Answer:
warmer
Explanation:
The law of conservation of energy tells us that energy cannot be created or destroyed, it can be transferred or converted from one from to another. In this question when the beer that is at room temperature is put in the fridge, it loses some heat energy. This heat energy is not destroyed, the fridge through multiple processes eventually releases this heat to the room through pipes at the back which is why they are normally warm. the heat from the food inside is expelled to the room. It is not lost.
A ray of light strikes the midpoint of one face of an equiangular crown glass prism (n=1.52) at an angle of incidence of 30.0.
(a.) Find the angle of reflection at the first surface of the prism.
(b.) Find the angle of refraction
Answer:
Angle of reflection at plane surface of prism = 30°
Explanation:
Angle of reflection = angle of incidence at any surface plane or curved .
Here angle of incidence at plane surface of prism = 30°
Angle of reflection at plane surface of prism = 30°
If i be the angle of incidence and r be angle of refraction
sin i / sin r = refractive index
= 1.52
sin r = sin i / 1.52
= sin 30 / 1.52
= .5 / 1.52
= .3289
r = 19.2 °
Lenses are described as convergent or divergent depending on how they refract light. What is the difference between these two types of lenses?
Answer:
Explanation:
Convergent lens is the lens which converges the rays of light falling on it. It is thicker at the middle and thinner at the edges. It is also known as convex lens.
Divergent lens is the lens which diverge the rays of light falling on it. It is thinner at the middle and thicker at the edges. It is also called as concave lens.
Convergent lenses, or convex lenses, thicken in the middle and focus light rays to a point, forming real or virtual images; divergent lenses, or concave lenses, are thinner in the middle, causing the light rays to spread out and only form virtual images. Snell’s law describes the refraction process in both lens types as being influenced by differences in the index of refraction.
Explanation:Difference Between Convergent and Divergent Lenses
Lenses are often categorized as either convergent (convex) or divergent (concave) depending on how they refract light. A converging lens is thicker at the middle and causes parallel light rays entering the lens to converge, or come together, at a single point known as the focal point. This lens can create both real and virtual images, the nature of which depends on the position of the object relative to the lens. In contrast, a divergent lens is thinner at the middle and causes parallel light rays to spread apart or diverge after passing through the lens. Divergent lenses only form virtual images.
Snell's law helps explain the refraction in both types of lenses, considering the difference in the index of refraction between the lens material and the surrounding air. The lens power is determined by its ability to bend light, described by the lens's focal length, with more powerful lenses having shorter focal lengths.
A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) the amplitude, (b) the frequency, (c) the total energy, and (d) the kinetic energy and potential energy when x = 0.360 m.
Answer:
(a) A = 0.650 m
(b) f = 1.3368 Hz
(c) E = 17.1416 J
(d) K = 11.8835 J
U = 5.2581 J
Explanation:
Given
m = 1.15 kg
x = 0.650 cos (8.40t)
(a) the amplitude,
A = 0.650 m
(b) the frequency,
if we know that
ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)
⇒ f = 1.3368 Hz
(c) the total energy,
we use the formula
E = m*ω²*A² / 2
⇒ E = (1.15)(8.40)²(0.650)² / 2
⇒ E = 17.1416 J
(d) the kinetic energy and potential energy when x = 0.360 m.
We use the formulas
K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)
and
U = (1/2)*m*ω²*x² (the potential energy)
then
K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)
⇒ K = 11.8835 J
U = (1/2)*(1.15)*(8.40)²*(0.360)²
⇒ U = 5.2581 J
The amplitude (A) of the oscillation is 0.650 m, and the frequency (f) is approximately 1.34 Hz. To determine the total energy and the kinetic and potential energy at the point when x = 0.360 m, the spring constant is needed.
Explanation:In order to answer your question about a 1.15-kg mass oscillating according to the equation x = 0.650 cos(8.40t), we can use the facts about simple harmonic motion.
(a) The amplitude is the coefficient of the cosine function, which is 0.650 m.
(b) The frequency (f) can be determined from the coefficient of the t inside the cosine function. In simple harmonic motion, the angular frequency (ω) equals 2πf. Therefore f = ω / 2π = 8.40 / 2π = 1.34 Hz.
(c) The total energy (E) of an oscillator is given by the formula E = 1/2 kA². However, without the spring constant (k) given in the question, this cannot be determined.
(d) When x = 0.360 m, the potential energy (PE) is given by PE = 1/2 kx², and the kinetic energy (KE) can be calculated by subtracting the potential energy from the total energy. But again, without the spring constant, these cannot be determined.
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The specific heat capacity of copper is three times the specific heat capacity of lead. Equal masses of copper and lead are heated from room temperature to the temperature of boiling water. To achieve, this the amount of heat added to the copper is______times the heat added to lead.
Answer:
Three times
Explanation:
The amount of heat added to a substance when its temperature is increased from T_1 to T_2 is given by
[tex]Q=mc(T_2-T_1)[/tex]
c= specific heat capacity
m= mass of the substance
⇒Q∝c
that is if c is increased three times the amount of heat added is also increased three times. Therefore, the amount of heat added to the copper is three times the heat added to lead.
An object of mass m is dropped from a height h above the surface of a planet of mass M and radius R. Find the speed of the object when it hits the planet surface. Enter your answer as an expression in terms of m, M, R, h and the universal constant of gravity G.
Final answer:
The speed of an object of mass m when it hits the planet's surface after being dropped from height h is found by using the conservation of energy, yielding the formula v = √[2GM(1/R - 1/(R+h))] where G is the gravitational constant, M the planet's mass, and R its radius.
Explanation:
To find the speed v of an object of mass m when it hits the surface of a planet, we will use the principles of energy conservation in the context of gravitational fields. The total mechanical energy (potential plus kinetic) at the beginning and at the end must be equal since we are ignoring air resistance and any other non-conservative forces.
The initial potential energy when the object is at height h above the planet is given by the gravitational potential energy formula U = -G(Mm)/(R+h) and the initial kinetic energy is zero as the object is initially at rest. When the object reaches the surface of the planet, its potential energy is U = -G(Mm)/R and its kinetic energy is K = (1/2)m*v^2. Conservation of energy dictates that the initial total energy equals the final total energy:
-G(Mm)/(R+h) = -G(Mm)/R + (1/2)m*v^2
Solving for v, the speed of the object at the planetary surface, we get:
v = √[2GM(1/R - 1/(R+h))]
This expression shows that the object's speed increases with a greater initial height h, a more massive planet M, and decreases with a larger planetary radius R.
Check all that apply. check all that apply. when two hydrogen atoms are very far apart, the potential energy approaches zero. when the distance between two hydrogen atoms is 0.74 å, a covalent bond is formed. when two hydrogen atoms that are far apart approach each other, the potential energy decreases. at a distance of 0.50 å the potential energy is less than that at 0.74 å. when the potential energy is zero, a covalent bond is formed.
The given parameters;
distance between the atoms of the hydrogen, r = 0.74 å = 0.74 x 10⁻¹⁸ mdistance between the atoms of the hydrogen, r =0.5 x 10 10⁻¹⁸ mThe potential energy of the hydrogen atoms is calculated as follows;
[tex]V = \frac{kq}{r}[/tex]
where;
k is Coulomb's constantq is the charge of the electronr is the distance between the atomsThe potential energy when the distance = 0.74 x 10⁻¹⁸ m
[tex]V = \frac{9\times 10^{9} \times 1.602\times 10^{-19}}{0.74\times 10^{-18}} = 1.95\times 10^9 \ N.m[/tex]
The potential energy when the distance = 0.5 x 10⁻¹⁸ m
[tex]V = \frac{9\times 10^{9} \times 1.602\times 10^{-19}}{0.5\times 10^{-18}} = 2.88 \times 10^9 \ N.m[/tex]
Thus, we can conclude that the following;
when two hydrogen atoms are very far apart, the potential energy approaches zero.when two hydrogen atoms that are far apart approach each other, the potential energy decreasesthe interatomic distance between hydrogen molecules is 0.74 å, thus when the distance between two hydrogen atoms is 0.74 å, a covalent bond is formed.Learn more here:https://brainly.com/question/18160141
When hydrogen atoms approach each other, their potential energy decreases and they begin to form a bond. At an optimal distance of 0.74 å, a stable covalent bond is formed. The potential energy being zero or less does not signify a bond formation.
Explanation:When two hydrogen atoms are far apart, the potential energy indeed approaches zero because there is no overlap of their atomic orbitals. As they approach each other, their atomic orbitals start overlapping, the energy of the system decreases, initiating the process of bond formation. When the distance between the two hydrogen atoms reaches 0.74 å (Angstroms), their energy is at its lowest point which signifies the formation of a covalent bond. This is because at this optimal distance, both repulsive and attractive forces combine to achieve the lowest possible energy configuration, leading to a stable covalent bond.
However, if the distance were to reduce to 0.50 å, the atoms would be too close and the energy of the system would increase due to stronger nuclear repulsions and electron confinement. In essence, a scenario where the potential energy is either less than or equals to zero does not correlate to the formation of a covalent bond. Instead, it's the optimal balance of forces at a specific distance that gives rise to a stable covalent bond in hydrogen atoms.
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coil of wire is connected to a power supply, and a current runs in the coil. A single loop of wire is located near the coil, with its axis on the same line as the axis of the coil. The radius of the loop is 2 cm.
At time t1 the magnetic field at the center of the loop, due to the coil, is 0.4 T, in the direction shown in the diagram; the current in the coil is constant.
(a) What is the absolute value of the magnetic flux through the loop at time t1?
Φ mag = T m2
(b) What approximations or assumptions did you make in calculating your answer to part (a)?
Check all that apply.
1.The magnitude of the magnetic field due to the coil is uniform over the area of the loop.
2.The magnetic field outside the loop is zero.
3. The magnetic field due to the coil is uniform in direction over the area of the loop.
(c) What is the direction of the "curly" electric field inside the wire of the loop at time t1? (Remember that at this time the current in the coil is constant.)
E = 0
At a later time t2, the current in the coil begins to decrease.
(d) Now what is the direction of the "curly" electric field in the loop?
counter-clockwise
At time t2 the rate of change of the magnetic field at the center of the loop, due to the coil, is -0.36 T/s.
(e) At this time, what is the absolute value of the rate of change of the magnetic flux through the loop?
|dΦmag/dt| = T m2/s
(f) At this time, what is the absolute value of the emf in the loop?
|emf| = V
(g) What is the magnitude of the electric field at location P, which is inside the wire?
|E| = V/m
(h) Now the wire loop is removed. Everything else remains as it was at time t2; the magnetic field is still changing at the same rate. What is the magnitude of the electric field at location P?
|E| = V/m
a) The magnetic flux is 0.0005Tm².
b) 1. The magnitude of the magnetic field due to the coil is uniform over the area of the loop.3. The magnetic field due to the coil is uniform in direction over the area of the loop.
c) The direction of the electric field inside the wire of the loop at time t1 is zero
d) The direction of the "curly" electric field in the loop at time when the current in the coil t₂ is counter clockwise
e) The absolute value of the rate of change of the magnetic flux through the loop is [tex]0.000045T^2[/tex]
f) The absolute value of the emf in the loop at previous instant is 0.000045V
g) The magnitude of the electric field at location P, which is inside the wire is 0.0036V/m
h) Now the wire loop is removed.
The question is investigating the interactions of current, magnetic fields, magnetic flux, and electromagnetic forces.
(a) The magnetic flux is the magnetic field times the area it is flowing through. So we have
[tex]\phi_{mag}= B * A = 0.4T * \pi*(0.02m)^2 = 0.0005 T m^2[/tex]
(b) The assumptions made are: 1. The magnetic field is uniform over the loop's area, 2. Magnetic field outside the loop is zero, 3. The magnetic field is uniform in direction over the loop's area.
(c) The direction of the electric field inside the wire of the loop at time t1 is zero, as the magnetic field and the current are constant and hence there’s no change in magnetic flux.
(d) If the current in the coil begins to decrease, Lenz's law tells us that an induced current will always counteract the change in magnetic flux that caused it. Therefore, the direction of the "curly" electric field in the loop will be counter-clockwise.
(e) At the later time the rate of change of the magnetic flux, |dΦmag/dt|, would be equal to the rate of change of the magnetic field times the area of the loop [tex]= |0.36 T/s * \pi*(0.02m)^2| = 0.000045 T m^2/s.[/tex]
(f) The absolute value of the emf in the loop would be equal to the absolute value of the rate of change of magnetic flux. Based on Faraday's law this means [tex]|emf| = |d\phi_{mag}/dt| = 0.000045 V.[/tex]
(g) [tex]\|E| = \frac{|\text{emf}|}{2 \pi r} = \frac{0.000452 \, \text{V}}{2 \pi (0.02 \, \text{m})} \approx 0.0036 \, \text{V/m}[/tex]
(h ) The magnitude of the electric field at location P, which is inside the wire, would depend on specifics not given in the question.
As a system increases in volume, it absorbs 51.0 J of energy in the form of heat from the surroundings. The piston is working against a pressure of 0.545 atm. The final volume of the system is 55.6 L. What was the initial volume of the system if the internal energy of the system decreased by 106.0 J?
Answer:52.72 L
Explanation:
Given
Energy absorb by system [tex]Q=51 J[/tex]
Piston is working against a pressure of [tex]P=0.545 atm[/tex]
Final volume of system [tex]V_f=55.6 L[/tex]
change in internal Energy [tex]\Delta U=-106 J[/tex]
Work done by system [tex]W=P(V_f-V_i)[/tex]
According to first law of thermodynamics
[tex]Q=dU+PdV[/tex]
[tex]51=-106+W[/tex]
[tex]W=157 J[/tex]
[tex]157=0.545\times 10^5\times 10^{-3}(55.6-V)[/tex]
[tex]157=54.5\times (55.6-V)[/tex]
[tex]55.6-V=2.88[/tex]
[tex]V=52.72 L[/tex]
Two speakers face each other, and they each emit a sound of wavelength (lambda). One speaker is 180 (degrees)
%u2218out of phase with respect to the other. If we separate the speakers by a distance1.5 (lambda)
,how far from the left-most speaker should we place a microphone in order to pick up the loudest sound? Ignore reflections from nearby surfaces. Select all that apply.
A. 3/4 lambda
B. 0 lambda
C. 1/2 lambda
D. 1/4 lambda
E. 1 lambda
Answer:
D. 1/4 lambda
Explanation:
Given:
Initially out of phase speakers by 180°.
Then on being separated by 1.5 λ.
The initial case of speakers being out of phase by 180°
In such a case we obtain destructive interference because the vectors of the waveform are equal and opposite in direction so they cancel out each other.On separating the speakers by a distance of 1.5 λ.
We obtain that the waves-forms are in-phase and form constructive interference leading to the highest value at the crest of the wave pattern, i.e. we obtain the loudest sound when we place a microphone at a distance of one-fourth of the wavelength.If the cost of this work input is 10¢/kW · h, how does its cost compare with the direct heat transfer achieved by burning natural gas at a cost of 86.0 cents per therm. (A therm is a common unit of energy for natural gas and equals 1.055 ✕ 108 J.) cost of heat pump cost of natural gas =
Answer:
Explanation:
One therm = 1.055 x 10⁸J
= 1.055 X 10⁵ kJ
= [tex]\frac{1.055\times10^5}{60\times60}[/tex]kWh
29.3055 kWh
cost of 29.3055 kWh is 86 cent
or cost of imput= [tex]\frac{86}{29.3055}[/tex]
= 2.93 cent /]kWh
compare it with cost rate quoted earlier which each was
= 10
Early on in the story the narrator mentions the sound of wind through the wires strung overhead: "The wind in the wires took up the story . . ."Taking the plot's rising action at this point into account, which of the following literary devices is the author most likely using?
A.creative license
B.irony
C.extended metaphor
D.personification
Answer:
option D.
Explanation:
The correct answer is option D.
The irony is the figure of speech which represents the contradiction between what is stated and what actually the scenario is.
the statement by the poet does not have any contradictory statements.
The extended metaphor is a part of speech which is used when there is a comparison between two unlike things in the paragraph.
in the given statement of the poem, there is no comparison.
Personification is the part of speech where human quality are given to non-living things.
in the give statement wind in the wire took up the story can be taken as the human quality.
so, the statement part of speech is personification.
How is the air found in soils different from the air in the atmosphere?
Answer:
Air found in the soil is trapped in it and is not exposed to the air currents, poor in oxygen and rich in moisture.
Explanation:
Air found in the soil is trapped in it and is not exposed to the air currents.
It contains more moisture than the atmospheric air.
It is rich in carbon dioxide and poor in oxygen.
The air found in the soil forms a part of the lithosphere and not the atmosphere.
Nitrogen trapped in the soil is used by the plants to make protein, carbon dioxide is used for the photosynthesis and oxygen is used for the respiration of roots and microorganisms.
As your hand moves back and forth to generate longitudinal pulses in a spiral spring, your hand completes 2.91 back-and-forth cycles every 3.67 s. The velocity of the pulse in the spring is 0.925 cm/s. What is the wavelength? Answer in units of m.
Answer:
Wavelength, [tex]\lambda=0.011\ m[/tex]
Explanation:
Given that,
Number of cycles in a spiral spring is 2.91 in every 3.67 s
The velocity of the pulse in the spring is 0.925 cm/s, v = 0.00925 m/s
To find,
Wavelength
Solution,
Number of cycles per unit time is called frequency of a wave. The frequency of the longitudinal pulse is,
[tex]f=\dfrac{2.91}{3.67}=0.79\ Hz[/tex]
The wavelength of a wave is given by :
[tex]\lambda=\dfrac{v}{f}[/tex]
[tex]\lambda=\dfrac{0.00925\ m/s}{0.79\ Hz}[/tex]
[tex]\lambda=0.011\ m[/tex]
So, the wavelength of the longitudinal pulse is 0.011 meters. Hence, this is the required solution.
To find the wavelength, the frequency is calculated by dividing the number of cycles by the total time. The velocity is then converted from cm/s to m/s. The wavelength is calculated using these values in the equation for wave speed to get approximately 0.012 meters.
Explanation:In the scenario described in your question, your hand completes 2.91 back-and-forth cycles every 3.67 seconds, resulting in longitudinal pulses in a spiral spring. Here, we have frequency and wave velocity, and we need to find the wavelength. We can determine the frequency by dividing the number of cycles by the total time and the wavelength using the formula for wave speed: v = fλ, where v is the velocity, f is the frequency, and λ is the wavelength.
Frequency (f) = number of cycles / total time = 2.91 cycles / 3.67 s = 0.793 Hz
Velocity (v) = 0.925 cm/s = 0.00925 m/s (since we need the answer in meters)
To find the wavelength, we can use the formula v = fλ, to rearrange this to solve for λ, we get λ = v/f. So, λ = 0.00925 m/s / 0.793 Hz = 0.01166 meters or approximately 0.012 meters.
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A force F~ = (1.8 N)ˆı + (3 N) ˆ is applied to an object that is pivoted about a fixed axis aligned along the z coordinate axis. The force is applied at the point R~ = (4.1 m)ˆı + (5.4 m) ˆ. Find the z-component of the net torque. Answer in units of N m.
Answer:
The z-component of the net torque is: 2.58 (N*m)
Explanation:
We need to apply the torque equation ([tex]T=RXF[/tex]), where T is the torque, R is the distance at the point where the force is applied and F is the force, and remembering the cross product ([tex]AXB=abSin(\beta)(n)[/tex]), where A and B are the vectors and ab are the magnitudes, (beta) the angle between A and B in the plane containing them, and (n) is the direction of the vector given by the right-hand rule. Knowing those things, we can get:[tex]T=RXF=4.1(i)+5.4(j)X1.8(i)+3(j)=0+4.1*3(k)+5.4*1.8(-k)+0=2.58(k)(N*m)[/tex] that is the z-component of the torque.
The torque on the object is -2.58Nm
Data;
F = (1.8i + 3j)Nr = (4.1i + 5.4j)mNet TorqueThe net torque is calculated as the product of the force and distance
[tex]t = r * F[/tex]
Let's substitute the values and solve
Since this is a dot product, we can easily multiply through
[tex]\tau = r * F\\\tau = (1.8*5.4) - (3 * 4.1) = -2.58Nm[/tex]
Using 2 by 2 matrix method, the torque on the object is -2.58Nm
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A body moves in circle cover half of revolution in 10 sec its linear distance became 10m along circumference of circle than its centripetal acceleration
The centripetal acceleration is [tex]0.63 m/s^2[/tex]
Explanation:
The centripetal acceleration for an object in circular motion is given by
[tex]a=\frac{v^2}{r}[/tex]
where
v is the linear speed
r is the radius of the circle
The body in the problem cover half of revolution in
t = 10 s
And the corresponding linear distance covered is
L = 10 m
which corresponds to half of the circumference, so
[tex]L=2\pi r[/tex]
From this equation we find the radius of the circle:
[tex]r=\frac{L}{2\pi}=\frac{10}{2\pi}=1.59 m[/tex] [m]
While the linear speed is:
[tex]v=\frac{L}{t}=\frac{10}{10}=1 m/s[/tex]
Therefore, the centripetal acceleration is
[tex]a=\frac{1^2}{1.59}=0.63 m/s^2[/tex]
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Io experiences tidal heating primarily because __________.
Io experiences tidal heating primarily because of the intense gravitational pull of Jupiter which creates friction inside Io and ultimately results in heating. This process is enhanced by Io's elliptical orbit.
Explanation:Io, one of Jupiter's moons, experiences tidal heating primarily due to the intense gravitational pull of Jupiter. These gravitational forces create friction within Io, resulting in heat. This process is also facilitated by Io's elliptical orbit which varies the gravitational forces acting on it during its orbit around Jupiter, thereby enhancing the frictional heating. Therefore, the main contributor to Io's tidal heating is its interaction with Jupiter's gravitational field.
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Are wind-driven currents found near the surface of the oceans or in the deep waters of the ocean?
Wind-driven currents are found NEAR THE SURFACE of the oceans
Explanation:
Surface currents are at the interphase between the hydrosphere and atmosphere. Therefore the feel the greatest effect of drag by wind currents, especially prevailing winds (that blow predominantly in one direction like westerlies and easterlies) within the lower atmosphere. The deep currents, on the other hand, are more influenced by Coriolis effect of the earth’s rotation. It is these differences in influences of surface and deeper currents that cause Ekman transport in oceans.
Learn More:
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A satellite has a mass of 5850 kg and is in a circular orbit 4.1 x10 to the 5th power m above the surface of a planet. The period of the orbit is two hours. The radius of the planet is 4.15 x 10 to the 6th power m. What is the true weight of the satellite when itis at rest on the planet's surface?
Answer:
W = 24.28 kN
Explanation:
given,
Mass of satellite = 5850 Kg
height , h = 4.1 x 10⁵ m
Radius of planet = 4.15 x 10⁶ m
Time period = 2 h
= 2 x 3600 = 7200 s
Time period of satellite
[tex]T = \dfrac{2\pi}{R}\sqrt{\dfrac{(R+h)^3}{g}}[/tex]
R is the radius of planet
h is the height of satellite
[tex]T^2 = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{g}}[/tex]
now calculation of acceleration due to gravity
[tex]g = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{T^2}}[/tex]
[tex]g = \dfrac{4\pi^2}{(4.15\times 10^6)^2}\ {\dfrac{(4.15\times 10^6+4.1\times 10^5)^3}{(7200)^2}}[/tex]
g = 4.15 m/s²
True weight of satellite
W = m g
W = 5850 x 4.15
W = 24277.5 N
W = 24.28 kN
True weight of the satellite is W = 24.28 kN
The true weight of the satellite, when the satellite is at rest on the surface of the planet, is 24.28 kN.
What is the time period of satellites?Time period of satellites is the total time taken by a satellite to complete a full orbit around a body. It can be given as,
[tex]T=\dfrac{2\pi}{R}\sqrt{\dfrac{(R+h)^3}{g}}[/tex]
Here, (R) is the radius of the body, and (g) is the gravitational acceleration force.
In a circular orbit 4.1 x10 to the 5th power m above the surface of a planet. The period of the orbit is 2 hours and the radius of the planet is 4.15 x 10 to the 6th power m.
To find the weight of the satellite, first find the value of gravitation acceleration using the time period formula as,
[tex]2=\dfrac{4\pi^2}{4.15\times10^6}\sqrt{\dfrac{(4.15\times10^6+4.1\times10^5)^3}{g}}\\g=4.15\rm m/s^2[/tex]
The weight of the body is mass time gravity. As the satellite has a mass of 5850 kg and value of g is 4.15 m/s². Thus, the weight of it is,
[tex]W=5850\times4.15\\W=24277.5\rm N\\W=24.28\rm \; kN[/tex]
Thus, the true weight of the satellite, when the satellite is at rest on the surface of the planet, is 24.28 kN.
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A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30 with the horizontal. His force is parallel to the slope. If the speed of the crate decreases at a rate of 1.5 m/s2, then the work done by the mand is:A) –200 JB) 61 JC) 140 JD) 200 JE) 260 J
Answer:
The work done by the man is 260 J.
(E) is correct option.
Explanation:
Given that,
Force = 80 N
Distance = 5.0 m
Angle = 30°
Acceleration = 1.5 m/s²
Let F be the man's force
We need to calculate the force
Using balance equation
[tex]F-mg\sin\theta=ma[/tex]
[tex]F-80\sin30=\dfrac{80}{9.8}\times1.5[/tex]
[tex]F=\dfrac{80}{9.8}\times1.5+80\sin30[/tex]
[tex]F=52\ N[/tex]
We need to calculate the work done by the man
Using formula of work done
[tex]W=F\dotc d[/tex]
[tex]W=52\times 5.0[/tex]
[tex]W=260\ J[/tex]
Hence, The work done by the man is 260 J.
If the speed of the crate decreases at a rate of 1.5 m/s2, then the work done by the man is E) 260 J
Explanation:Work is the quantity of energy transferred by displacement. The following is formula of work done
[tex]\Delta KE = F*d[/tex]
Whereas the following is formula for balance equation:
[tex]F-mg\sin\theta=ma[/tex]
The equation above is come from Newton's Second Law. It applies to a wide range of physical phenomena, but it is not a fundamental principle like the Conservation Laws and also it applies only if the force is the net external force.
Force = 80 N, thereforeMass = Force/g = 80 N / 10 = 8 kgDistance = 5.0 mAngle = 30°Acceleration = 1.5 m/s²For the Case of fictionless slope, the speed at the bottom of a frictionless incline does not depend upon the angle of the incline
Then the work done by the man is...
[tex]w = mgd* sin\theta + \Delta KE\\w = 80 N *5.0 m *sin 30 + (F*d)\\w = 80 N *5.0 m *\frac{1}{2} + (F*d)\\w = 40 N *5.0 m+ (F*d)\\w = 40 N *5.0 m+ (m*a*d)\\w = 200 Nm+ (8*1.5*5)\\w = 200 Nm+ 60\\w = 260 J\\[/tex]
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The main greenhouse gases in the atmospheres of the terrestrial planets are
Answer:
Answered
Explanation:
Terrestrial Planets are the planets that have solid surface and are smaller in size. Mercury, Venus, Earth and Mars are the terrestrial planets of our solar system.
The main greenhouse gases in the atmospheres of the terrestrial planets are, Carbon dioxide( the atmosphere of Venus has mainly CO_2 in it making it hottest planet of the solar system because of greenhouse effect). Water vapor and Cloroflorocarbons in the Earth atmosphere. Traces of methane and nitrous oxide are also present.
To get off a frozen lake, a 70 kg person removes his shoe of mass 0.175 kg and throws it horizontally away from the shore at a velocity of 3.2 m/s. If the person is 5.15 m from the shore, how long do they take to reach the shore?
Answer:
the person will be in the shore at 10.73 minutes after launch the shoe.
Explanation:
For this we will use the law of the lineal momentum.
[tex]L_i = L_f[/tex]
Also,
L = MV
where M is de mass and V the velocity.
replacing,
[tex]M_i V_i = M_{fp}V_{fp} + M_{fz}V_{fz}[/tex]
wher Mi y Vi are the initial mass and velocity, Mfp y Vfp are the final mass and velocity of the person and Mfz y Vfz are the final mass and velocity of the shoe.
so, we will take the direction where be launched the shoe as negative. then:
(70)(0) = (70-0.175)([tex]V_fp[/tex]) + (0.175)(-3.2m/s)
solving for [tex]V_fp[/tex],
[tex]V_fp[/tex] = [tex]\frac{(3.2)(0.175)}{69.825}[/tex]
[tex]V_fp[/tex] = 0.008m/s
for know when the person will be in the shore we will use the rule of three as:
1 second -------------- 0.008m
t seconds-------------- 5.15m
solving for t,
t = 5.15m/0.008m
t = 643.75 seconds = 10.73 minutes
The time to reach the shore is 643.75 seconds.
The problem presented involves conservation of momentum on a frictionless surface, which is a physics concept. To solve for the time it takes the person to reach the shore, we can use the principle that the momentum before throwing the shoe is equal to the momentum after throwing the shoe, as there are no external forces acting on the system (since friction is ignored).
The initial momentum of the system is zero because the person is not moving. After throwing the shoe, the momentum of the shoe can be calculated using the formula: p = m × v, where p is momentum, m is mass, and v is velocity. The momentum of the shoe is 0.175 kg × 3.2 m/s, which must be equal and opposite to the momentum of the person. Thus, the velocity of the person, v_p, can be found through the equation m_shoe × v_shoe = m_person × v_person.
Once the velocity of the person is calculated, the time, t, taken to cover the distance to the shore can be found using the formula: t = d / v_p, where d is the distance to the shore.
Carrying out the calculations gives us:
p_shoe = 0.175 kg × 3.2 m/s = 0.56 kg m/s
v_person = p_shoe / m_person = 0.56 kg m/s / 70 kg = 0.008 m/s
t = d / v_person = 5.15 m / 0.008 m/s = 643.75 seconds
The person will take approximately 643.75 seconds to reach the shore.
A person runs up several flights of stairs and is exhausted at the top. Later the same person walks up the same stairs and does not feel as tired. Ignoring air resistance, does it take more work to run up the stairs than to walk up?
It takes the same amount of work to move up the stairs whether running or walking, as work is related to the change in potential energy, which does not depend on the speed of climbing. However, running up the stairs requires additional work to be done against gravity due to the higher kinetic energy, leading to a feeling of greater exhaustion.
Explanation:The question concerns whether it takes more work to run up stairs than to walk up. In physics, the term 'work' refers to the amount of energy transferred by a force through a distance. In the context of climbing stairs, work is done against gravity to elevate a person's body to a higher potential energy level. According to the work-energy principle, work is equal to the change in kinetic energy plus the change in potential energy.
Using the given information, the person does 1764 J of work to move up the stairs, regardless of whether they run or walk.
This is the work done to overcome gravity. Only 120 J of work is used to increase the person's kinetic energy when running, which is additional work that occurs due to the higher speed.
Therefore, the total work done when running will be the sum of these two amounts.
However, the amount of work done to raise the person's body to a higher elevation (the potential energy) does not change whether the person runs or walks.
The difference is in the kinetic energy.
The feeling of exhaustion may be greater when running due to the higher pace requiring more power output (rate of doing work) and possibly higher energy expenditure related to physiological effects, but the work done against gravity, which depends on the change in elevation and mass of the person, remains the same in both cases.
A tennis player swings her 1000 g racket with a speed of 15.0 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 18.0 m/s. The ball rebounds at 40.0 m/s. How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision.
_________m/s
If the tennis ball and racket are in contact for 7.00 , what is the average force that the racket exerts on the ball?
Answer:
[tex]v_r=13.68\ m.s^{-1}[/tex] is the final velocity of the racket.
[tex]F=18.86\ N[/tex]
Explanation:
Given:
mass of the racket, [tex]m_r=1000\ g[/tex]mass of ball, [tex]m_b=60\ g[/tex]initial speed of racket, [tex]u_r=15\ m.s^{-1}[/tex]initial speed of ball, [tex]u_b=18\ m.s^{-1}[/tex]final speed of ball, [tex]v_b=40\ m.s^{-1}[/tex]time of contact of racket with the ball, [tex]t=0.07\ s[/tex]By the law of conservation of momentum:
[tex]m_r.u_r+m_b.u_b=m_r.v_r+m_b.v_b[/tex]
where: [tex]v_r=[/tex] final velocity of the racket
[tex]1000\times 15+60\times 18=1000\times v_r+60\times 40[/tex]
[tex]v_r=13.68\ m.s^{-1}[/tex] is the final velocity of the racket.
By the Newton's second law of motion:
[tex]F=\frac{dp}{dt}[/tex] ............................(1)
where:
dp = change in momentum
dt = change in time
Change in momentum of ball:
[tex]\Delta p_b=m_b.v_b-m_b.u_b[/tex]
[tex]\Delta p_b=60\times 10^{-3}\times (40-18)[/tex]
[tex]\Delta p_b=1.32\ kg.m.s^{-1}[/tex]
Now, using eq.(1):
[tex]F=\frac{1.32}{0.07}[/tex]
[tex]F=18.86\ N[/tex]
The tennis racket will be moving at 11.52 m/s immediately after the impact. The average force exerted on the ball by the racket, during the brief 7 ms contact time, is approximately 497 N.
The problem can be solved using the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision if no external forces act on the system. The formula for momentum is p = mv, where p is momentum, m is mass, and v is velocity.
The momentum before the collision (pi) for both the racket and the ball is:
pi,racket = mracket \\times vi,racket = 1.0 kg \\times 15.0 m/s = 15 kg\\cdot m/s
pi,ball = mball \\times vi,ball = 0.060 kg \\times (-18.0 m/s) = -1.08 kg\\cdot m/s
The negative sign indicats that the ball is moving in the opposite direction to the racket. Total initial momentum Pi,total = pi,racket + pi,ball.
The momentum after the collision (pf) for the ball is:
pf,ball = mball \\times vf,ball = 0.060 kg \\times 40.0 m/s = 2.4 kg\\cdot m/s
Using the conservation of momentum, Pi,total = Pf,total, i.e.,
15 kg\\cdot m/s - 1.08 kg\\cdot m/s = pf,racket + 2.4 kg\\cdot m/s
Solving for pf,racket, the momentum of the racket after collision:
pf,racket = 15 - 1.08 - 2.4 = 11.52 kg\\cdot m/s
The velocity of the racket after the collision (vf,racket) can be found by dividing pf,racket by the mass of the racket:
vf,racket = pf,racket / mracket = 11.52 kg\\cdot m/s / 1.0 kg = 11.52 m/s
To calculate the average force exerted on the ball by the racket, we can use the change in momentum (impulse) and divide it by the time of contact. The change in momentum of the ball (\\Delta p) is the momentum after minus the momentum before:
\\Delta p = pf,ball - pi,ball = 2.4 kg\\cdot m/s - (-1.08 kg\\cdot m/s) = 3.48 kg\\cdot m/s
The average force (F) exerted on the ball is:
F = \\Delta p / \\Delta t = 3.48 kg\\cdot m/s / 0.007 s = 497.14 N
The racket will be moving at 11.52 m/s immediately after the impact, and the average force that the racket exerts on the ball is approximately 497 N.
The box now rests on a frictionless ramp angled at 15◦ . The mover pulls up on a rope attached to the box to move it up the incline. 33 kg F 29 ◦ 15◦ If the rope makes an angle of 29 ◦ with the horizontal, what is the smallest force the mover would have to exert to move the box up the ramp? The acceleration due to gravity is 9.81 m/s 2 .
Answer:
F = 84.61 N
Explanation:
As in the figure, since there is no friction so if component of Force applied along the incline is greater than the component of weight along the incline, then the object will move up the incline.
component of Force along the incline = F cos(23° - 15°) = F cos(8°)
component of weight along the incline = 33*g*sin(15°) = 33*9.81*sin(15°)
Equating the above two components of forces will give the minimum Force required.
F cos(8°) = 33*9.81*sin(15°)
F = 33*9.81*sin(15°) / cos(8°) (calculate the value using a scientific calculator)
F = 84.61 N
What is the difference between an sbb galaxy and an sb galaxy?
Answer:
One of the most important difference between the two galaxies" is that the sbb galaxy has a lot of material and bar structures passing thourgh the center and the sb galaxy does not have this".
Explanation:
Basically we have 3 types of galaxies (ellipticals, spirals, and irregulars)
Sb galaxies
On this type of galaxies the "intermediate type of spiral typically has a medium-sized nucleus. They can contain stars, star clouds, and interstellar gas and dust. Usually show wide dispersions in terms of their shape. Other members of this subclass have arms that begin tangent to a bright, while still others reveal a small, bright spiral pattern inset into the nuclear bulge. In any of these cases, the spiral arms may be set at different pitch angles. is important to mention that a pitch angle is defined as the angle between an arm and a circle centred on the nucleus and intersecting the arm".
SB galaxies
"For these types of galaxies the luminosities, dimensions, spectra, and distributions of the barred spirals tend to be indistinguishable from those of normal spirals".
One special case occuers at the SBb systems, that "have a smooth bar as well as relatively smooth and continuous arms. In some galaxies of this type, the arms start at or near the ends of the bar, with conspicuous dust lanes along the inside of the bar that can be traced right up to the nucleus"
One of the most important difference between the two galaxies" is that the sbb galaxy has a lot of material and bar structures passing thourgh the center and the sb galaxy does not have this".
A small car of mass 643 kg is parked behind a small truck of mass 1415 kg on a level road. The brakes of both the car and the truck are off so that they are free to roll with negligible friction. A 43 kg woman sitting on the tailgate of the truck shoves the car away by exerting a constant force on the car with her feet. The car accelerates at 2 m/s². What is the acceleration of the truck? Answer in units of m/s².
Answer:
0.882 m/s²
Explanation:
m = Mass of car = 643 kg
a = Acceleration of the car = 2 m/s²
[tex]a_2[/tex] = Acceleration of the truck
Force exerted on the car
[tex]F=ma\\\Rightarrow F=643\times 2\\\Rightarrow F=1286\ N[/tex]
Mass of the truck with the woman = [tex]M=1415+43=1458\ kg[/tex]
From Newton's second law
[tex]F=Ma_2\\\Rightarrow a_2=\frac{1286}{1458}\\\Rightarrow a=0.882\ m/s^2[/tex]
The acceleration of the truck is 0.882 m/s²
Final answer:
The acceleration of the truck can be found by using Newton's second and third laws. Equal and opposite forces act on the car and the truck, and with the mass of each known, calculations can provide the truck's acceleration.
Explanation:
To find the acceleration of the truck when the woman pushes the car, we use Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. When the woman pushes the car with a certain force, the car pushes back on the woman (and hence the truck she is sitting on) with an equal and opposite force. This means the force causing the car to accelerate is the same force causing the truck to accelerate, but in the opposite direction.
Let's denote F as the force with which the woman pushes the car. Given the mass of the car (mcar = 643 kg) and its acceleration (acar = 2 m/s²), we can find the force using Newton's second law of motion: F = mcar × acar.
The same force is responsible for the acceleration of the truck in the opposite direction. Therefore, for the truck with mass mtruck = 1415 kg and acceleration atruck, Newton's second law gives us the relationship: F = mtruck × atruck. By substituting the force F from the previous calculation, we can solve for atruck, which is the acceleration of the truck.
A proton of mass m moving with a speed of 3.0 × 106 m/s undergoes a head on elastic collision with an alpha particle of mass 4m at rest. What are the velocities of the two particles after the collision?
Answer:6.0×10^5m/s
Explanation:
According to the law of conservation of momentum, sum of the momenta of the bodies before collision is equal to the sum of their momenta after collision.
After their collision, the two bodies will move with a common velocity (v)
Momentum = mass × velocity
Let m1 be the mass of the proton = m
Let m2 be the mass of the alpha particle = m2
Let v1 be the velocity of the proton = 3.0×10^6m/s
Let v2 be the velocity of the alpha particle = 0m/s (since the body is at rest).
Using the law,
m1v1 + m2v2 = (m1 + m2)v
m(3.0×10^6) + 4m(0) = (m + 4m)v
m(3.0×10^6) = 5mv
Canceling 'm' at both sides,
3.0×10^6 = 5v
v = 3.0×10^6/5
The common velocity v = 6.0×10^5m/s
A ball is thrown upward from the top of a building 160 ft high with an initial velocity of 64 ft/sec. Its height in feet above the ground is given by s(t)=−16t2+64t+160 where t
is in seconds.
a) When does the ball reach its maximum height?
b) What is its maximum height?
c) When does it hit the ground?
d) With what velocity does it hit the ground?
e) What was its acceleration at time t=1?
Answer:
2 seconds
5.74165 seconds
-32 ft/s²
119.73 ft/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 32 ft/s²
[tex]s=-16t^2+64t+160[/tex]
[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-64}{-32}\\\Rightarrow t=2\ s[/tex]
The maximum height will be reached in 2 seconds
From the given equation
[tex]s=-16(2)^2+64\times 2+160\\\Rightarrow s=224\ ft[/tex]
The maximum height is 224 ft from the ground
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 224=0t+\frac{1}{2}\times 32\times t^2\\\Rightarrow t=\sqrt{\frac{224\times 2}{32}}\\\Rightarrow t=3.74165\ s[/tex]
Time taken to fall from the maximum height is 3.74165 seconds
Time taken from the point in time the ball is thrown is 2+3.74165=5.74165 seconds
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 32\times 224+0^2}\\\Rightarrow v=119.73\ ft/s[/tex]
Velocity with which the ball will hit the ground is 119.73 ft/s
At t = 1
Acceleration of a body thrown is always i.e., a body in free fall is always 32 ft/s². Here as the ball is thrown up the it will be negative so -32 ft/s²
Light of wavelength λ travels through a medium with an index of refraction n1before striking a thin film with an index of refraction n2 at an angle of incidence of 0. Some of the light is reflected off of and some is transmitted through the n2 thin film. The transmitted light travels a distance t through the thin film before encountering the n1 medium again. Some of the light reflects off of the n1 medium.What is the effective path length difference between the light that reflected off of the n2 medium and the light that reflected off the n1 medium, given that n1>n2?
A. t+λ/n2
B. 2t+λ/n2
C. 2t+λ/(2n2)
D. t+λ/(2n2)
Answer:
option C
Explanation:
The correct answer is option C
A light that transmits through n₂ travels t distance before reflection off the n₁ medium and again travels distance t before reaching the point from where it entered n₂ medium. Hence it travels 2 t distance more than the light that is reflected off n₂.
It( light entering n₂) also travels an additional distance equal to, half of the wavelength, when reflected off n₁ ( as n₁ is greater than n₂).
Wavelength in n₂ is = [tex]\dfrac{\lambda}{n_2}[/tex]
Hence, path length difference = [tex]2t +\dfrac{\lambda}{2 n_2}[/tex]
The effective path length difference between the light that reflected off of the n2 medium and the light that reflected off the n1 medium is
Path length difference = 2t+λ/(2n2)
What is wavelength?Generally, Wavelength is simply defined as the distance between points in the adjacent cycles of a waveform signal channeled through a space.
In conclusion, A light that travels through n2, travels in a time t, a distance before reflection from the n1 channel, travels a distance d before reaching the point from where it entered n2 channel.
Path length difference = 2t+λ/(2n2)
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Technician A says that maintenance-free batteries use a large amount of water. Technician B says that maintenance-free batteries use materials that reduce the amount of water used by the battery. Which technician is correct
a. Technician A
b. Technician B
c. Both A and B
d. Neither A nor B
Answer:
Option B
Explanation:
Maintenance-free battery uses Recombination Electrolyte in order to limit the oxygen and hydrogen formation during charging.
Each plate of the battery uses a glass micro-fiber separator that absorbs the whole liquid electrolyte in its pore thus resulting in no free acid.
As the battery charges to its fullest, the oxygen produced at the positive plate moves via the pores of the separator to the negative plate.
Initially it reacts and forms [tex]PbSO_{4}[/tex], i.e., Lead Sulphate, then on further charging it gets converted to Pb, i.e., lead.
Due to this, the negative plate is not able to reach the correct potential in order to liberate hydrogen, and hence no water formation takes place.
Thus technician B is correct.