Answer:
According to Newton's 2nd law
The force acting on a body produces acceleration in its direction which is directly propotional to the force but inversly propotinal to the mass of tbe body.
Explanation:
a = F/m
F = ma
Where( F) is force (m) is mass and (a) is acceleration.
A girl and her bicycle have a total mass of 40 kg. At the top of the hill her speed is 5.0 m/s. The hill is 10 m high and 100 m long. If the force of friction as she rides down the hill is 20 N, what is her speed at the bottom
Answer:
v = 11 m/s is her final speed
Explanation:
work done by gravity = m g Δh = 40×9.8×10 = 3920 Joules
Work done by friction = - force×distance = - 20×100 = - 2000 Joules
[minus sign because friction force is opposite to the direction of motion]
Initial K.E. = (1/2) m u^2 = (1/2) × 40 × 5^2 = 500 Joules
Now, by work energy theorem
Work done = change in kinetic energy.
Final K.E. = initial K.E. + total work = 500 + 3920 - 2000 = 2420 Joules
Now, Final K.E. = (1/2) m v^2 [final speed being v= speed at the bottom]
⇒ 2420 = (1/2)×40×v^2
⇒ 121 = v^ 2
v = 11 m/s is her final speed
Holding force constant, what will be the effect of increasing the Moment arm?
(a) Depends on the direction of the force.
(b) Torque will increase.
(c) Torque is constant.
(d) Torque will decrease.
(e) The direction of rotation will change.
Answer:
(b) Torque will increase.
Explanation:
Torque is given as the product of force and moment arm (radius).
τ = F x r
F = τ / r
where;
F is force
τ is torque
r is radius (moment arm)
Keeping force constant, we will have the following;
τ ∝ r
This shows that torque is directly proportional moment arm (radius), thus increase in moment arm, will cause increase in torque.
For instance;
let the constant force = 5 N
let the initial moment arm, r = 2m
Torque, τ = 5 N x 2m = 10 Nm
When the moment arm is increased to 4 m
Torque, τ = 5 N x 4m = 20 Nm
Therefore, at a constant force, increasing in the Moment arm, will cause increase in torque.
Coorect option is "(b) Torque will increase."
disadvantage of vb language
Answer:
visual basics
Explanation:
not suited for programming, slower than the other languages. hard to translate to other operating systems
A 20-kg object sitting at rest is struck elastically in a head-on collision with a 10-kg object initially moving at 3.0 m/s. Find the final velocity of the 10-kg object after the collision.
Answer:1m/s
Explanation: As the stationary ball is hit by the moving ball ,the two moves together after collision, with a single velocity. The attached photo further explains how the answer is calculated
When stationary ball is hit by the moving ball, both the balls moves together after collision. The final velocity of the object after collision is 1 m/s.
When stationary ball is hit by the moving ball, both the balls moves together after collision
The conservation of momentum,
[tex]\bold {m_1 u_1 + m_2u_2 = (m_1+m_2) V}\\[/tex]
Where,
m1 - initial mass = 20 kg
m2 - final mass =10 kg
u1 - initial velocity = 0 m/s (object at rest)
u2 - final velocity = 3 m/s
V- velocity after collision = ?
Put the values int he formula and calculate for V2,
[tex]\bold { 10 \times 0 + 20 \times 3 = (10+20) V}\\\\\bold {V = \dfrac {30}{30}}\\\\\bold {V = 1\ m/s}[/tex]
Therefore, final velocity of the object after collision is 1 m/s.
To know more about velocity,
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The energy band gap of GaAs is 1.4ev. calculate the optimum wavelength of the light for photovoltaic generation in a GaAs solar cell
Answer:
The wavelength is [tex]\lambda = 886 \ nm[/tex]
Explanation:
From the question we are told that
The energy band gap is [tex]E = 1.4 eV[/tex]
Generally the energy of a single photon of light emitted for an electron jump in a GaAS solar cell is mathematically represented as
[tex]E = \frac{hc}{\lambda }[/tex]
Where h is the Planck's constant with values
[tex]h = 4.1357 * 10^{-15} eV[/tex]
and c is the speed of light with values [tex]c = 3*10^{8} \ m/s[/tex]
So
[tex]\lambda = \frac{hc}{E}[/tex]
substituting values
[tex]\lambda = \frac{4.1357 *10^{-15} * 3.0 *10^{8}}{1.4}[/tex]
[tex]\lambda = 886 \ nm[/tex]
g A mass of 2 kg is attached to a spring whose constant is 7 N/m. The mass is initially released from a point 4 m above the equilibrium position with a downward velocity of 10 m/s, and the subsequent motion takes place in a medium that imparts a damping force numerically equal to 10 times the instantaneous velocity. What is the differential equation for the mass-spring system.
Answer:
mass 20 times of an amazing and all its motion
A car traveling with velocity v is decelerated by a constant acceleration of magnitude a. It takes a time t to come to rest. If its initial velocity were doubled, the time required to stop would
Answer:
If the initial speed is doubled the time is also doubled
Explanation:
You have that a car with velocity v is decelerated by a constant acceleration a in a time t.
You use the following equation to establish the previous situation:
[tex]v'=v-at[/tex] (1)
v': final speed of the car = 0m/s
v: initial speed of the car
From the equation (1) you solve for t and obtain:
[tex]t=\frac{v-v'}{a}=\frac{v}{a}[/tex] (2)
To find the new time that car takesto stop with the new initial velocity you use again the equation (1), as follow:
[tex]v'=v_1-at'[/tex] (3)
v' = 0m/s
v1: new initial speed = 2v
t': new time
You solve the equation (3) for t':
[tex]0=2v-at'\\\\t'=\frac{2v}{a}=2t[/tex]
If the initial speed is doubled the time is also doubled
A rock falls from a vertical cliff that is 4.0 m tall and experiences no significant air resistance as it falls. At what speed will its gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy
Answer:
About 6.26m/s
Explanation:
[tex]mgh=\dfrac{1}{2}mv^2[/tex]
Divide both sides by mass:
[tex]gh=\dfrac{1}{2}v^2[/tex]
Since the point of equality of kinetic and potential energy will be halfway down the cliff, height will be 4/2=2 meters.
[tex](9.8)(2)=\dfrac{1}{2}v^2 \\\\v^2=39.4 \\\\v\approx 6.26m/s[/tex]
Hope this helps!
The gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.
Given data:
The height of vertical cliff is, h = 4.0 m.
Since, we are asked for speed by giving the condition for gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy. Then we can apply the conservation of energy as,
Kinetic energy = Gravitational potential energy
[tex]\dfrac{1}{2}mv^{2}=mgh[/tex]
Here,
m is the mass of rock.
v is the speed of rock.
g is the gravitational acceleration.
Solving as,
[tex]v=\sqrt{2gh}\\\\v=\sqrt{2 \times 9.8 \times 4.0}\\\\v =8.85 \;\rm m/s[/tex]
Thus, we can conclude that the gravitational potential energy (relative to the base of the cliff) be equal to its kinetic energy for speed of rock of 8.85 m/s.
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A disk of mass m and moment of inertia of I is spinning freely at 6.00 rad/s when a second identical disk, initially not spinning, is dropped onto it so that their axes coincide. In a short time, the two disks are corotating. What is the angular speed of the new system
Answer:
The angular speed of the new system is [tex]3\,\frac{rad}{s}[/tex].
Explanation:
Due to the absence of external forces between both disks, the Principle of Angular Momentum Conservation is observed. Since axes of rotation of each disk coincide with each other, the principle can be simplified into its scalar form. The magnitude of the Angular Momentum is equal to the product of the moment of inertial and angular speed. When both disks begin to rotate, moment of inertia is doubled and angular speed halved. That is:
[tex]I\cdot \omega_{o} = 2\cdot I \cdot \omega_{f}[/tex]
Where:
[tex]I[/tex] - Moment of inertia of a disk, measured in kilogram-square meter.
[tex]\omega_{o}[/tex] - Initial angular speed, measured in radians per second.
[tex]\omega_{f}[/tex] - Final angular speed, measured in radians per second.
This relationship is simplified and final angular speed can be determined in terms of initial angular speed:
[tex]\omega_{f} = \frac{1}{2}\cdot \omega_{o}[/tex]
Given that [tex]\omega_{o} = 6\,\frac{rad}{s}[/tex], the angular speed of the new system is:
[tex]\omega_{f} = \frac{1}{2}\cdot \left(6\,\frac{rad}{s} \right)[/tex]
[tex]\omega_{f} = 3\,\frac{rad}{s}[/tex]
The angular speed of the new system is [tex]3\,\frac{rad}{s}[/tex].
Using only the trainiris dataset, for each feature, perform a simple search to find the cutoff that produces the highest accuracy, predicting virginica if greater than the cutoff and versicolor otherwise. Use the seqfunction over the range of each feature by intervals of 0.1 for this search. Which feature produces the highest accuracy?
A. Sepal. Length
B. Sepal. Width
C. Petal. Length
D. Petal. Width
Answer: C. Petal. Length
Explanation: Petal are unit of Corolla which are usually brightly colored. This part of a plant or flower, helps attracts insects to the plant for pollination. And also provide protection to the reproductive parts of the plant or flower.
Examples of flowers with petals is the Sun Flower, which coincidentally is the flower plant with most petals.
A 50-kg block is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 150 N. Fp is parallel to the displacement of the block. The coefficient of kinetic friction is 0.25.
a) What is the total work done on the block?
b) If the box started from rest, what is the final speed of the block?
Answer:
a) WT = 137.5 J
b) v2 = 2.34 m/s
Explanation:
a) The total work done on the block is given by the following formula:
[tex]W_T=F_pd-F_fd=(F_p-F_f)d[/tex] (1)
Fp: force parallel to the displacement of the block = 150N
Ff: friction force
d: distance = 5.0 m
Then, you first calculate the friction force by using the following relation:
[tex]F_f=\mu_k N=\mu_k Mg[/tex] (2)
μk: coefficient of kinetic friction = 0.25
M: mass of the block = 50kg
g: gravitational constant = 9.8 m/s^2
Next, you replace the equation (2) into the equation (1) and solve for WT:
[tex]W_T=(F_p-\mu_kMg)d=(150N-(0.25)(50kg)(9.8m/s^2))(5.0m)\\\\W_T=137.5J[/tex]
The work done over the block is 137.5 J
b) If the block started from rest, you can use the following equation to calculate the final speed of the block:
[tex]W_T=\Delta K=\frac{1}{2}M(v_2^2-v_1^2)[/tex] (3)
WT: total work = 137.5 J
v2: final speed = ?
v1: initial speed of the block = 0m/s
You solve the equation (3) for v2:
[tex]v_2=\sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(137.5J)}{50kg}}=2.34\frac{m}{s}[/tex]
The final speed of the block is 2.34 m/s
A traveling electromagnetic wave in a vacuum has an electric field amplitude of 62.5 V/m. Calculate the intensity S of this wave. Then, determine the amount of energy ???? that flows through area of 0.0231 m2 over an interval of 14.9 s, assuming that the area is perpendicular to the direction of wave propagation.
Answer:
a) 5.19 W/m²
b) 1.79 J
Explanation:
For the calculation of intensity, I. We have
I = E(rms)² / (cμ), where
c = speed of light
μ = permeability of free space
I = (62.5 / √2)² / [(2.99 x 10^8) (1.26 x 10^-6)]
I = 1954 / 376.74
I = 5.19 W/m²
Therefore, the intensity, I = 5.19 W/m²
t = 14.9 s
A = 0.0231 m²
Amount if energy flowing, U = IAt
U = (5.19) (0.0231) (14.9) J
U = 1.79 J
In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with:
1. yellow light.
2. red light.
3. blue light.
4. green light.
5. The separation is the same for all wavelengths.
Answer:
we see that the lights with the most extreme wavelength are blue and red
we see that the separation between the interference lines (y) increases linearly with the wavelength for which the phenomenon is best observed in the RED response 2
Explanation:
In Young's double-slit experiment, constructive interference is written by the equation
d sin θ = m λ
where you give the gap separation, lam the length of the donda used and m the order of interference
in many he uses trigonometry to express the synth in confusing the distances on a very distant screen
so θ = y / L
in this experiment the angles are generally very small, so
tan θ = sin θ / cos θ = sin θ
sint θ = y / L
let's replace
d y / L = mλ
y = (m L / d) λ
now let's examine the effect of changing the wavelength
1 yellow lam = 600 10⁻⁹ m
2) red lam = 750 10⁻⁹m
3) blue lam = 450 10⁻⁸ nm
4) green lam = 550 10⁻⁹ nm
we see that the lights with the most extreme wavelength are blue and red
we see that the separation between the interference lines (y) increases linearly with the wavelength for which the phenomenon is best observed in the RED response 2
A 3.10-mm-long, 430 kgkg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 69.0 kgkg construction worker stands at the far end of the beam.What is the magnitude of the gravitational torque about the point where the beam is bolted into place?
Answer:
Explanation:
Given that,
The length of the beam L = 3.10m
The mass of the steam beam [tex]m_1[/tex] = 430kg
The mass of worker [tex]m_2[/tex] = 69.0kg
The distance from the fixed point to centre of gravity of beam = [tex]\frac{L}{2}[/tex]
and our length of beam is 3.10m
so the distance from the fixed point to centre of gravity of beam is
[tex]\frac{3.10}{2}=1.55m[/tex]
Then the net torque is
[tex]=-W_sL'-W_wL\\\\=-(W_sL'+W_wL)[/tex]
[tex]W_s[/tex] is the weight of steel rod
[tex]=430\times9.8=4214N[/tex]
[tex]W_w[/tex] is the weight of the worker
[tex]=69\times9.8\\\\=676.2N[/tex]
Torque can now be calculated
[tex]-(4214\times1.55+676.2\times3.9)Nm\\\\-(6531.7+2637.18)Nm\\\\-(9168.88)Nm[/tex]
≅ 9169Nm
Therefore,the magnitude of the torque is 9169Nmwhen their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.2525 N. What were the initial charges on the spheres
Answer:
q1 = 7.6uC , -2.3 uC
q2 = 7.6uC , -2.3 uC
( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )
Explanation:
Solution:-
- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.
- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:
[tex]F = k\frac{|q_1|.|q_2|}{r^2}[/tex]
Where,
k: The coulomb's constant = 8.99*10^9
- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.
- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.
- Therefore, the force of attraction between the spheres would be:
[tex]\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1[/tex] ... Eq 1
- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).
- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,
[tex]q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}[/tex]
- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:
[tex]\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6[/tex] .. Eq2
- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:
[tex]-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123[/tex]
[tex]q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\[/tex]
A bag is gently pushed off the top of a wall at A and swings in a vertical plane at the end of a rope of length l. Determine the angle θ for which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the bag.
Answer:
Dear user,
Answer to your query is provided below
The angle for which the rope will break θ = 41.8°
Explanation:
Explanation of the same is attached in image
A bag is gently pushed off the top of a wall at A and swings in a vertical plane at the end of a rope of length l. The angle θ for which the rope will break, is 41.81°
What is tension?The tension is a kind of force which acts on linear objects when subjected to pull.
The maximum tension Tmax =2W
From the work energy principle,
T₂ = 1/2 mv²
Total energy before and after pushing off
0+mglsinθ = 1/2 mv²
v² = 2gflsinθ..............(1)
From the equilibrium of forces, we have
T= ma +mgsinθ = mv²/l +mgsinθ
2mg = mv²/l +mgsinθ
2g = v²/l +gsinθ
Substitute the value of v² ,we get the expression for θ
θ = sin⁻¹(2/3)
θ =41.81°
Hence, the angle θ for which the rope will break, is 41.81°
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The uniform dresser has a weight of 91 lb and rests on a tile floor for which μs = 0.25. If the man pushes on it in the horizontal direction θ = 0∘, determine the smallest magnitude of force F needed to move the dresser. Also, if the man has a weight of 151 lb , determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip.
Answer:
F = 22.75 lb
μ₁ = 0.15
Explanation:
The smallest force required to move the dresser must be equal to the force of friction between the man and the dresser. Therefore,
F = μR
F = μW
where,
F = Smallest force needed to move dresser = ?
μ = coefficient of static friction = 0.25
W = Weight of dresser = 91 lb
Therefore,
F = (0.25)(91 lb)
F = 22.75 lb
Now, for the coefficient of static friction between shoes and floor, we use the same formula but with the mas of the man:
F = μ₁W₁
where,
μ₁ = coefficient of static friction between shoes and floor
W₁ = Weight of man = 151 lb
Therefore,
22.75 lb = μ₁ (151 lb)
μ₁ = 22.75 lb/151 lb
μ₁ = 0.15
Which one of the following is closely related to the law of conservation of
energy, which states that energy can be transformed in different ways but can
never be created or destroyed?
O A. Charles's Law
B. Boyle's Law
C. Second law of thermodynamics
O D. First law of thermodynamics
Answer:
D
Explanation:
Answer:
It is D
Explanation: No cap
Which circuits are parallel circuits?
Answer:
The bottom two lines.
Explanation:
They need their own line of voltage quantity. A parallel circuit has the definition of 'two or more paths for current to flow through.' The voltage does stay the same in each line.
An object is dropped from a tower, 576576 ft above the ground. The object's height above ground t seconds after the fall is s(t)equals=576 minus 16 t squared576−16t2. Determine the velocity and acceleration of the object the moment it reaches the ground.
Answer: 192 ft/s
Explanation:
The initial height of the object is:
576ft above the ground.
The position equation is:
p(t) = -16*t^2 + 576
in the position equation, we only can see the therm of the initial height and the term of the acceleration (that is equal to the gravitational acceleration g = 32 ft/s^2 over 2)
So we have no initial velocity, this means that at the beginning we only have potential energy:
U = m*g*h
where m is the mass of the object, g = 32m/s^2 and h = 576 ft.
Now, as the object starts to fall down, the potential energy is transformed into kinetic energy, and when the object is about to hit the ground, all the potential energy has become kinetic energy.
The kinetic energy equation is:
K = (m/2)*v^2
where v is the velocity of the object, then the maximum kinetic energy (when the object reaches the ground) is equal to the initial potential energy:
m*g*h = (m/2)*v^2
now we can solve this for v.
v = √(2*g*h) = √(2*32ft/s^2*576ft) = 192 ft/s
A 73 kg swimmer dives horizontally off a 462 kg raft initially at rest. If the diver's speed immediately after leaving the raft is 5.54 m/s, what is the corresponding raft speed
Answer:
Corresponding raft speed = -0.875 m/s (the minus sign indicates that the raft moves in the direction opposite to the diver)
Explanation:
Law of conservation of momentum gives that the momentum of the diver and the raft before the dive is equal to the momentum of the diver and the raft after the dive.
And since the raft and the diver are initially at rest, the momentum of the diver after the dive is equal and opposite to the momentum experienced by the raft after the dive.
(Final momentum of the diver) + (Final momentum of the raft) = 0
Final Momentum of the diver = (mass of the diver) × (diving velocity of the diver)
Mass of the diver = 73 kg
Diving velocity of the diver = 5.54 m/s
Momentum of the diver = 73 × 5.54 = 404.42 kgm/s
Momentum of the raft = (mass of the raft) × (velocity of the raft)
Mass of the raft = 462 kg
Velocity of the raft = v
Momentum of the raft = 462 × v = (462v) kgm/s
404.42 + 462v = 0
462v = -404.42
v = (-404.42/462) = -0.875 m/s (the minus sign indicates that the raft moves in the direction opposite to the diver)
Hope this Helps!!!
Question 9(Multiple Choice Worth 4 points) (05.03 LC) What most likely happens when water vapor cools? It changes into gas. It changes into liquid. Its temperature increases. Its temperature remains constant.
Answer:
it changes into liquid
Answer:
It changes in to liquids
Explanation:
This is because the water vapor cools down and condenses it attaches it self to dust forming water droplets. Those water droplets are water.
An airplane is flying on a bearing of N 400 W at 500 mph. A strong jet-stream speed wind of 100 mph is blowing at S 500 W.
Required:
a. Find the vector representation of the plane and of the wind.
b. Find the resultant vector that represents the actual course of the plane.
c. Give the resulting speed and bearing of the plane.
Answer:
A. a (-321.393, 383.022) b (-76.40, -64.278)
B. (-397.991, 318.744)
C. a. resulting speed 509.9mph b. bearing of the plane = 51.6°
Explanation:
An ideal, or Carnot, heat pump is used to heat a house to a temperature of 294 K (21 oC). How much work must the pump do to deliver 3000 J of heat into the house (a) on a day when the outdoor temperature is 273 K (0 oC) and (b) on another day when the outdoor temperature is 252 K (-21 oC)
Answer:
a) [tex]W_{in} = 214.286\,J[/tex], b) [tex]W_{in} = 428.571\,J[/tex]
Explanation:
a) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.
[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.
Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 273\,K[/tex]. The Coefficient of Performance is:
[tex]COP_{HP} = \frac{294\,K}{294\,K-273\,K}[/tex]
[tex]COP_{HP} = 14[/tex]
Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.
[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]
The input work to deliver a determined amount of heat to the house:
[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]
If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 14[/tex], the input work that is needed is:
[tex]W_{in} = \frac{3000\,J}{14}[/tex]
[tex]W_{in} = 214.286\,J[/tex]
b) The performance of a Carnot heat pump is determined by the Coefficient of Performance, which is equal to the following ratio:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
Where:
[tex]T_{L}[/tex] - Temperature of surroundings, measured in Kelvin.
[tex]T_{H}[/tex] - Temperature of the house, measured in Kelvin.
Given that [tex]T_{H} = 294\,K[/tex] and [tex]T_{L} = 252\,K[/tex]. The Coefficient of Performance is:
[tex]COP_{HP} = \frac{294\,K}{294\,K-252\,K}[/tex]
[tex]COP_{HP} = 7[/tex]
Besides, the performance of real heat pumps are determined by the following form of the Coefficient of Performance, that is, the ratio of heat received by the house to input work.
[tex]COP_{HP} = \frac{Q_{H}}{W_{in}}[/tex]
The input work to deliver a determined amount of heat to the house:
[tex]W_{in} = \frac{Q_{H}}{COP_{HP}}[/tex]
If [tex]Q_{H} = 3000\,J[/tex] and [tex]COP_{HP} = 7[/tex], the input work that is needed is:
[tex]W_{in} = \frac{3000\,J}{7}[/tex]
[tex]W_{in} = 428.571\,J[/tex]
Explain how a refrigerator works to cool down warm objects that would otherwise be room temperature
Answer: evaporation
Explanation:
Refrigerators work by causing the refrigerant circulating inside them to change from a liquid into a gas. This process, called evaporation, cools the surrounding area and produces the desired effect.
A high-jumper clears the bar and has a downward velocity of - 5.00 m/s just before landing on an air mattress and bouncing up at 1.0 m/s. The mass of the high-jumper is 60.0 kg. What is the magnitude and direction of the impulse that the air mattress exerts on her
-- As she lands on the air mattress, her momentum is (m v)
Momentum = (60 kg) (5 m/s down) = 300 kg-m/s down
-- As she leaves it after the bounce,
Momentum = (60 kg) (1 m/s up) = 60 kg-m/s up
-- The impulse (change in momentum) is
Change = (60 kg-m/s up) - (300 kg-m/s down)
Magnitude of the change = 360 km-m/s
The direction of the change is up /\ .
The direction of a body or object's movement is defined by its velocity.In its basic form, speed is a scalar quantity.In essence, velocity is a vector quantity.It is the speed at which distance changes.It is the displacement change rate.
Solve the problem ?
Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is travelling along a path.In other words, velocity is a vector, whereas speed is a scalar value. We discuss the conceptive impulse in this puzzle.A high jumper weighing 60.0 kg sprints on the matrix at minus 6 meters per second in the downhill direction before falling to the mattress.her admirer.Speed drops to 0 meters/second.We must determine the impulse's size and presumed direction, which is upward and positive.The change in momentum is then equal to the impulse.The impulse therefore equals m times.the end velocity less the starting velocity.60.0kg times 0 minus minus 6 meters per second is the impulse, therefore.The impulse is 360 kilogram meters per second, or 360 newtons, to put it another way.The second is upward, and the direction.To learn more about magnitude refer
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Your roommate is working on his bicycle and has the bike upside down. He spins the 68.0 cm -diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. A. What is the pebble's speed? B. What is the pebble's acceleration?
Answer:
a. 6.41 m/s
b. 120.85 m/s^2
Explanation:
The computation is shown below:
a. Pebble speed is
As we know that according to the tangential speed,
[tex]v = r \times \omega[/tex]
[tex]= \frac{0.68}{2} \times 18.84[/tex]
= 6.41 m/s
The 18.84 come from
[tex]= 2 \times 3.14 \times 3[/tex]
= 18.84
b. The pebble acceleration is
[tex]a = \frac{v^2}{r}[/tex]
[tex]= \frac{6.41^2}{0.34}[/tex]
= 120.85 m/s^2
We simply applied the above formulas so that the pebble speed and the pebble acceleration could come and the same is to be considered
A horizontal 790-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 45 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 4.0 s. (Assume it is a solid cylinder. Also assume the force is applied at the outside edge.)
Answer:
404.3 J
Explanation:
Given that
Weight of the merry go round = 790 N
Radius if the merry go round = 1.6 m
Horizontal force applied = 45 N
Time taken = 4 s
To find the mass of the merry go round, we divide the weight by acceleration due to gravity. Thus,
m = F/g
m = 790 / 9.8
m = 80.6 kg
We know that the moment of inertia is given as
I = ½mr², on substitution, we have
I = ½ * 80.6 * 1.6²
I = 103.17 kgm²
Torque = Force applied * radius, so
τ = 45 * 1.6
τ = 72 Nm
To get the angular acceleration, we have,
α = τ / I
α = 72 / 103.17
α = 0.70 rad/s²
Then, the angular velocity is
ω = α * t
ω = 0.7 * 4
ω = 2.8 rad/s
Finally, to get the Kinetic Energy, we have
K.E = ½ * Iω², on substituting, we get
K.E = ½ * 103.17 * 2.8²
K.E = 404.3 J
Therefore, the kinetic energy is 404.3 J
One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the floor. A man weighing 700 N could climb up to 7.0 m before slipping. What is the coefficient of static friction between the floor and the ladder
Answer:
μ = 0.37
Explanation:
For this exercise we must use the translational and rotational equilibrium equations.
We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive
let's write the rotational equilibrium
W₁ x/2 + W₂ x₂ - fr y = 0
where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances
cos 60 = x / L
where L is the length of the ladder
x = L cos 60
sin 60 = y / L
y = L sin60
the horizontal distance of man is
cos 60 = x2 / 7.0
x2 = 7 cos 60
we substitute
m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0
fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60
let's calculate
fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)
fr = (735 + 2450) / 8.66
fr = 367.78 N
the friction force has the expression
fr = μ N
write the translational equilibrium equation
N - W₁ -W₂ = 0
N = m₁ g + W₂
N = 30 9.8 + 700
N = 994 N
we clear the friction force from the eucacion
μ = fr / N
μ = 367.78 / 994
μ = 0.37
when a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N.m^2/c. when the paper is turned 25 degree with respect to the field the flux through it is:
Answer:
22.66Nm²/C
Explanation:
Flux through an electric field is expressed as ϕ = EAcosθ
When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25N.m^2/c. If the paper is turned 25 degree with respect to the field the flux through it can be calculated using the formula.
From the formula above where:
EA = 25N.m^2/C
θ = 25°
ϕ = 25cos 25°
ϕ = 22.66Nm²/C