Answer:
M = 1433.5 kg
Explanation:
This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,
B = ρ g V
with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition
Σ F = 0
B-W = 0
B = W
body weight
W = M g
the volume is
V = l to h
rho_liquid g (l to h) = M g
M = rho_liquid l a h
we calculate
M = 1000 4.7 6.10 0.05
M = 1433.5 kg
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.
Two blocks of masses m1 and m2 are placed in contact with each other on a smooth, horizontal surface. Block m1 is on the left of block m2 . A constant horizontal force F to the right is applied to m1 . What is the horizontal force acting on m2?
Answer:
The horizontal force acting on m2 is F + 9.8m1
Explanation:
Given;
Block m1 on left of block m2
Make a sketch of this problem;
F →→→→→→→→→→→-------m1--------m2
Apply Newton's second law of motion;
F = ma
where;
m is the total mass of the body
a is the acceleration of the body
The horizontal force acting on block m2 is the force applied to block m1 and force due to weight of block m1
F₂ = F + W1
F₂ = F + m1g
F₂ = F + 9.8m1
Therefore, the horizontal force acting on m2 is F + 9.8m1
The force acting on the block of mass m₂ is [tex]\frac{m_2F}{m_1+m_2}[/tex]
Force acting on the block:Given that there are two blocks of mass m₁ and m₂.
m₁ is on the left of block m₂. They are in contact with each other.
A force F is applied on m₁ to the right.
According to Newton's laws of motion:
The equation of motion of the blocks can be written as:
F = (m₁ + m₂)a
here, a is the acceleration.
so, acceleration:
a = F / (m₁ + m₂)
Now, the force acting on the block of mass m₂ is:
f = m₂a
[tex]f = \frac{m_2F}{m_1+m_2}[/tex]
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Wind erosion can be reduced by _____.
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]
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
Two small charged spheres are 7.59 cmcm apart. They are moved, and the force on each of them is found to have been tripled. How far apart are they now?
Answer:
The two small charged spheres are now 4.382 cm apart
Explanation:
Given;
distance between the two small charged sphere, r = 7.59 cm
The force on each of the charged sphere can be calculated by applying Coulomb's law;
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
where;
F is the force on each sphere
q₁ and q₂ are the charges of the spheres
r is the distance between the spheres
[tex]F = \frac{kq_1q_2}{r^2} \\\\kq_1q_2 = Fr^2 \ \ (keep \ kq_1q_2 \ constant)\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\r_2 = \sqrt{\frac{F_1r_1^2}{F_2}} \\\\r_2 = r_1\sqrt{\frac{F_1}{F_2}}\\\\(r_1 = 7.59 \ cm, \ F_2 = 3F_1)\\\\r_2 = 7.59cm\sqrt{\frac{F_1}{3F_1}}\\\\r_2 = 7.59cm\sqrt{\frac{1}{3}}\\\\r_2 = 7.59cm *0.5773\\\\r_2 = 4.382 \ cm[/tex]
Therefore, the two small charged spheres are now 4.382 cm apart.
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 student slides her 80.0-kg desk across the level floor of her dormitory room a distance 4.40 m at constant speed. If the coefficient of kinetic friction between the desk and the floor is 0.400, how much work did she do
Answer:
F = umg where u is coefficient of dynamic friction
Explanation:
F = 0.4 x 80 x 9.81 = 313.92 N
You have a 160-Ω resistor and a 0.430-H inductor. Suppose you take the resistor and inductor and make a series circuit with a voltage source that has a voltage amplitude of 30.0 V and an angular frequency of 220 rad/s .
Part A: What is the impedance of the circuit? ( Answer: Z = ? Ω )
Part B: What is the current amplitude? ( Answer: I = ? A )
Part C: What is the voltage amplitude across the resistor? ( Answer: VR = ? V )
Part D: What is the voltage amplitudes across the inductor? ( Answer: VL = ? V )
Part E: What is the phase angle ϕ of the source voltage with respect to the current? ( Answer: ϕ = ? degrees )
Part F: Does the source voltage lag or lead the current? ( Answer: the voltage lags the current OR the voltage leads the current )
Answer:
A. Z = 185.87Ω
B. I = 0.16A
C. V = 1mV
D. VL = 68.8V
E. Ф = 30.59°
Explanation:
A. The impedance of a RL circuit is given by the following formula:
[tex]Z=\sqrt{R^2+\omega^2L^2}[/tex] (1)
R: resistance of the circuit = 160-Ω
w: angular frequency = 220 rad/s
L: inductance of the circuit = 0.430H
You replace in the equation (1):
[tex]Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega[/tex]
The impedance of the circuit is 185.87Ω
B. The current amplitude is:
[tex]I=\frac{V}{Z}[/tex] (2)
V: voltage amplitude = 30.0V
[tex]I=\frac{30.0V}{185.87\Omega}=0.16A[/tex]
The current amplitude is 0.16A
C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:
[tex]V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV[/tex] (3)
D. The voltage across the inductor is:
[tex]V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V[/tex]
E. The phase difference is given by:
[tex]\phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°[/tex]
A horizontal spring with spring constant 290 N/m is compressed by 10 cm and then used to launch a 300 g box across the floor. The coefficient of kinetic friction between the box and the floor is 0.23. What is the box's launch speed?
Answer:
Explanation:
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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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A rigid tank A of volume 0.6 m3 contains 5 kg air at 320K and the rigid tank B is 0.4 m3 with air at 600 kPa, 360 K. They are connected to a piston cylinder initially empty with closed valves. The pressure in the cylinder should be 800 kPa to float the piston. Now the valves are slowly opened and the entire process is adiabatic. The internal energy of the mixture at final state is:_____.
a. 229 k/kg.
b. 238 kJ/kg
c. 257 kg
d. cannot be determined.
Answer:
the internal energy of the mixture at final state = 238kJ/kg
Explanation:
Given
V= 0.6m³
m=5kg
R=0.287kJ/kg.K
T=320 K
from ideal gas equation
PV = nRT
where P is pressure, V is volume, n is number of mole, R is ideal gas constant , T is the temperature.
Recall, mole = mass/molar mass
attached is calculation of the question.
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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A street light is at the top of a pole that has a height of 17 ft . A woman 5 ft tall walks away from the pole with a speed of 8 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40ft from the pole?
Answer:
8 ft/s
Explanation:
This is a straight forward question without much ado.
It is given from the question that she walks with a speed of 8 ft/s
3. Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
Object 2 has the larger drag coefficient
Explanation:
The drag force, D, is given by the equation:
[tex]D = 0.5 c \rho A v^2[/tex]
Object 1 has twice the diameter of object 2.
If [tex]d_2 = d[/tex]
[tex]d_1 = 2d[/tex]
Area of object 2, [tex]A_2 = \frac{\pi d^2 }{4}[/tex]
Area of object 1:
[tex]A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2[/tex]
Since all other parameters are still the same except the drag coefficient:
For object 1:
[tex]D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2[/tex]
For object 2:
[tex]D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2[/tex]
Since the drag force for the two objects are the same:
[tex]0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2[/tex]
Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient
The relationship between the Period (T) caused by the oscillation of the mass on the end of a hanging spring and the mass (m) is:
Answer:
T= 2p√m/k
Explanation:
This is because the period of oscillation of the mass of spring system is directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.
The period of a mass on a spring is given by the equation
T=2π√m/k.
Where T is the period,
M is mass
K is spring constant.
An increase in mass in a spring increases the period of oscillation and decrease in mass decrease period of oscillation.
When there is the relationship between the Period (T) caused by the oscillation of the mass should be considered as the T= 2p√m/k.
Oscillation of the mass:The mass of the spring system with respect to period of oscillation should be directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.
So the following equation should be considered
T=2π√m/k.
Here,
T is the period,
M is mass
K is spring constant.
An increase in mass in a spring rises the period of oscillation and reduce in mass decrease period of oscillation.
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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.
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 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
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
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
Which characteristic gives the most information about what kind of element an atom is ?
Answer:
The atomic number
Explanation:
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 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.
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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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
A copper transmission cable 180 km long and 11.0 cm in diameter carries a current of 135 A.
Required:
a. What is the potential drop across the cable?
b. How much electrical energy is dissipated as thermal energy every hour?
Answer:
a) 43.98 V
b) E = 21.37 MJ
Explanation:
Parameters given:
Length of cable = 180 km = 180000 m
Diameter of cable = 11 cm = 0.11 m
Radius = 0.11 / 2 = 0.055 m
Current, I = 135 A
a) To find the potential drop, we have to find the voltage across the wire:
V = IR
=> V = IρL / A
where R = resistance
L = length of cable
A = cross-sectional area
ρ = resistivity of the copper wire = 1.72 * 10^(-8) Ωm
Therefore:
V = (135 * 1.72 * 10^(-8) * 180000) / (π * 0.055^2)
V = 43.98 V
The potential drop across the cable is 43.98 V
b) Electrical energy is given as:
E = IVt
where t = time taken = 1 hour = 3600 s
Therefore, the energy dissipated per hour is:
E = 135 * 43.98 * 3600
E = 21.37 MJ (mega joules, 10^6)
If 62.9 cm of copper wire (diameter = 1.15 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of 8.43 mT/s, at what rate is thermal energy generated in the loop?
Answer:
The answer is "[tex]\bold{7.30 \times 10^{-6}}[/tex]"
Explanation:
length of the copper wire:
L= 62.9 cm
r is the radius of the loop then:
[tex]r=\frac{L}{2 \pi}\\[/tex]
[tex]=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01\\[/tex]
area of the loop Is:
[tex]A_L= \pi r^2[/tex]
[tex]=100.2001\times 3.14\\\\=314.628[/tex]
change in magnetic field is:
[tex]=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}[/tex]
then the induced emf is: [tex]e = A_L \times \frac{dB}{dt}[/tex]
[tex]=314.628 \times 0.01\\\\=3.14\times 10^{-5}V[/tex]
resistivity of the copper wire is: [tex]\rho =[/tex] 1.69 × 10-8Ω·m
diameter d = 1.15mm
radius (r) = 0.5mm
[tex]= 0.5 \times 10^{-3} \ m[/tex]
hence the resistance of the wire is:
[tex]R=\frac{\rho L}{\pi r^2}\\[/tex]
[tex]=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}\\[/tex]
Power:
[tex]P=\frac{e^2}{R}[/tex]
[tex]=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}[/tex]
The final answer is: [tex]\boxed{7.30 \times 10^{-6} \ W}[/tex]
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.
A cyclotron operates with a given magnetic field and at a given frequency. If R denotes the radius of the final orbit, then the final particle energy is proportional to which of the following?
A. 1/RB. RC. R^2D. R^3E. R^4
Answer:
C. R^2
Explanation:
A cyclotron is a particle accelerator which employs the use of electric and magnetic fields for its functioning. It consists of two D shaped region called dees and the magnetic field present in the dee is responsible for making sure the charges follow the half-circle and then to a gap in between the dees.
R is denoted as the radius of the final orbit then the final particle energy is proportional to the radius of the two dees. This however translates to the energy being proportional to R^2.