Answer:
T₂ = 95.56°C
Explanation:
The final resistance of a material after being heated is given by the relation:
R' = R(1 + αΔT)
where,
R' = Final Resistance = 207.4 Ω
R = Initial Resistance = 154.9 Ω
α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹
ΔT = Change in Temperature = ?
Therefore,
207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]
207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT
1.34 - 1 = (0.0045°C⁻¹)ΔT
ΔT = 0.34/0.0045°C⁻¹
ΔT = 75.56°C
but,
ΔT = Final Temperature - Initial Temperature
ΔT = T₂ - T₁ = T₂ - 20°C
T₂ - 20°C = 75.56°C
T₂ = 75.56°C + 20°C
T₂ = 95.56°C
When you release the mass, what do you observe about the energy?
Explanation:
Mass and energy are closely related. Due to mass–energy equivalence, any object that has mass when stationary (called rest mass) also has an equivalent amount of energy whose form is called rest energy, and any additional energy (of any form) acquired by the object above that rest energy will increase the object's total mass just as it increases its total energy. For example, after heating an object, its increase in energy could be measured as a small increase in mass, with a sensitive enough scale.
Julie is playing with a toy car and is pushing it around on the floor. The little car has a mass of 6.3 g. The car has a velocity of 2.5 m/s. What is the car's momentum?
Answer:
Momentum of the car = [tex]1.575\times 10^{-2}[/tex] kg meter per second
Explanation:
Julie is playing with a car which has mass = 6.3 g = [tex]6.3\times 10^{-3}[/tex] kg
Velocity of the car is 2.5 meter per second
Since formula to calculate the momentum of an object is,
p = mv
Where, p = momentum of the object
m = mass of the object
v = velocity of the object
By substituting these values in the formula,
p = [tex](6.3\times 10^{-3})\times 2.5[/tex]
= [tex]1.575\times 10^{-2}[/tex] Kg meter per second
Therefore, momentum of the car will be [tex]1.575\times 10^{-2}[/tex] Kg meter per second.
A motorcyclist changes his speed from 20 km / h to 100 km / h in 3 seconds, maintaining a constant acceleration in that time interval. If the mass of the motorcycle is 200 kg and that of its rider is 80 kg, what is the value of the net force to accelerate the motorcycle? Help!
Answer:
2000 N
Explanation:
20 km/h = 5.56 m/s
100 km/h = 27.78 m/s
F = ma
F = m Δv/Δt
F = (200 kg + 80 kg) (27.78 m/s − 5.56 m/s) / (3 s)
F = 2074 N
Rounded to one significant figure, the force is 2000 N.
A 25-kilogram object is placed on a compression spring, and it creates a displacement of 0.15 meters. What is the weight of an object that creates a displacement of 0.23 m on the same spring? Enter your answer as a number rounded to the nearest tenth, such as: 42.5
Answer:
I hope it is correct ✌️
What is the one single most important reason that human impact on the planet has been so great?
Answer:
Increasing population
Explanation:
As we can see that the death rate is decreasing while at the same time the birth rate is increasing due to which it increased the population that directly impact the planet so great
Day by day the population of the villages, cities, states, the country is increasing which would create a direct human impact on the planet
Therefore the increasing population is the one and single most important reason
A lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is
The lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffraction standpoint is: in the near-ultraviolet.
What is diffraction?The act of bending light around corners such that it spreads out and illuminates regions where a shadow is anticipated is known as diffraction of light. In general, since both occur simultaneously, it is challenging to distinguish between diffraction and interference. The diffraction of light is what causes the silver lining we see in the sky. A silver lining appears in the sky when the sunlight penetrates or strikes the cloud.
Longer wavelengths of light are diffracted at a greater angle than shorter ones, with the amount of diffraction being dependent on the wavelength of the light. Hence, among the light waves of the visible, near-infrared, and near-ultraviolet range, near-ultraviolet waves have the shortest wavelengths. So, The best resolution of this lens from a diffraction standpoint is in the near-ultraviolet, where diffraction is minimum.
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When separated by distance d, identically charged point-like objects A and B exert a force of magnitude F on each other. If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, what will be the new force that they exert on each other in terms of force F
Answer:
F ’= F 0.25
Explanation:
This problem refers to the electric force, which is described by Coulomb's law
F = k q₁ q₂ / r²
where k is the Coulomb constant, q the charges and r the separation between them.
The initial conditions are
F = k q_A q_B / d²
they indicate that the loads are reduced to ¼ q and the distance is reduced to ½ d
F ’= k (q / 4 q / 4) / (0.5 d)²
F ’= k q / 16 / 0.25 d²
F ’= k q² / d² 0.0625 / 0.25
F ’= F 0.25
Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.
Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. This can be explained through Coulomb's law.
What is Coulomb's law?Coulomb's law is a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.
[tex]F = k \frac{q_Aq_B}{d^{2} } = k \frac{q^{2} }{d^{2} } [/tex]
where,
[tex]q_A [/tex] and [tex]q_B[/tex] are the charges of A and B (and equal to q).k is the Coulomb's constant.If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, the new force will be:
[tex]F_2 = k \frac{(0.25q_A)(0.25q_B)}{(0.5d)^{2} } = 0.25k\frac{q^{2} }{d^{2} } = 0.25 F[/tex]
Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.
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A student in her physics lab measures the standing-wave modes of a tube. The lowest frequency that makes a resonance is 30 Hz. As the frequency is increased, the next resonance is at 90 Hz.
What will be the next resonance after this?
Answer:
The next resonance will be 150 Hz.
Explanation:
The frequency of the sound produced by a tube, both open and closed, is directly proportional to the speed of propagation. Hence, to produce the different harmonics of a tube, the wave propagation speed must be increased.
The frequency of the sound produced by a tube, both open and closed, is inversely proportional to the length of the tube. The greater the length of the tube, the frequency is lower.
Frecuency of the standing sound wave modes in a open-closed tube is:
fₙ=n*f₁ where m is an integer and f₁ is the first frecuency (30 Hz)
The next resonance is at 90 Hz. This means that it occurs when n = 3:
f₃=3*30 Hz= 90 Hz
This means that the next resonance occurs when n = 5:
f₅=5*30 Hz= 150 Hz
The next resonance will be 150 Hz.
An object of mass 2 kg has a speed of 6 m/s and moves a distance of 8 m. What is its kinetic energy in joules?
Answer:
36 JoulesExplanation:
Mass ( m ) = 2 kg
Speed of the object (v) = 6 metre per second
Kinetic energy =?
Now,
We have,
Kinetic Energy = [tex] \frac{1}{2} \times m \times {v}^{2} [/tex]
Plugging the values,
[tex] = \frac{1}{2} \times 2 \times {(6)}^{2} [/tex]
Reduce the numbers with Greatest Common Factor 2
[tex] = {(6)}^{2} [/tex]
Calculate
[tex] = 36 \: joule[/tex]
Hope this helps...
Good luck on your assignment...
The Kinetic energy of the object will be "36 joules".
Kinetic energyThe excess energy of moving can be observed as that of the movement of an object, component, as well as the group of components. There would never be a negative (-) amount of kinetic energy.
According to the question,
Mass of object, m = 2 kg
Speed of object, v = 6 m/s
As we know the formula,
→ Kinetic energy (K.E),
= [tex]\frac{1}{2}[/tex] × m × v²
By substituting the values, we get
= [tex]\frac{1}{2}[/tex] × 2 × (6)²
= [tex]\frac{1}{2}[/tex] × 2 × 36
= 36 joule
Thus the above answer is appropriate.
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An empty parallel plate capacitor is connected between the terminals of a 9.0-V battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor
Answer:
The new voltage between the plates of the capacitor is 18 V
Explanation:
The charge on parallel plate capacitor is calculated as;
q = CV
Where;
V is the battery voltage
C is the capacitance of the capacitor, calculated as;
[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]
[tex]q = \frac{\epsilon _0A V}{d}[/tex]
where;
ε₀ is permittivity of free space
A is the area of the capacitor
d is the space between the parallel plate capacitors
If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;
[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]
Therefore, the new voltage between the plates of the capacitor is 18 V
The voltage between the cathode and the screen of a television set is 30 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed (m/s) just before it hits the screen
Answer:
The speed is [tex]v =10.27 *10^{7} \ m/s[/tex]
Explanation:
From the question we are told that
The voltage is [tex]V = 30 kV = 30*10^{3} V[/tex]
The initial velocity of the electron is [tex]u = 0 \ m/s[/tex]
Generally according to the law of energy conservation
Electric potential Energy = Kinetic energy of the electron
So
[tex]PE = KE[/tex]
Where
[tex]KE = \frac{1}{2} * m* v^2[/tex]
Here m is the mass of the electron with a value of [tex]m = 9.11 *10^{-31} \ kg[/tex]
and
[tex]PE = e * V[/tex]
Here e is the charge on the electron with a value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]e * V = \frac{1}{2} * m * v^2[/tex]
=> [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]
substituting values
[tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]
[tex]v =10.27 *10^{7} \ m/s[/tex]
distributed uniformly over the surface of a metal sphere with a radius 24.0 cm. If the potential is zero at a point at infinity, find the value of the pote my jobntA total electric charge of 3.50 nC is distributed uniformly over the surface of a metal sphere with a radius 24.0 cm. If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) 48.0 cm (b) 2ial at the following distances from the center of the sphere: (a) 48.0 cm (b) 24.0 cm (c) 12.0 cm
Answer:
(a) V = 65.625 Volts
(b) V = 131.25 Volts
(c) V = 131.25 Volts
Explanation:
Recall that:
1) in a metal sphere the charges distribute uniformly around the surface, and the electric field inside the sphere is zero, and the potential is constant equal to:
[tex]V=k\frac{Q}{R}[/tex]
2) the electric potential outside of a charged metal sphere is the same as that of a charge of the same value located at the sphere's center:
[tex]V=k\frac{Q}{r}[/tex]
where k is the Coulomb constant ( [tex]9\,\,10^9\,\,\frac{N\,m^2}{C^2}[/tex] ), Q is the total charge of the sphere, R is the sphere's radius (0.24 m), and r is the distance at which the potential is calculated measured from the sphere's center.
Then, at a distance of:
(a) 48 cm = 0.48 m, the electric potential is:
[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.48} =65.625\,\,V[/tex]
(b) 24 cm = 0.24 m, - notice we are exactly at the sphere's surface - the electric potential is:
[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]
(c) 12 cm (notice we are inside the sphere, and therefore the potential is constant and the same as we calculated for the sphere's surface:
[tex]V=k\frac{Q}{R}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]
Answer:
c) a difference in electric potential
Explanation:
my insta: priscillamarquezz
A 100 kg lead block is submerged in 2 meters of salt water, the density of which is 1096 kg / m3. Estimate the value of the hydrostatic pressure.
Answer:
21,920 Pascals
Explanation:
P = ρgh
P = (1096 kg/m³) (10 m/s²) (2 m)
P = 21,920 Pa
A car has a mass of 1200 kg and an acceleration of 4 m/s^2. If the friction on the car is 200 N, how much force is the thrust providing?
Answer:
5000N
Explanation:
According to Newton's second law of motion, the net force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of the body caused by the force. i.e
∑F = m x a -------------(i)
From the question, the net force is the combined effect of the thrust (F) and the friction force (Fₓ). i.e
∑F = F + Fₓ -------------(ii)
Where;
Fₓ = -200N [negative sign because the friction force opposes motion]
Combine equations(i) and (ii) together to get;
F + Fₓ = m x a
F = ma - Fₓ -------------(iii)
Where;
m = mass of car = 1200kg
a = acceleration of the car = 4m/s²
Now substitute the values of m, a and Fₓ into equation (iii) as follows;
F = (1200 x 4) - (-200)
F = 4800 + 200
F = 5000N
Therefore, the force the thrust is providing is 5000N
A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What are the magnitude and direction of the net force on the car
Answer:
9,375
Explanation:
Data provided
The mass of the car m = 1500 Kg.
The diameter of the circular track D = 200 m.
For the computation of magnitude and direction of the net force on the car first we need to find out the radius of the circular path which is shown below:-
The radius of the circular path is
[tex]R = \frac{D}{2}[/tex]
[tex]= \frac{200}{2}[/tex]
= 100 m
after the radius of the circular path we can find the magnitude of the centripetal force with the help of below formula
[tex]Force F = \frac{mv^2}{R}[/tex]
[tex]= \frac{1500\times (25)^2}{100}[/tex]
= 9,375
Therefore for computing the magnitude of the centripetal force we simply applied the above formula.
a ring with a clockwise current is situated with its center directly above another ring. The current in the top ring is decreasing. What is the directiong of the induced current in the bottom ring
Answer:
clockwise
Explanation:
when current flows through a ring in a clockwise direction, it produces the equivalent magnetic effect of a southern pole of a magnet on the coil.
Since the current is decreasing, there is a flux change on the lower ring; generating an induced current on the lower ring. According to Lenz law of electromagnetic induction, "the induced current will act in such a way as to oppose the motion or the action producing it". In this case, the induced current will have to be the same polarity to the polarity of the current change producing it so as to repel the two rings far enough to stop the electromagnetic induction. The induced current will then be in the clockwise direction on the lower ring.
The direction of the induced current in the bottom ring is in the clockwise direction.
The given problem is based on the concept and fundamentals of the induced current and the direction of flow of the induced current.
When current flows through a ring in a clockwise direction, it produces the equivalent magnetic effect of a southern pole of a magnet on the coil. Since the current is decreasing, there is a flux change on the lower ring; generating an induced current on the lower ring. According to Lenz law of electromagnetic induction, "the induced current will act in such a way as to oppose the motion or the action producing it". In this case, the induced current will have to be the same polarity to the polarity of the current change producing it so as to repel the two rings far enough to stop the electromagnetic induction. The induced current will then be in the clockwise direction on the lower ring.Thus, we can conclude that the direction of the induced current in the bottom ring is in the clockwise direction.
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Which characteristic gives the most information about what kind of element an atom is ?
Answer:
The atomic number
Explanation:
Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time moving with a total distance of 0.12 m. Samantha pushes the sandpaper against the surface with a normal force of 2.6 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the normal force during the sanding process
Answer:
W = 12.96 J
Explanation:
The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:
F = μR
where,
F = Friction Force = ?
μ = 0.92
R = Normal Force = 2.6 N
Therefore,
F = (0.92)(2.6 N)
F = 2.4 N
Now, the displacement is given as:
d = (0.12 m)(45)
d = 5.4 m
So, the work done will be:
W = F d
W = (2.4 N)(5.4 m)
W = 12.96 J
A 54.0 kg ice skater is moving at 3.98 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.802 m around the pole.
(a) Determine the force exerted by the horizontal rope on her arms.N
(b) What is the ratio of this force to her weight?(force from part a / her weight)
Answer:
(a) force is 1066.56N
Explanation:
(a) MV²/R
A uniform thin rod of mass ????=3.41 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m=0.249 kg , are attached to the ends of the rod. What must the length L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is ????=0.929 kg·m2 ?
Answer:
The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]
Explanation:
The Diagram for this question is gotten from the first uploaded image
From the question we are told that
The mass of the rod is [tex]M = 3.41 \ kg[/tex]
The mass of each small bodies is [tex]m = 0.249 \ kg[/tex]
The moment of inertia of the three-body system with respect to the described axis is [tex]I = 0.929 \ kg \cdot m^2[/tex]
The length of the rod is L
Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as
[tex]I = I_r + 2 I_m[/tex]
Where [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as
[tex]I_r = \frac{ML^2 }{12}[/tex]
And [tex]I_m[/tex] the moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented as
[tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]
Thus [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]
Hence
[tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]
=> [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]
substituting vales we have
[tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]
[tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]
[tex]L = 1.5077 \ m[/tex]
A square copper plate, with sides of 50 cm, has no net charge and is placed in a region where there is a uniform 80 kN / C electric field directed perpendicular to the plate. Find a) the charge density of each side of the plate and b) the total load on each side.
Answer:
a) ±7.08×10⁻⁷ C/m²
b) 1.77×10⁻⁷ C
Explanation:
For a conductor,
σ = ±Eε₀,
where σ is the charge density,
E is the electric field,
and ε₀ is the permittivity of space.
a)
σ = ±Eε₀
σ = ±(8×10⁴ N/C) (8.85×10⁻¹² F/m)
σ = ±7.08×10⁻⁷ C/m²
b)
σ = q/A
7.08×10⁻⁷ C/m² = q / (0.5 m)²
q = 1.77×10⁻⁷ C
Zuckerman’s test for sensation seeking measures which of the following characteristics?
dangerousness, antisocial traits, “letting loose,’ and intolerance for boredom
thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom
adventurousness, physical prowess, creative morality, and charisma
dangerousness, adventurousness, creativity, and thrill and adventure seeking
The correct answer is B. thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom
Explanation:
Marvin Zuckerman was an important American Psychologists mainly known for his research about personality and the creation of a model to study this aspect of human psychology. This model purposes five factors define personality, these are the thrill and adventure-seeking that involves seeking for adventures and danger; experience seeking that implies a strong interest in participating in new activities; disinhibition that implies being open and extrovert; and susceptibility to boredom that implies avoiding boredom or repetition. Thus, option B correctly describes the characteristics used in Zuckerman's test.
A box with an initial speed of 15 m/s slides along a surface where the coefficient of sliding friction is 0.45. How long does it take for the block to come to rest
Answer:
t = 3.4 s
The box will come to rest in 3.4 s
Explanation:
For the block to come to rest, the friction force must become equal to the unbalanced force. Therefore:
Unbalanced Force = Frictional Force
but,
Unbalanced Force = ma
Frictional Force = μR = μW = μmg
Therefore,
ma = μmg
a = μg
where,
a = acceleration of box = ?
μ = coefficient of sliding friction = 0.45
g = 9.8 m/s²
Therefore,
a = (0.45)(9.8 m/s²)
a = -4.41 m/s² (negative sign due to deceleration)
Now, for the time to stop, we use first equation of motion:
Vf = Vi + at
where,
Vf = Final Speed = 0 m/s (since box stops at last)
Vi = Initial Speed = 15 m/s
t = time to stop = ?
Therefore,
0 m/s = 15 m/s + (-4.41 m/s²)t
(-15 m/s)/(-4.41 m/s²) = t
t = 3.4 s
The box will come to rest in 3.4 s
Copper wire of diameter 0.289 cm is used to connect a set of appliances at 120 V, which draw 1850 W of power total. The resistivity of copper is 1.68×10−8Ω⋅m.
A. What power is wasted in 26.0 m of this wire?
B. What is your answer if wire of diameter 0.417 cm is used?
Answer:
(a) The power wasted for 0.289 cm wire diameter is 15.93 W
(b) The power wasted for 0.417 cm wire diameter is 7.61 W
Explanation:
Given;
diameter of the wire, d = 0.289 cm = 0.00289 m
voltage of the wire, V = 120 V
Power drawn, P = 1850 W
The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m
Area of the wire;
A = πd²/4
A = (π x 0.00289²) / 4
A = 6.561 x 10⁻⁶ m²
(a) At 26 m of this wire, the resistance of the is
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 6.561 x 10⁻⁶
R = 0.067 Ω
Current in the wire is calculated as;
P = IV
I = P / V
I = 1850 / 120
I = 15.417 A
Power wasted = I²R
Power wasted = (15.417²)(0.067)
Power wasted = 15.93 W
(b) when a diameter of 0.417 cm is used instead;
d = 0.417 cm = 0.00417 m
A = πd²/4
A = (π x 0.00417²) / 4
A = 1.366 x 10⁻⁵ m²
Resistance of the wire at 26 m length of wire and 1.366 x 10⁻⁵ m² area;
R = ρL / A
R = (1.68 x 10⁻⁸ x 26) / 1.366 x 10⁻⁵
R = 0.032 Ω
Power wasted = I²R
Power wasted = (15.417²)(0.032)
Power wasted = 7.61 W
In an oscillating LC circuit, the total stored energy is U and the maximum current in the inductor is I. When the current in the inductor is I/2, the energy stored in the capacitor is
Answer:
The definition of that same given problem is outlined in the following section on the clarification.
Explanation:
The Q seems to be endless (hardly any R on the circuit). So energy equations to describe and forth through the inducer as well as the condenser.
Presently take a gander at the energy stored in your condensers while charging is Q.
⇒ [tex]U =\frac{Qmax^2}{C}[/tex]
So conclude C doesn't change substantially as well as,
When,
⇒ [tex]Q=\frac{Qmax}{2}[/tex]
⇒ [tex]Q^2=\frac{Qmax^2}{4}[/tex]
And therefore only half of the population power generation remains in the condenser that tends to leave this same inductor energy at 3/4 U.
The bulldog and skateboard have a combined mass of 20 kg. In case B (the middle of the three pictures of the bulldog and the well), the bulldog and skateboard have a KE of 380 J at the bottom of the well. How deep is the well in meters?
Answer:
h = 1.94 m
Explanation:
When the bull dog and skate board reach the bottom of the well, all of its potential energy is converted to the kinetic energy:
Kinetic Energy Gained by Bull Dog and Skate Board = Potential Energy Lost by Bull Dog and Skate Board
K.E = P.E
K.E = mgh
h = K.E/mg
where,
h = depth of well = ?
K.E = Kinetic Energy at bottom = 380 J
m = mass of bull dog and skate board = 20 kg
g = 9.8 m/s²
Therefore,
h = 380 J/(20 kg)(9.8 m/s²)
h = 1.94 m
A hungry 177 kg lion running northward at 81.8 km/hr attacks and holds onto a 32.0 kg Thomson's gazelle running eastward at 59.0 km/hr. Find the final speed of the lion–gazelle system immediately after the attack.
Answer:
The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.
Explanation:
Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:
[tex]\vec p_{L} + \vec p_{G} = \vec p_{F}[/tex]
Where:
[tex]\vec p_{L}[/tex] - Linear momentum of the lion, measured in kilograms-meters per second.
[tex]\vec p_{G}[/tex] - Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.
[tex]\vec p_{F}[/tex] - Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.
After using the definition of momentum, the system is expanded:
[tex]m_{L}\cdot \vec v_{L} + m_{G}\cdot \vec v_{G} = (m_{L} + m_{G})\cdot \vec v_{F}[/tex]
Vectorially speaking, the final velocity of the lion-gazelle system is:
[tex]\vec v_{F} = \frac{m_{L}}{m_{L}+m_{G}}\cdot \vec v_{L} + \frac{m_{G}}{m_{L}+m_{G}}\cdot \vec v_{G}[/tex]
Where:
[tex]m_{L}[/tex], [tex]m_{G}[/tex] - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.
[tex]\vec v_{L}[/tex], [tex]\vec v_{G}[/tex], [tex]\vec v_{F}[/tex] - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.
If [tex]m_{L} = 177\,kg[/tex], [tex]m_{G} = 32\,kg[/tex], [tex]\vec v_{L} = 81.8\cdot j\,\left[\frac{km}{h} \right][/tex] and [tex]\vec v_{G} = 59.0\cdot i\,\left[\frac{km}{h} \right][/tex], the final velocity of the lion-gazelle system is:
[tex]\vec v_{F} = \frac{177\,kg}{177\,kg+32\,kg}\cdot \left(81.8\cdot j\right)\,\left[\frac{km}{h} \right] + \frac{32\,kg}{177\,kg+32\,kg}\cdot \left(59.0\cdot i\right)\,\left[\frac{km}{h} \right][/tex]
[tex]\vec v_{F} = 9.033\cdot i + 69.276\cdot j\,\left[\frac{km}{h} \right][/tex]
The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:
[tex]\|\vec v_{F}\| = \sqrt{\left(9.033\frac{km}{h} \right)^{2}+\left(69.276\frac{km}{h} \right)^{2}}[/tex]
[tex]\|\vec v_{F}\| \approx 69.862\,\frac{km}{h}[/tex]
The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.
A helicopter rotor blade is 3.40m long from the central shaft to the rotor tip. When rotating at 550rpm what is the radial acceleration of the blade tip expressed in multiples of g?
Answer:
a = 1.15 10³ g
Explanation:
For this exercise we will use the relations of the centripetal acceleration
a = v² / r
where is the linear speed of the rotor and r is the radius of the rotor
let's use the relationships between the angular and linear variables
v = w r
let's replace
a = w² r
let's reduce the angular velocity to the SI system
w = 550 rev / min (2pi rad / 1 rev) (1 min / 60 s)
w = 57.6 rad / s
let's calculate
a = 57.6² 3.4
a = 1.13 10⁴ m / s²
To calculate this value in relation to g, let's find the related
a / g = 1.13 10⁴ / 9.8
a = 1.15 10³ g
What is unique about the c-ray that is not about other rays? Note: Refer to the concave mirror video Select one: a. only ray whose angle of incidence = angle of reflection b. only ray that reflects back in the same direction it came from c. both the above statements are true d. none of the above
Answer:
b. only ray that reflects back in the same direction it came from
Explanation:
C-rays can be said to be a ray that comes from the center of the curvature. It is known that any ray that comes from the center of the curvature reflects back in the same direction it came from, this is because the line joining from the center of the curvature to any point in the mirror is perpendicular to the mirror.
Correct answer is option B.
C-ray is the only ray that reflects back in the same direction it came from.
Option A is incorrect because for other rays, angle of incidence = angle of reflection. This is not a property of c-ray.
Cass is walking her dog (Oreo) around the neighborhood. Upon arriving at Calina's house (a friend of Oreo's), Oreo turns part mule and refuses to continue on the walk. Cass yanks on the chain with a 67 N force at an angle of 30° above the horizontal. Determine the horizontal and vertical components of the tension force.
Answer:
Horizontal component: [tex]F_x = 58\ N[/tex]
Vertical component: [tex]F_y = 33.5\ N[/tex]
Explanation:
To find the horizontal and vertical components of the force, we just need to multiply the magnitude of the force by the cosine and sine of the angle with the horizontal, respectively.
Therefore, for the horizontal component, we have:
[tex]F_x = F * cos(angle)[/tex]
[tex]F_x = 67 * cos(30)[/tex]
[tex]F_x = 58\ N[/tex]
For the vertical component, we have:
[tex]F_y = F * sin(angle)[/tex]
[tex]F_y = 67 * sin(30)[/tex]
[tex]F_y = 33.5\ N[/tex]
So the horizontal component of the tension force is 58 N and the vertical component is 33.5 N.