a window is calculated by using Fourier's law. Substituting the given values into Fourier's law yields a heat transfer rate of around 1.97 × 10⁴ W. This represents a significant amount of heat loss through the windows.
Explanation:The heat conduction rate can be calculated using Fourier's law, which states that the rate of heat transfer per time, or heat flux, is proportional to the temperature gradient. The formula for this law is Q = k*A*(T1-T2)/d, where Q is the heat transfer rate in watts, k is the thermal conductivity, A is the surface area in square meters, T1 and T2 are the temperatures of the two surfaces, and d is the thickness of the material. Given that glass has a thermal conductivity of about 0.84 W/mºC, when we substitute these values into Fourier's law, we obtain:
Q = 0.84 W/mºC * 3.00 m² * (5.00ºC - (-10.0ºC))/0.635 cm, whereby this yields around 1.97 × 10⁴ W.
This indicates that a significant amount of heat is being lost through the windows, contributing to the chilling of the air in the room.
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Calculate the minimum thickness of an oil slick on water that appears blue when illuminated by white light perpendicular to its surface. Take the blue wavelength to be 470 nm and the index of refraction of oil to be 1.40.
Answer:
t = 83.93 nm
Explanation:
given,
blue light wavelength (λ)= 470 nm
refractive index of oil (μ)= 1.40
minimum thickness of an oil slick on water = ?
using constructive interference formula
now,
[tex]2 \mu t = (m + \dfrac{1}{2})\lambda[/tex]
where, t is the thickness of the oil slick
m = 0,1,2
for minimum thickness m = 0
[tex]2\times 1.40\times t = (0 + 0.5)\times 470[/tex]
[tex]2.8\times t = 235[/tex]
[tex]t = \dfrac{235}{2.8}[/tex]
t = 83.93 nm
minimum thickness of an oil slick on water = t = 83.93 nm
The minimum thickness of an oil slick should be 83.93 nm
Important information:Take the blue wavelength to be 470 nm and the index of refraction of oil to be 1.40
Calculation of the minimum thickness:[tex]2\times 1.40 \times t= (0 + 0.5) \times 470\\\\28 \times t = 235[/tex]
t = 83.93 nm
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Most cells cannot harness heat to perform work because
Answer:
temperature is usually uniform throughout a cell or temperature is usually uniform to do work.
Explanation:
Heat (thermal energy) is a kinetic energy. It s connected with the random movement of the atoms or molecules. The temperature is usually uniform throughout a cell, so most of the cells cannot harness heat to perform work.
Cells are unable to harness heat to perform work due to the second law of thermodynamics, which results in energy being lost in a form that is unusable, often as heat, during transfers and transformations. This heat energy is essentially lost to the cell for performing work. And as entropy increases, less energy becomes available for work.
Explanation:Most cells cannot harness heat to perform work because of the second law of thermodynamics. This law states that all energy transfers and transformations are never completely efficient, with some amount of energy being lost in a form that is unusable, commonly as heat energy. Strictly speaking, heat energy is defined as the energy transferred from one system to another that is not doing work.
For example, during cellular metabolic reactions, a portion of the energy is lost as heat energy. Despite it contributing to maintaining the body temperature of warm-blooded creatures, it is essentially lost to the cell for performing work. This demonstrates how the second law of thermodynamics makes the tasks of cells obtaining, transforming, and using energy to do work more difficult than they appear.
Taking into account the entropy, less and less energy in the universe is available to do work as entropy increases. Eventually, as all fuels are exhausted and temperatures equalize, it will become impossible for heat engines to function, or for work to be done. Hence, cells can't harness heat to perform work due to the inefficiencies described by the second law of thermodynamics.
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The early workers in spectroscopy (Fraunhofer with the solar spectrum, Bunsen and Kirchhoff with laboratory spectra) discovered what very significant fact about the spectra produced by hot gases, such as elements heated in a flame?
The hot gases produce their own characteristic pattern of spectral lines, which remain fixed as the temperature increases moderately.
Explanation:A continuous light spectrum emitted by excited atoms of a hot gas with dark spaces in between due to scattered light of specific wavelengths is termed as an atomic spectrum. A hot gas has excited electrons and produces an emission spectrum; the scattered light forming dark bands are called spectral lines.
Fraunhofer closely observed sunlight by expanding the spectrum and a huge number of dark spectral lines were seen. "Robert Bunsen and Gustav Kirchhoff" discovered that when certain chemicals were burnt using a Bunsen burner, atomic spectra with spectral lines were seen. Atomic spectral pattern is thus a unique characteristic of any gas and can be used to independently identify presence of elements.
The spectrum change does not depend greatly on increasing temperatures and hence no significant change is observed in the emitted spectrum with moderate increase in temperature.
A copper-constantan thermocouple generates a voltage of 4.75 x 10-3 volts when the temperature of the hot junction is 110 °C and the reference junction is kept at 0 °C. If the voltage is proportional to the difference in temperature between the junctions, what is the temperature in degrees Celsius of the hot junction when the voltage is 1.76 x 10-3 volts?
Answer:
The temperature of the hot junction is 40.7°C.
Explanation:
Given that,
Voltage [tex]V=4.75\times10^{-3}\ volts[/tex]
Voltage [tex]V'=1.76\times10^{-3}\ volt[/tex]
Temperature of hot junction = 110°C
If the voltage is proportional to the difference in temperature between the junctions,
We need to calculate the temperature of the hot junction
Using formula of temperature
[tex]\dfrac{V}{V'}=\dfrac{\Delta T}{\Delta T}[/tex]
[tex]\dfrac{V}{V'}=\dfrac{T_{2}-T_{1}}{T_{2}-T_{1}}[/tex]
Here,T₁=0°C
[tex]\dfrac{V}{V'}=\dfrac{110}{T_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{4.75\times10^{-3}}{1.76\times10^{-3}}=\dfrac{110}{T_{2}}[/tex]
[tex]T_{2}=\dfrac{110\times1.76\times10^{-3}}{4.75\times10^{-3}}[/tex]
[tex]T_{2}=40.7^{\circ}C[/tex]
Hence, The temperature of the hot junction is 40.7°C.
Final answer:
To determine the temperature of the hot junction in a copper-constantan thermocouple when the voltage is 1.76 x 10-3 volts, we can set up a proportion between the voltages and the temperature difference.
Explanation:
To determine the temperature of the hot junction when the voltage is 1.76 x 10-3 volts, we can use the proportionality between voltage and temperature difference. We know that the voltage generated by the thermocouple is directly proportional to the difference in temperature between the junctions. So, we can set up a proportion:
(4.75 x 10-3 volts) / (110 °C - 0 °C) = (1.76 x 10-3 volts) / (x °C - 0 °C)
By cross-multiplying and solving for x, we can find the temperature of the hot junction. The result is x = 41.5 °C.
A wave pulse travels down a slinky. The mass of the slinky is m = 0.87 kg and is initially stretched to a length L = 6.8 m. The wave pulse has an amplitude of A = 0.23 m and takes t = 0.478 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.48 Hz.
1) What is the speed of the wave pulse? m/s2) What is the tension in the slinky?N3) What is the wavelength of the wave pulse?mPLEASE, CLEARLY WORK IT OUT CORRECTLY..THNKS
Answer:
1. [tex]v=14.2259\ m.s^{-1}[/tex]
2. [tex]F_T=25.8924\ N[/tex]
3. [tex]\lambda=29.6373\ m[/tex]
Explanation:
Given:
mass of slinky, [tex]m=0.87\ kg[/tex]length of slinky, [tex]L=6.8\ m[/tex]amplitude of wave pulse, [tex]A=0.23\ m[/tex]time taken by the wave pulse to travel down the length, [tex]t=0.478\ s[/tex]frequency of wave pulse, [tex]f=0.48\ Hz=0.48\ s^{-1}[/tex]1.
[tex]\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel[/tex]
[tex]v=\frac{6.8}{0.478}[/tex]
[tex]v=14.2259\ m.s^{-1}[/tex]
2.
Now, we find the linear mass density of the slinky.
[tex]\mu=\frac{m}{L}[/tex]
[tex]\mu=\frac{0.87}{6.8}\ kg.m^{-1}[/tex]
We have the relation involving the tension force as:
[tex]v=\sqrt{\frac{F_T}{\mu} }[/tex]
[tex]14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }[/tex]
[tex]202.3774=F_T\times \frac{6.8}{0.87}[/tex]
[tex]F_T=25.8924\ N[/tex]
3.
We have the relation for wavelength as:
[tex]\lambda=\frac{v}{f}[/tex]
[tex]\lambda=\frac{14.2259}{0.48}[/tex]
[tex]\lambda=29.6373\ m[/tex]
A 500 kg horse can provide a steady output power of 750 W (that is, 1 horsepower) when pulling a load; how about a 38 kg sled dog? Data shows that a 38 kg dog can pull a sled that requires a pulling force of 60 N at a steady 2.2 m/s.
1. What is the specific power for the dog and the horse?
2. What is the minimum number of dogs needed to provide the same power as one horse?
Answer:
1.5 W/kg, 3.47 W/kg
6 dogs
Explanation:
[tex]m_d[/tex] = Mass of dog = 38 kg
[tex]v_d[/tex] = Pulling speed of dog = 2.2 m/s
[tex]F_d[/tex] = Force on sled = 60 N
[tex]P_h[/tex] = Power of horse = 750 W
[tex]m_h[/tex] = Mass of horse = 500 kg
Power is given by
[tex]P_d=F_dv_d\\\Rightarrow P_d=60\times 2.2\\\Rightarrow P_d=132\ W[/tex]
Specific power is given by
[tex]P_{sh}=\frac{P_h}{m_h}\\\Rightarrow P_{sh}=\frac{750}{500}\\\Rightarrow P_{sh}=1.5\ W/kg[/tex]
Specific power of horse is 1.5 W/kg
[tex]P_{sd}=\frac{P_d}{m_d}\\\Rightarrow P_{sd}=\frac{132}{38}\\\Rightarrow P_{sh}=3.47\ W/kg[/tex]
Specific power of dog is 3.47 W/kg
We have the relation
[tex]nP_d=P_h\\\Rightarrow n=\frac{P_h}{P_d}\\\Rightarrow n=\frac{750}{132}=5.68\ dogs\approx 6\ dogs[/tex]
The number of dogs is 6
(1) The specific power of the horse is 1.5 W/kg and the specific power of the dog is 3.5 W/kg.
(2) The minimum number of dogs needed to provide the same power as one horse is 6 dogs.
Specific power of the horseThe specific power of the horse is calculated as follows;
Cp = P/m
Cp = 750/500
Cp = 1.5 W/kg
Specific power of the horseThe specific power of the dog is calculated as follows;
Cp = P/m
Cp = (Fv)/m
Cp = (60 x 2.2)/38
Cp = 3.5 W/kg
Minimum number of dogs required to provide same power as horsen(Fv) = 750
n = (750)/(60 x 2.2)
n = 6 dogs
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