Answer:
Explanation:
A )
At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy
1/2 m V² = mg x 2r + 1/2 mv²
m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top
1/2 V² = g x 2r + 1/2 v²
V² = g x 4r + v²
V² = 9.8 x 4 + 8²
V² = 103.2
V = 10.16 m/s
B )
If T be the tension at the top
Net downward force
= mg + T . This force provides centripetal force for the circular motion
mg +T = mv² / r
T = mv²/r -mg
= m ( v²/r - g )
= .005 ( 8²/1 -g )
= .005 x 54.2
= .27 N .
C ) At the bottom
Net force = T - mg , T is tension at the bottom , V is velocity at bottom
T-mg = mV²/r
T = m ( V²/r +g )
= .005 ( 10.16²/1 +9.8)
= .005 x 113
= .56 N .
Nowdothesameproblemwiththepivotatthe toes. A Ballet dancer puts all her weight on the toes of one foot. If her mass is 60 kg, what is the force that has to be exerted by her leg muscle to hold that pose? Assume the pivot is at the toes.
Answer:
The force is [tex]F = 2400 \ N[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The mass of the dancer is [tex]m_d = 60 \ kg[/tex]
From the diagram the
The first distance is [tex]l_1 = 20 \ cm[/tex]
The second distance is [tex]l_2 = 5 \ cm[/tex]
At equilibrium the moment about the center of the dancers feet is mathematically represented as
[tex]F * l_2 - (mg* l_1)[/tex]
Where [tex]g= 10 \ m/s^2[/tex]
substituting values
[tex]F * 5 - (60* 9.8 * 20)[/tex]
=> [tex]F = \frac{60 * 10 * 30}{5}[/tex]
=> [tex]F = 2400 \ N[/tex]
A proton moving along the x axis has an initial velocity of 4.0 × 106 m/s and a constant acceleration of 6.0 × 1012 m/s2. What is the velocity of the proton after it has traveled a distance of 80 cm? Group of answer choices
Answer:
5.06*10^6 m/s
Explanation:
Given that
Initial velocity, u = 4*10^6 m/s
Acceleration, a = 6*10^12 m/s²
Distance traveled, s = 80 cm
Final velocity, v = ?
We can find the final velocity by using one of the equations of motion.
v² = u² + 2as
On substituting the values, we have
v² = (4*10^6)² + 2 * 6*10^12 * 0.8
v² = 2.56*10^13
v = √2.56*10^13
v = 5.06*10^6 m/s
Therefore, the final velocity of the proton is adjudged to be 5.06*10^6 m/s
The final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex].
The given parameters;
initial velocity of the proton, u = 4 x 10⁶ m/sacceleration of the proton, a = 6 x 10¹² m/s²distance traveled by the proton, s = 80 cm = 0.8 mThe final velocity of the proton over the given distance is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v^2 = (4\times 10^6)^2 \ + \ 2(6.0 \times 10^{12})(0.8)\\\\v^2 = 2.56 \times 10^{13} \\\\v = \sqrt{2.56 \times 10^{13} } \\\\v = 5.06 \times 10^6 \ m/s[/tex]
Thus, the final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex]
Learn more here:https://brainly.com/question/13613973
What is the resistance of a circuit with a voltage of 10 V in a current of 5 A use almond law to create the resistance
Answer:
2Ω
Explanation:
Ohm's law:
V = IR
10 V = (5 A) R
R = 2 Ω
When a fuel is burned in a cylinder fitted with a piston, the volume expands from an initial value of 0.250 L against an external pressure of 2.00 atm. The expansion does 288 J of work on the surroundings. What is the final volume of the cylinder
Answer:
Vf = 0.0017 m³ = 1.7 L
Explanation:
The work done by the system on the surrounding at constant pressure is given by the following formula:
W = PΔV
W = P(Vf - Vi)
where,
W = Work done = 288 J
P = Constant Pressure = (2 atm)(101325 Pa/atm) = 202650 Pa
Vf = Final Volume f Cylinder = ?
Vi = Initial Volume of Cylinder = (0.25 L)(0.001 m³/ 1 L) = 0.00025 m³
Therefore,
288 J = (202650 Pa)(Vf - 0.00025 m³)
Vf = 288 J/202650 Pa + 0.00025 m³
Vf = 0.0017 m³ = 1.7 L
The friends now feel ready to try a problem. Suppose an Atwood machine has a mass of m1 = 2.5 kg and another mass of m2 = 8.5 kg hanging on opposite sides of the pulley. Assume the pulley is massless and frictionless, and the cord is ideal. Determine the magnitude of the acceleration of the two objects and the tension in the cord.
Answer:
a = 5.34 m/s²
T = 37.86 N
Explanation:
This is the case where two masses are hanging vertically on sides of the pulley. In such case, the formula for acceleration of objects is derived to be:
a = g(m₂ - m₁)/(m₂ + m₁)
where,
a = acceleration of both masses = ?
g = 9.8 m/s²
m₂ = heavier mass = 8.5 kg
m₁ = lighter mass = 2.5 kg
Therefore,
a = (9.8 m/s²)(8.5 kg - 2.5 kg)/(8.5 kg + 2.5 kg)
a = (9.8 m/s²)(6 kg)/(11 kg)
a = 5.34 m/s²
The formula for tension in cable is derived to be:
T = 2m₁m₂g/(m₁ + m₂)
T = (2)(2.5 kg)(8.5 kg)(9.8 m/s²)/(2.5 kg + 8.5 kg)
T = 37.86 N
A low C (f=65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and
its mass density is 5.0 g/m2, determine the tension of the wire.
Answer:
T = 676 N
Explanation:
Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g[tex]/m^{2}[/tex] = 0.005 kg
A stationary wave that is set up in the string has a frequency of;
f = [tex]\frac{1}{2L}[/tex][tex]\sqrt{\frac{T}{M} }[/tex]
⇒ T = 4[tex]L^{2}[/tex][tex]f^{2}[/tex]M
Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.
But M = L × ρ = (2 × 0.005) = 0.01 kg/m
T = 4 × [tex]2^{2}[/tex] ×[tex]65^{2}[/tex] × 0.01
= 4 × 4 ×4225 × 0.01
= 676 N
Tension of the wire is 676 N.
An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (constant) acceleration of the vehicle during this time? Group of answer choices
Answer:
Dear Kaleb
Answer to your query is provided below
Acceleration of the vehicle is 12m/s^2
Explanation:
Explanation for the same is attached in image
A person sits on a freely spinning lab stool that has no friction in its axle. When this person extends her arms, A. her moment of inertia decreases and her angular speed decreases B. her moment of inertia decreases and her angular speed increases C. her moment of inertia increases and her angular speed decreases D. her moment of inertia increases and her angular speed decreases E. her moment of inertia increases and her angular speed remains the same.
Answer:
C. her moment of inertia increases and her angular speed decreases
D. her moment of inertia increases and her angular speed decreases
Explanation:
The moment of inertia of a body is the sum of the products of an increment of mass and the square of its distance from the center of rotation. When a spinning person extends her arms, part of her mass increases its distance from the center of rotation, so increases the moment of inertia.
The kinetic energy of a spinning body is jointly proportional to the moment of inertia and the square of the angular speed. Hence an increase in moment of inertia will result in a decrease in angular speed unless there is a change in the rotational kinetic energy.
This effect is used by figure skaters to increase their spin rate by drawing their arms and legs closer to the axis of rotation. Similarly, they can slow the spin by extending arms and legs.
When the person extends her arms, her moment of inertia increases and her angular speed decreases.
_____
Note to those looking for a letter answer
Both choices C and D have identical (correct) wording the way the problem is presented here. You will need to check carefully the wording in any problem you may think is similar.
To open a door, you apply a force of 10 N on the door knob, directed normal to the plane of the door. The door knob is 0.9 meters from the hinge axis, and the door swings open with an angular acceleration of 5 radians per second squared. What is the moment of inertia of the door
Answer:
The moment of inertia is [tex]I = 1.8 \ kg m^2[/tex]
Explanation:
From the question we are told that
The force applied is [tex]F = 10 \ N[/tex]
The distance of the knob to the hinge is [tex]d = 0.9 \ m[/tex]
The angular acceleration is [tex]a = 5 \ rad/s[/tex]
The moment is mathematically represented as
[tex]I = \frac{d Fsin(\theta)}{a}[/tex]
Here [tex]\theta = 90^o[/tex] This is because the force direction is perpendicular to the plane of the door
substituting values
[tex]I = \frac{0.9 * 10 * sin (90)}{5}[/tex]
[tex]I = 1.8 \ kg m^2[/tex]
Which nucleus completes the following equation?
Se+?
O A. Ga
B. P
C. 31P
D. CI
Answer:First option
Explanation:
hope it helped
A silver rod having a length of 83.0 cm and a cross-sectional diameter of 2.40 cm is used to conduct heat from a reservoir at a temperature of 540 oC into an otherwise completely thermally insulated chamber that contains 1.43 kg of ice at 0 oC. How much time is required for the ice to melt completely
Answer:
3985 s or 66.42 mins
Explanation:
Given:-
- The length of the rod, L = 83.0 cm
- The cross sectional diameter of rod , d = 2.4 cm
- The temperature of reservoir, Tr = 540°C
- The amount of ice in chamber, m = 1.43 kg
- The temperature of ice, Ti = 0°C
- Thermal conductivity of silver, k = 406 W / m.K
- The latent heat of fusion of water, Lf = 3.33 * 10^5 J / kg
Find:-
How much time is required for the ice to melt completely
Solution:-
- We will first determine the amount of heat ( Q ) required to melt 1.43 kg of ice.
- The heat required would be used as latent heat for which we require the latent heat of fusion of ice ( Lf ). We will employ the first law of thermodynamics assuming no heat is lost from the chamber ( perfectly insulated ):
[tex]Q = m*L_f\\\\Q = ( 1.43 ) * ( 3.33 * 10 ^5 )\\\\Q = 476190 J[/tex]
- The heat is supplied from the hot reservoir at the temperature of 540°C by conduction through the silver rod.
- We will assume that the heat transfer through the silver rod is one dimensional i.e along the length ( L ) of the rod.
- We will employ the ( heat equation ) to determine the rate of heat transfer through the rod as follows:
[tex]\frac{dQ}{dt} = \frac{k.A.dT}{dx}[/tex]
Where,
A: the cross sectional area of the rod
dT: The temperature difference at the two ends of the rod
dx: The differential element along the length of rod ( 1 - D )
t: Time ( s )
- The integrated form of the heat equation is expressed as:
[tex]Q = \frac{k*A*( T_r - T_i)}{L}*t[/tex]
- Plug in the respective parameters in the equation above and solve for time ( t ):
[tex]476190 = \frac{406*\pi*0.024^2 * ( 540 - 0 ) }{0.83*4}*t \\\\t = \frac{476190}{119.49619} \\\\t = 3985 s = 66.42 mins[/tex]
Answer: It would take 66.42 minutes to completely melt the ice
A camera takes a picture that is the correct brightness and the correct zoom level, but the depth-of-focus is too small. One way to increase the depth-of-focus is to increase the f-number. Assuming that we will make changes that have the overall effect to:
1. increase the f-number, and
2. keep the brightness and the zoom level the same, which changes should we make to the aperture diameter and to the shutter time? (keep in mind we're talking about the time the shutter is open; we aren't talking about the shutter speed)
a. Increase the aperture diameter, decrease the shutter time
b. Decrease the aperture diameter, increase the shutter time
c. Increase both the aperture diameter as well as the shutter time
d. Decrease both the aperture diameter as well as the shutter time
What is the speed at which a spaceship shoots up from earth ?
Answer:
Once at a steady cruising speed of about 16,150mph (26,000kph
Explanation:
Some types of spiders build webs that consist of threads made of dry silk coated with a solution of a variety of compounds. This coating leaves the threads, which are used to capture prey, hygroscopic - that is, they attract water from the atmosphere. It has been hypothesized that this aqueous coating makes the threads good electrical conductors. To test the electrical properties of coated thread, researchers placed a 5-mm length of thread between two electrical contacts. The researchers stretched the thread in 1-mm increments to more than twice its original length, and then allowed it to return to its original length, again in 1-mm increments. Some of the resistance measurements are shown.If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area A of the coating compare when the thread is 13 mm long versus the starting length of 5 mm? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread.If the conductivity of the thread results from the aqueous coating only, how does the cross-sectional area of the coating compare when the thread is 13 long versus the starting length of 5 ? Assume that the resistivity of the coating remains constant and the coating is uniform along the thread.A13mm is about 1/10 A5mm.A13mm is about 1/4 A5mm. === correct answer... I figured it out. R = pL/A. L is 2.5 times. Therefore, A must be 1/4 times.A13mm is about 2/5 A5mm.A13mm is the same as A5mm.
Answer:
A13 mm is about 1/4 A5 mm
Explanation:
Find the attachment
Two very large parallel sheets a distance d apart have their centers directly opposite each other. The sheets carry equal but opposite uniform surface charge densities. A point charge that is placed near the middle of the sheets a distance d/2 from each of them feels an electrical force F due to the sheets. If this charge is now moved closer to one of the sheets so that it is a distance d/4 from that sheet, what force will feel
Answer:
the force we will feel is F
Explanation:
According to the Gauss law, electric field due to very large sheet of charge is as follows.
[tex]E = \frac{\sigma}{2 \times \epsilon_{o}}[/tex]
where,
[tex]\sigma[/tex] = charge per unit area
Since, it is given that there are two sheets of equal and opposite charge. Therefore, electric field between the plates will be as follows. [tex]E = \frac{\sigma}{2 \times \epsilon_{o}} + \frac{\sigma}{2 \times \epsilon_{o}}[/tex]
Also, we know that relation between force and electric field is as follows.
F = qE
Hence, force felt by the charge present inside the plates will be as follows.
[tex]F = q \times \frac{\sigma}{2 \times \epsilon_{o}}[/tex]
This depicts that force is not dependent on the distance and the charge is kept from one of the plate. Therefore, force F felt by the charge is same when it is placed at a distance d/2 and at a distance d/4 from one of the plate.
A projectile is launched on the Earth with a certain initial velocity and moves without air resistance. Another projectile is launched with the same initial velocity on the Moon, where the acceleration due to gravity is one-sixth as large. How does the maximum altitude of the projectile on the Moon compare with that of the projectile on the Earth?
With smaller gravitational forces and therefor less vertical acceleration, the projectile launched on the moon ... with the same initial speed and direction ...
-- climbs faster,
-- spends more time climbing,
-- reaches a higher peak,
-- falls slower,
-- spends more time falling, and
-- covers more horizontal distance
than the projectile launched on the Earth.
This is not because of air resistance. It would be true even if there were no air resistance on the Earth. It's entirely a gravity thing.
Which of the followings is true about EMF?
a. an induced emf is caused by a changing magnetic flux.
b. an emf can only be induced in a conducting loop by moving the loop through an area that has a constant magnetic field.
c. an induced emf can be observed by measuring the current that is created.
d. an induced emf and conventional induced current are in opposite directions.
Answer:
a. TRUTH
b. FALSE
c. TRUTH
d. FALSE
Explanation:
The emf (electromagnetic force) is generated in a loop or solenoid by the change in the magnetic flux in a closed conductor path (for example, a wire).
This can be noted in the following formula, which is known as the Lenz's law:
[tex]emf=-N\frac{d\Phi_B}{dt}=-N\frac{d(AB)}{dt}[/tex] (1)
Then, the change, in time, of the area of the conductor, or the change in the magnitude of the magnetic field, the induced emf acquires different values. Furthermore, the loops have a resistance, then, a current can be measured when an emf is induced.
Based on this information you have:
a. an induced emf is caused by a changing magnetic flux. TRUTH
b. an emf can only be induced in a conducting loop by moving the loop through an area that has a constant magnetic field. FALSE
c. an induced emf can be observed by measuring the current that is created. TRUTH
d. an induced emf and conventional induced current are in opposite directions. TRUTH (the minus sing in the equation (1) )
Astronaut Flo wishes to travel to a star 20 light years away and return. Her husband Malcolm, who was the same age as Flo when she departs, stays home (baking cookies). If Flo travels at a constand speed of 80% of the speed of light (except for a short time to turn around), how much younger than Malcolm will Flo be when she returns? How long does Malcolm sit around baking cookies? How far is the distance to Flo?
Answer:
a. about 20 years younger
b. Malcolm sits around for 49.94 years
c. 2.268x[tex]10^{17}[/tex] m
Explanation:
light travels 3x[tex]10^{8}[/tex] m in one seconds
in 20 years that will be 3x[tex]10^{8}[/tex] x 20 x 60 x 60 x 24 x 365 = 1.89x[tex]10^{17}[/tex] m
for the to and fro journey, total distance covered will be 2 x 1.89x[tex]10^{17}[/tex] = 3.78x[tex]10^{17}[/tex] m
Flo's speed = 80% of speed of light = 0.8 x 3x[tex]10^{8}[/tex] = 2.4x[tex]10^{8}[/tex] m/s
time that will pass for Malcolm will be distance/speed = 3.78x[tex]10^{17}[/tex] /2.4x[tex]10^{8}[/tex]
= 1575000000 s = 49.94 years
the relativistic time t' will be
t' = t x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]
t' = 49.94 x [tex]\sqrt{1 - 0.8^{2} }[/tex]
t' = 49.94 x 0.6 = 29.96 years this is the time that has passed for Flo
this means that Flo will be about 20 years younger than Malcolm when she returns
relativistic distance is
d' = d x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]
d' = 3.78x[tex]10^{17}[/tex] x [tex]\sqrt{1 - 0.8^{2} }[/tex]
d' = 3.78x[tex]10^{17}[/tex] x 0.6
d' = 2.268x[tex]10^{17}[/tex] m this is how far it is to Flo
Now consider a different electromagnetic wave, also described by: Ex(z,t) = Eocos(kz - ω t + φ) In this equation, k = 2π/λ is the wavenumber and ω = 2π f is the angular frequency. In this case, though, assume φ = +30o and Eo = 1 kV/m. What is the value of Ex(z,t) when z/λ = 0.25 and ft = 0.125?
Answer:
Explanation:
Ex(z,t) = Eocos(kz - ω t + φ)
k = 2π/λ , ω = 2π f
φ = +30° , E₀ = 10³ V .
z/λ = 0.25 , ft = 0.125
Ex(z,t) = Eocos(2πz/λ - 2πf t + φ)
Putting the values given above
Ex(z,t) = 10³ cos ( 2π / 4 - 2π x .125 + 30⁰ )
= 1000cos (90⁰ - 45+30)
= 1000 cos 75
=258.8 V .
An electron and a positron collide head on, annihilate, and create two 0.804 MeV photons traveling in opposite directions. What was the initial kinetic energy of an electron? What was the initial kinetic energy of a positron?
Answer:
Ke- = Ke+ = 0.294MeV
Explanation:
To fins the kinetic energy of both electron and positron you use the following formula, for the case of annihilation of one electron an positron:
2[tex]E_p=2E_o+K_{e^-}+K_{e^+}[/tex] (1)
Ep: photon energy = 0.804MeV
Eo: rest energy of one electron (and positron) = 0.51MeV
Ke-: kinetic energy of electron
Ke+: kinetic energy of positron
You replace the values of Ep and Eo in the equation (1):
[tex]K_{e^-}+K_{e^+}=2E_p-2E_o=2(0.804MeV-0.51MeV)=0.588MeV[/tex]
Iy you assume both positron and electron have the same speed, then, the kinetic energy of them are equal, and the kinetic energy of each one is:
[tex]K_{e^-}=K_{e^+}=\frac{0.588MeV}{2}=0.294MeV[/tex]
Find the equivalent resistance from the indicated terminal pair of the networks in the attached doc
Answer:
a) R = 2.5 Ω, b) R = 1 Ω, c) R = 2R / 3 Ω
Explanation:
The resistance configuration can be in series or in parallel, for each one the equivalent resistance can be calculated
series, the equivalent resistance is the sum of the resistances
parallel, the inverse of the equivalent resistance is the inverse of the sum of the resistances
let's apply these principles to each case
case a)
equivalent series resistance
R₁ = 1 +4 = 5 ohm
R₂ = 2 +3 = 5 ohn
these two are in parallel
1 / R = 1/5 +1/5
1 / R = 2/5
R = 2.5 Ω
case B
we solve the parallel
1 / R₁ = ½ + ½ = 1
R₁ = 1 Ω
we solve the resistors in series
R₂ = 1 + 1
R₂ = 2 Ω
finally we solve the last parallel
1 / R = ½ +1/2 = 1
R = 1 Ω
case C
we solve house resistance pair in series
R₁ = R + 2R = 3R
we go to the next mesh
R₂ = R + 2R = 3R
R₃ = R + 2R = 3R
last mesh
R₄ = R + R = 2R
now we solve the parallel of this equivalent resistance
1 / R = 1 / R₁ + 1 / R₂ + 1 / R₃ + 1 / R₄
1 / R = 1 / 3R + 1 / 3R + 1 / 3R + 1 / 2R
1 / R = 3 / 3R + 1 / 2R = 1 / R + 1 / 2R
1 / R = 3 / 2R
R = 2R / 3 Ω
When Marcel finds the distance L from the previous part, it turns out to be greater than Lend, the distance from the pivot to the end of the seesaw. Hence, even with Jacques at the very end of the seesaw, the twins Gilles and Jean exert more torque than Jacques does. Marcel now elects to balance the seesaw by pushing sideways on an ornament (shown in red) that is at height h above the pivot. (Figure 3)With what force in the rightward direction, Fx, should Marcel push? If your expression would give a negative result (using actual values) that just means the force should be toward the left.Express your answer in terms of W, Lend, w, L2, L3, and h.
Answer:
Fx = - (1/h)( wL2 + wL3 - wLend )
Explanation:
Assuming The twins Gilles and Jean has a weight ( w ) each
The torque that would balance the equation would be = wL2 + wL3 -------- 1
THEREFORE the ccw torques are = wLend + Fh ----------- 2
hence equation 2 equals equation 1
= wLend + Fh = wL2 + wL3 --------- 3
equation 3 can as well be represented as
F = ( 1/h) ( wL2 + wL3 - wLend )---------- 4
From equation 4 it can be seen that F is on the left hand side therefore the value of Fx is negative
therefore equation 4 is represented as
Fx = - (1/h)( wL2 + wL3 - wLend )
the part of the brain stem called the has been shown to related to arousal in lab animals. when this part is stimulated the animal is awake when it is severed cut the animal goes into coma
Answer:
Its called PSY
Explanation: I so do not know why they named it this way but, hope i helped.
An unstrained horizontal spring has a length of 0.36 m and a spring constant of 320 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.033 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.
Answer:
1.been both -ve charged or both +be charged particles
2. 3.52mC
Explanation:
For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.
Now the Force sustain by the extended string is
F = Ke;
Where K is the force constant of the string, 320 N/m
e is the extension,0.033 m
F = 320 × 0.033 =10.56N
2.But according to columns law of charge;
F = kQ1 Q2
But Q1=Q2{ since the charge are of the same magnitude}.
Hence F = KQ^2
Where K is columns constant =9×10^9F/m
Hence Q=√F/K
Q= √10.56/9×10^9
=3.52×10^-3C
= 3.52mC
Explain why it is necessary to have a high voltage
Answer:
SO THAT
EACH APPLIANCE CAN GET SUFFICIENT POTENTIAL DIFF. TO RUNThe 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 windows conduct heat, calculate the rate of conduction in watts through a 2.82 m2 window that is 0.675 cm thick if the temperatures of the inner and outer surfaces are 5.00°C and −10.0°C, respectively. This rapid rate will not be maintained — the inner surface will cool, and frost may even form. The thermal conductivity of glass is 0.84 J/(s · m · °C).
Answer:
Q = - 5264 W = - 5.26 KW
Here, negative sign indicates the outflow of heat
Explanation:
Fourier's Law of heat conduction, gives the following formula:
Q = - KAΔT/t
where,
Q = Rate of Heat Conduction out of window = ?
K = Thermal Conductivity of Glass = 0.84 W/m.°C
A =Surface Area of window = 2.82 m²
ΔT = Difference in Temperature of both sides of surface
ΔT = Inner Surface Temperature - Outer Surface Temperature= 5°C - (- 10°C)
ΔT = 15°C
t = thickness of window = 0.675 cm = 0.00675 m
Therefore,
Q = - (0.84 W/m.°C)(2.82 m²)(15°C)/0.00675 m
Q = - 5264 W = - 5.26 KW
Here, negative sign indicates the outflow of heat.
Thana reminds Alston that because the electric field is uniform, a constant electric force is exerted on the electron. Alston recognizes that, in this case, they can use the kinematic equations to describe the motion of the charged particle while it is inside the region containing the electric field. He asks Thana to write down an equation they can use to calculate the acceleration of the particle while it is inside the region containing a uniform electric field. Which of These equations is correct?
Answer:
a = - e E / m
a = - 1,758 10¹¹ E
Explanation:
For this exercise we can use Newton's second law
F = m a
where the force is electric
the forces given by the product of the charge by the electric field
F = q E
in this case tell us that the charge is the charge of the electron
q = -e = - 1.6 10⁻¹⁹ C
we substitute
- e E = m a
a = - e E / m
we calculate
a = - 1.6 10⁻¹⁹ / 9.1 10⁻³¹ E
a = - 1,758 10¹¹ E
The negative sign indicates that the acceleration is in the opposite direction to the electric field
Jason takes off from rest across level water on his jet-powered skis. The combined mass of Jason and his skis is 75 kg (the mass of the fuel is negligible). The skis have a thrust of 200 N and a coefficient of kinetic friction on water of 0.10. Unfortunately, the skis run out of fuel after only 75 s. What is Jason's top speed?
Answer:
v = 126 m / s
Explanation:
Let's analyze this exercise a little, they give us the thrust that is the applied force and the time that it lasts, and they ask us for the final speed, so we can use the Impulse ratio and the variation of the amount of movement
I = F t = Dp
F t = pf -p₀
Now let's use Newton's second law to find the net thrust
F = E - fr
the friction force has the formula
fr = μ N
let's write Newton's second law on the y-axis
N-W = 0
N = W
we substitute
fr = μ mg
we look for the net out
F = 200 - μ mg
With the skater starting from rest, the initial speed is zero (vo = 0)
we substitute
(200 - very m g) t = m v
v = (200 µm - very g) t
let's calculate
v = (200/75 - 0.10 9.8) 75
v = 126 m / s
A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 45.0kg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.700 kg, traveling perpendicular to the door at 12.0m/s just before impact
A) Find the final angular speed of the door.
answer in rad/s
B) Does the mud make a significant contribution to the moment of inertia?
Yes or No
Answer:
0.19rad/s and Yes
Explanation:
From the principle of conservation of momentum it means momentum before and after collision is the same.
Momentum before collision is 0.700 kg×12 = 8.4Ns
Momentum of the door = mass of door × velocity of door
8.4Ns = mass of door × velocity of door
Velocity of door = 8.4Ns/45 =0.19m/s
But velocity V= w×r ;
w-angular velocity
r- raduis = width
w= 0.19/1m = 0.19rad/s
2. Yes it did because it resisted The moment of inertia and ensued the locking of the door.
Penny is adjusting the position of a stand up piano of mass mp = 150 kg in her living room. The piano is lp = 1.35 m in length. The piano is currently at an angle of θp = 36 degrees to the wall. Penny wants to rotate the piano across the carpeted floor so that it is flat up against the wall. To move the piano, Penny pushes on it at the point furthest from the wall. This piano does not have wheels, so you can assume that the friction between the piano and the rug acts at the center of mass of the piano.
Required:
a. Write an expression for the minimum magnitude of the force FS in N Penny needs to exert on the piano to get it moving. Assume the corner of the piano on the wall doesn't slide and the static friction between the rug and the piano is µs.
b. The coefficient of kinetic friction between the carpet and the piano is uk = 0.27. Once the piano starts moving, calculate the torque τ in N·m that Penny needs to apply to keep moving the piano at a constant angular velocity.
c. Calculate the amount of work Wp, in J Penny does on the piano as she rotates it.
Answer:
a) Fs = (μs*mp*g)/2 | b) τ = Fs*lp | c) Wτ,constant = τΘ
Explanation:
a) Fs = (μs*mp*g)/2
b) τ = Fs*lp
c) Wτ,constant = τΘ
