As shown in (Figure 1), a layer of water covers a slab of material X in a beaker. A ray of light traveling upwards follows the path indicated.
a) Using the information on the figure, find the index of refraction of material X .
b) Find the angle the light makes with the normal in the air.
Answers
Answer: the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.
Explanation: To find the answer, we need to know the Snell's law.
What is Snell's law of refraction? Using this, how to solve the problem?The Snell's law for refraction can be written as,[tex]\frac{sin (i)}{sin(r)} =\frac{n_r}{n_i}[/tex]
where, i is the incident angle, r is the refracted angle, n is the refractive index.
As we know that the refractive index of water is 1.33For the first case, incident angle from the picture is 65°, and the refracted angle is 48°. Thus, the refractive index of the medium X will be,[tex]\frac{sin 65}{sin48} =\frac{n_w}{n_X} \\n_X=\frac{1.33*sin48}{sin65} =1.09\\[/tex]
In the second case, incident angle is 48° and we have to find the refracted angle r for the air.As we know that the refractive index of air is 1.Thus, the refracted angle will be,[tex]\frac{sin 48}{sin r}=\frac{1}{1.33} \\\\ sin(r)=\frac{1.33*sin 48}{1}=0.988\\\\r=sin^{-1}(0.988)=81.25 degrees.[/tex]
Thus, we can conclude that, the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.
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The light makes an angle of 81.25° with the normal in the air and the index of refraction of material X will be 1.09.
In order to determine the solution, we must understand Snell's law.
What is the refraction law of Snell? How can the issue be resolved with this?One way to express Snell's law for refraction is as follows:[tex]\frac{sin(i)}{sin(r)}=\frac{n_r}{n_i}[/tex]
where the refractive index is n and the incidence angle is i. The refracted angle is r.
As is well known, water has a refractive index of 1.33.The incidence angle from the image in the first case is 65°, while the refracted angle is 48°. Consequently, the medium X's refractive index will be,[tex]n_X=\frac{n_w*sin 48}{sin65} =1.09[/tex]
The incident angle in the second scenario is 48°, and we must determine the refracted angle r for the air.Since we now know that air has a refractive index of 1, so that the refracted angle is,[tex]sin(r)=n_w*sin48=0.988\\r=sin^{-1}(0.988)=81.25 degrees.[/tex]
As a result, we can infer that the material X will have an index of refraction of 1.09 and that the angle the light makes with the normal in the air is 81.25°.
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Related Questions
An object weighs 63.8 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 16.8 N.
1. Determine the density of the object.
2. When the object is immersed in oil, the force scale reads 35.6 N. Calculate the density of the oil.
Answers
Answer:
The density of this object is approximately [tex]1.36\; {\rm kg \cdot L^{-1}}[/tex].
The density of the oil in this question is approximately [tex]0.600\; {\rm kg \cdot L^{-1}}[/tex].
(Assumption: the gravitational field strength is [tex]g =9.806\; {\rm N \cdot kg^{-1}}[/tex])
Explanation:
When the gravitational field strength is [tex]g[/tex], the weight [tex](\text{weight})[/tex] of an object of mass [tex]m[/tex] would be [tex]m\, g[/tex].
Conversely, if the weight of an object is [tex](\text{weight})[/tex] in a gravitational field of strength [tex]g[/tex], the mass [tex]m[/tex] of that object would be [tex]m = (\text{weight}) / g[/tex].
Assuming that [tex]g =9.806\; {\rm N \cdot kg^{-1}}[/tex]. The mass of this [tex]63.8\; {\rm N}[/tex]-object would be:
[tex]\begin{aligned} \text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{63.8\; {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 6.506\; {\rm kg}\end{aligned}[/tex].
When an object is immersed in a liquid, the buoyancy force on that object would be equal to the weight of the liquid that was displaced. For instance, since the object in this question was fully immersed in water, the volume of water displaced would be equal to the volume of this object.
When this object was suspended in water, the buoyancy force on this object was [tex](63.8\; {\rm N} - 16.8\; {\rm N}) = 47.0\; {\rm N}[/tex]. Hence, the weight of water that this object displaced would be [tex]47.0 \; {\rm N}[/tex].
The mass of water displaced would be:
[tex]\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{47.0\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 4.793\; {\rm kg}\end{aligned}[/tex].
The volume of that much water (which this object had displaced) would be:
[tex]\begin{aligned}\text{volume} &= \frac{\text{mass}}{\text{density}} \\ &\approx \frac{4.793\; {\rm kg}}{1.00\; {\rm kg \cdot L^{-1}}} \\ &\approx 4.793\; {\rm L}\end{aligned}[/tex].
Since this object was fully immersed in water, the volume of this object would be equal to the volume of water displaced. Hence, the volume of this object is approximately [tex]4.793\; {\rm L}[/tex].
The mass of this object is [tex]6.50\; {\rm kg}[/tex]. Hence, the density of this object would be:
[tex]\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{6.506\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 1.36\; {\rm kg \cdot L^{-1}} \end{aligned}[/tex].
(Rounded to [tex]\text{$3$ sig. fig.}[/tex])
Similarly, since this object was fully immersed in oil, the volume of oil displaced would be equal to the volume of this object: approximately [tex]4.793\; {\rm L}[/tex].
The weight of oil displaced would be equal to the magnitude of the buoyancy force: [tex]63.8\; {\rm N} - 35.6\; {\rm N} = 28.2\; {\rm N}[/tex].
The mass of that much oil would be:
[tex]\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{28.2\: {\rm N}}{9.806\; {\rm N \cdot kg^{-1}}} \\ &\approx 2.876\; {\rm kg}\end{aligned}[/tex].
Hence, the density of the oil in this question would be:
[tex]\begin{aligned} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &\approx \frac{2.876\; {\rm kg}}{4.793\; {\rm L}} \\ &\approx 0.600\; {\rm kg \cdot L^{-1}} \end{aligned}[/tex].
(Rounded to [tex]\text{$3$ sig. fig.}[/tex])
Calculate the amount of air in a room 6m long, 5m wide and 3mm high.
Answers
Answer:
0.09kg of air
Explanation:
The dimensions of the room are given
change the height to meters by dividing it by thousand.
For the volume multiplying the length,width and height (all should be in the same unit most suitable being meters).
Volume refers to the amount of space inside a box or a object.
The amount of air is equal to the volume.
Answer:
90 m^3
Explanation:
Volume of the room:
6 m * 5 m * 3 m = 90 m^3 <=====( I changed 3mm to 3 m)
if 3mm is not a typo mistake
volume becomes ( 3 mm = .003 m)
6 m * 5 m * .003 m = .09 m^3 ( though unlikely )
The thermal emission of the human body has maximum intensity at a wavelength of approximately 9.5 μm.What photon energy corresponds to this wavelength?
Answers
Answer:
Explanation:
2.1 x 10^2 - 20J
In our solar system, which celestial object is known as the dwarf planet?
Answers
Answer:
unfournatletly
Explanation:
i have no clue sorry to waste ur time ill rather not say a answer that will be incorrect.
Answer:
Pluto
Explanation:
Pluto was part of our solar system till 2006In 2006 international scientific committee removed it from planets listIt's known as dwarf planet nowA long uniform board weighs 52.8 N (10.6 lbs) rests on a support at its mid point. Two children weighing 206.0 N (41.2 lbs) and 272.0 N (54.4 lbs) stand on the board so that the board is balanced.
What is the upward force exerted on the board by the support?
Answers
The upward force exerted on the board by the support is 530.8 N.
Upward force exerted on the board by the supportThe upward force exerted on the board by the support is calculated as follows;
F(up) = 52.8 N + 206.0 N + 272.0 N
F(up) = 530.8 N
Thus, the upward force exerted on the board by the support is 530.8 N.
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Given that the acceleration of gravity at the surface of Mars is 0.38 of what it is on Earth, and that Mars' radius is 3400 km , determine the mass of Mars.
Answers
The mass of the planet Mar, given the data from the question is 6.45×10²³ Kg
Data obtained from the questionAcceleration due to gravity of Earth = 9.8 m/s²Acceleration due to gravity of Mar (g) = 0.38 × 9.8 = 3.724 m/s²Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²Radius (r) = 3400 Km = 3400 × 1000 = 3400000 mMass (M) =? How to determine the mass of Marg = GM / r²
Cross multiply
GM = gr²
Divide both sides by G
M = gr² / G
M = (3.724 × 3400000²) / 6.67×10¯¹¹
M = 6.45×10²³ Kg
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What force causes a bike to move forward?
A. Air resistance
B. Thrust
C. Friction
D. Gravity
Answers
The Answer is Option C. Friction.
Explanation:
The friction force acts in the forward direction on the rear wheel and it acts in the backward direction on the front wheel. The magnitude of friction force on the rear wheel can be more, equal or less than that on the front wheel.
Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A + B to be larger than the magnitude of A - B by the factor n, what must be the angle between them?
Answers
Answer:
[tex]\alpha=arccos[\frac{(a^2+b^2)(n-1)}{2ab(n+1)}].[/tex]
Explanation:
1) for A+B: a²+b²-2abcos(π-α);
2) for A-B: a²+b²-2abcos(α);
3) according to the condition (A+B):(A-B)=n, then
[tex]n=\frac{a^2+b^2-2abcos(\pi-a)}{a^2+b^2-2abcos(a)}; \ = > \ n=\frac{a^2+b^2+2abcos(a)}{a^2+b^2-2abcos(a)}; \ = > \ cos(\alpha)=\frac{(a^2+b^2)(n-1)}{2ab(n-1)}.[/tex]
A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of ρ0 is placed in a container of water. Initially the sphere floats and the water level is marked on the side of the container. What happens to the water level, when the original sphere is replaced with a new sphere which has different physical parameters? Notation: r means the water level rises in the container, f means falls, s means stays the same. Combination answers like 'f or s' are possible answers in some of the cases.
The new sphere has a density of ρ > ρ0 and a mass of m = m0.
The new sphere has a mass of m > m0 and a radius of r = r0.
The new sphere has a radius of r = r0 and a density of ρ > ρ0.
Answers
The correct response for each of the condition given in the questions are,
[tex]d > d_0 , m=m_0[/tex] ⇒ f or s [tex]m > m_0, r=r_0[/tex]⇒ r [tex]r=r_0,d > d_0[/tex] ⇒ rTo find the answer, we have to know about the Archimedes principle.
How to solve the problem for different conditions?The Archimedes principle states that the upthrust F on a body is equal to the weight W of the displaced liquid.The sum of forces must be zero for the sphere to be in equilibrium.[tex]F-W=0\\F=W\\W=mg, where.\\m=density*volume=d*V\\V=\frac{4}{3} \pi r^3[/tex]
Let's apply the idea of density to the body and water now. We are taking d instead of ρ.[tex]d_wVg = d_0(\frac{4}{3}\pi r^3 ) g \\d_wV = d_0(\frac{4}{3}\pi r^3 )[/tex] (1)
Let's examine each example for the initial condition with d₀, m₀, and r₀, where the height of the water is h.Case 1:
The new sphere's mass is m = m₀ .The new sphere's density, d > d₀.Here, the smaller, denser sphere with the same mass, If the sphere floats, the amount of water it displaces will be equal to its mass, which will be the same as the amount of water the original sphere displaces.Consequently, the water level is unchanged.But if the sphere descends, the water displaced is less than the sphere's mass, m = m0, and the level drops, f.Therefore, f or s is the appropriate response.Case 2:
The new sphere's radius, r = r₀, and mass, m > m₀.Consequently, the new sphere is denser than the old one.The right answer is r, because the mass of the water displaced where the sphere floats is m > m0, which is greater than the water displaced for the initial sphere.Case 3:
sphere with the same radius but a higher density.When the right side of equation (1) rises, the left side must also rise in order for the volume to rise and the height to rise as a result (r)Thus, we can conclude that,
[tex]d > d_0 , m=m_0[/tex] ⇒ f or s [tex]m > m_0, r=r_0[/tex]⇒ r [tex]r=r_0,d > d_0[/tex] ⇒ rLearn more about Archimedes principle here:
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Your friend's 10.8 g graduation tassel hangs on a string from his rearview mirror. When he accelerates from a stoplight, the tassel deflects backward toward the rear of the car at an angle of 5.23 ∘ relative to the vertical.
A) Find the tension in the string holding the tassel.
B) At what angle to the vertical will the tension in the string be twice the weight of the tassel?
Answers
The tension in the string holding the tassel and the vertical will the tension in the string
T = 0.1953 NФ = 34.4 °What is the tension in the string holding the tassel. ?Generally, the equation for Tension is mathematically given as
[tex]TCos\theta = mg[/tex]
Therefore
[tex]TCos6.58^{o} = 19.8*10^{-3}*9.8[/tex]
T = 0.1953 N
b).
Where
[tex]T* sin \theta = ma[/tex]
[tex]0.1953*Sin6.58 \textdegree = 19.8*10^{-3}*a[/tex]
a = 1.13 m/s^2
In conclusion
T* sinФ = ma
2msinФ = ma
2sinФ = a
[tex]sin\theta = \frac{a}{2}[/tex]
[tex]\theta = sin^{-1}\frac{a}{2} \\\\\theta= sin^{-1}\frac{1.13}{2}[/tex]
Ф = 34.4 °
In conclusion, The tension in the string holding the tassel and the vertical will the tension in the string
T = 0.1953 N
Ф = 34.4 °
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The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×10^3 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10^-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×10^5 m around the planet Glob. Calculate the speed of the satellite.
Answers
Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Explanation: To find the answer, we need to know about the different equations of planetary motion.
How to find the initial speed of the rock as it left the astronaut's hand?We have the expression for the initial velocity as,[tex]v=\sqrt{2gh}[/tex]
Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,[tex]g_g=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2}= 0.132[/tex]
Now, the velocity will become,[tex]v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s[/tex]
How to find the speed of the satellite?As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,[tex]v=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10^{-11}*7.88*10^{18}}{1.45*10^5} } =3.624km/s[/tex]
Thus, we can conclude that,
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
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The tiny planet known as "Glob" has a radius of 6.32× 10^4 meters and a mass of 7.88× 10^18 kg. On Glob's surface, an astronaut launches a rock straight upward. Before falling back down, the rock rises to a maximum height of 1.44×10^3 m above the planet's surface.
1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
Understanding the planetary motion equations is necessary in order to determine the solution.
How to determine the rock's original speed when it left the astronaut's hand?The starting velocity's expression is as follows:[tex]V=\sqrt{2gh}[/tex]
So, in order to determine v, we must determine the acceleration of glob caused by gravity. We already have,[tex]a=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} \\a=0.132m/s^2[/tex]
The velocity will now change to,[tex]V=\sqrt{2*0.132*1.44*10^3} =19.46m/s[/tex]
How can I determine the satellite's speed?As we are aware, the centripetal force and gravitational force are equivalent, and thus leads to the following satellite speed equation:[tex]v=\sqrt{\frac{GM}{r} } =3,624km/s\\where,\\M=7.88*10^{18}kg[/tex]
Consequently, we can say that
1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
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An object 10mm in height is located 20cm from a crown glass spherical surface whose power is +10.00DS. Locate the image
Answers
The image is present at 20cm from the crown glass spherical surface.
To find the answer, we need to know about the lens formula.
What's the lens formula?It's (1/V)-(1/U)= (1/f)V= image distance from the lens, U= object distance, f= focal length of the lensWhat's the image distance, if object is present at 20cm from crown glass of power 10DS?Focal length (f)= 1/ power = 1/10 = 0.1 m U= -20cm = -0.2m (-ve sign due to sign convection)(1/V)-(-1/0.2)= (1/0.1)=> (1/V)+5=10
=> 1/V= 5
=> V=0.2m = 20cm
Thus, we can conclude that the image is present at 20cm.
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. A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weight to be 7.84 N. She then weighs the crown while it is immersed in water, as shown below in Figure, and now the scale reads 6.86 N. Is the crown made of pure gold? Find the density of the crown and compare it to the density of gold. wag PODECID LE B water
Answers
Answer:
Average density of the crown: approximately [tex]8\; {\rm g \cdot mL^{-1}}[/tex].
Hence, if this crown contains no empty space, this crown is not made of pure gold.
Explanation:
Let [tex]m(\text{crown})[/tex] and [tex]V(\text{crown})[/tex] denote the mass and volume of this crown. Let [tex]g[/tex] denote the gravitational field strength.
Since this crown is fully immersed in water, the volume of water displaced [tex]V(\text{water, displaced})[/tex] is equal to the volume of this crown:
[tex]V(\text{water, displaced}) = V(\text{crown})[/tex].
The mass of water displaced would be:
[tex]\begin{aligned}m(\text{water, displaced}) &= \rho(\text{water}) \, V(\text{water, displaced}) \\ &= \rho(\text{water}) \, V(\text{crown})\end{aligned}[/tex].
The weight of water displaced would be [tex]m(\text{water, displaced})\, g = \rho(\text{water}) \, V(\text{crown})\, g\end{aligned}[/tex].
The buoyancy force on this crown is equal to the weight of water that this crown displaced:
[tex]F(\text{buoyancy}) = \rho(\text{water}) \, V(\text{crown})\, g[/tex].
The magnitude of this buoyancy force is [tex]7.84\; {\rm N} - 6.86\; {\rm N} = 0.98\; {\rm N}[/tex]. Rearrange the equation for buoyancy to find [tex]V(\text{crown})[/tex]:
[tex]\begin{aligned} V(\text{crown}) &= \frac{F(\text{buoyancy}) }{\rho(\text{water}) \, g}\end{aligned}[/tex].
Since the weight of this crown is [tex]\text{weight}(\text{crown}) = m(\text{crown})\, g[/tex], the mass of this crown would be [tex]m(\text{crown})= \text{weight}(\text{crown}) / g[/tex].
The average density of this crown would be:
[tex]\begin{aligned}\rho(\text{crown}) &= \frac{m(\text{crown})}{V(\text{crown})} \\ &= \frac{\text{weight}(\text{crown}) / g}{F(\text{buoyancy}) / (\rho(\text{water})\, g)} \\ &= \frac{\text{weight}(\text{crown})}{F(\text{buoyancy})}\, \rho(\text{water}) \\ &= \frac{7.84\; {\rm N}}{0.98\; {\rm N}}\times 1.000 \; {\rm g\cdot mL^{-1}} \\ &= 8.0\; {\rm g \cdot mL^{-1}}\end{aligned}[/tex].
The density of pure gold is significantly higher than [tex]8.0\; {\rm g\cdot mL^{-1}}[/tex]. Hence, if this crown contains no empty space (i.e., no air bubble within the crown), the crown would not be made of pure gold.
On planet Zog, Mr. Spock measures that it takes 1.38 s for a mass of 0.5 kg to hit the ground when released from rest from a height of 2.85 m.
1. Calculate the size of the acceleration of gravity on that planet.
2. He decides to repeat the experiment. Calculate the work he must do to move the mass from the ground back up to its initial height.
Answers
The acceleration due to gravity is 3 m/s^2 and the work done is -4.3 J.
What is the acceleration due to gravity?Now we must use the formula;
h = ut + 1/2gt^2
Since it was dropped from a height u = 0 m/s
h = height
u = initial velocity
g = acceleration due to gravity
t = time
h = 1/2gt^2
g = 2h/t^2
g = 2 * 2.85 /(1.38)^2
g = 5.7/1.9
g = 3 m/s^2
The work that must be done is against gravity hence;
W = -(mgh)
W = - (0.5 kg * 3 m/s^2 * 2.85 m)
W = - 4.3 J
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A pendulum in motion can either swing from side to side or turn in a continuous circle. The point at which it goes from one type of motion to the other is called the separatrix, and this can be calculated in most simple situations. When the pendulum is prodded at an almost constant rate though, the mathematics falls apart. Is there an equation that can describe that kind of separatrix?
Answers
The kind of equation that can be used to differentiate the kind of separatrix that shows change on motion is
H = 2g/l.
What is simple pendulum?A simple pendulum can be defined as the equipment that displays an oscillatory motion when a mass is tied on a rope and is suspended from it.
The various movements that occur using a simple pendulum is translational ( side to side) or continuous circle (oscillatory motion).
The equation that show that a change from one type of motion to another is H = 2g/l.
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A man exerts a horizontal force of 123 N on a crate with a mass of 40.2 kg.
(A) If the crate doesn't move, what's the magnitude of the static friction force (in N)?
______N
(B) What is the minimum possible value of the coefficient of static friction between the crate and the floor? (Assume the crate remains stationary.)?
Answers
The magnitude of the static friction force = 123 N
The coefficient of static friction = 0.31
What is static friction?Static friction is the frictional force that must be overcome in other for a body to that moving over another.
Since the crate does not move, the magnitude of the static frictional force is equal to the applied force.
The magnitude of the static frictional force = 123 N
The coefficient of static friction = frictional force/normal reaction
The coefficient of static friction = 123/(40* 9.8)
The coefficient of static friction = 0.31
In conclusion, the static frictional force on the on the crate is equal to the applied force on the crate.
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A space craft is moving relative to the earth , an observer on the earth finds that, between 1pm and 2pm according to her clock, 3601 seconds elapse on the space craft clock . What is the space craft speed relative to the earth?c=2.998×10^8ms
Answers
The speed of the space craft relative to the earth is given as: 0.024c. This is solved using the the equation for time dilation.
What is time dilation?
Time dilation is the "slowing down" of a clock as determined by an observer in relative motion with regard to that clock under the theory of special relativity.
The formula is given as :
Δt = [Δr]/ √ 1 - (v²/c²)
Thus,
v = c√1 - (Δr/Δt)²
= c √(1 - (3600/3601)²
v = 0.024c
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Calculate the magnitude of the electric field at one corner of a square 1.10 m on a side if the other three corners are occupied by 4.05×10−6 C charges. What is the direction of the electric field?
Answers
The direction of magnetic field is south-east and the magnitude is
[tex]23.66[/tex] × [tex]10^{3} N/C[/tex].
Here, the magnitude of CD and BC will be cancelled, as they both are in the opposite direction and equal to each other.
the magnitude, towards the diagonal AC will result in CP, which is the direction of the electric field.
magnitude of electric field can be defined as :- The force per charge on the test charge is a straight forward way to define the size of the electric field.
To find the magnitude of the electric field use the formula
[tex]E = kq/ r^{2}[/tex]
inserting the values,
[tex]E = 9. 10^{9}[/tex] × [tex]4.05[/tex] × [tex]10^{-6} / 1.1 \sqrt{2}[/tex]
[tex]E= 36.45[/tex] × [tex]10^{3}[/tex] / [tex]1.54[/tex]
[tex]E = 23.66[/tex] × [tex]10^{3}[/tex] N/C
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A 1300 kg steel beam is supported by two ropes. (Figure
1)
What is the tension in rope 1?
What is the tension in rope 2?
Answers
Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.
By Newton's second law,
the net horizontal force acting on the beam is[tex]R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0[/tex]
where [tex]R_1,R_2[/tex] are the magnitudes of the tensions in ropes 1 and 2, respectively;
the net vertical force acting on the beam is[tex]R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0[/tex]
where [tex]m=1300\,\rm kg[/tex] and [tex]g=9.8\frac{\rm m}{\mathrm s^2}[/tex].
Eliminating [tex]R_2[/tex], we have
[tex]\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)[/tex]
[tex]R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2[/tex]
[tex]R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2[/tex]
[tex]-R_1 \sin(50^\circ) = -\dfrac{mg}2[/tex]
[tex]R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}[/tex]
Solve for [tex]R_2[/tex].
[tex]\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0[/tex]
[tex]\dfrac{R_2}2 = -mg\cot(110^\circ)[/tex]
[tex]R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}[/tex]
Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N.
Calculate the magnitude of the tension in the string marked A. (You'll need to get the various positions from the graph. The ends of the strings are exactly on one of the tic marks.)
Answers
The magnitude of the tension in the string marked A is 52.5N
Generally, the equation for is mathematically given as
Let's take θ be an angle at A
So, tanθ = 3/8
Let's take α be an angle at B (Below X)
tanα = 5/4
Let's take β be an angle at C (Below x)
tanβ = 1/6
First we take the Horizontal Components
74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)
By solving the equation, we get
A = 78.9 - 0.668B … (1)
Now, we take the vertical components
74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)
By solving the equation, we get
40.07 = 1.015B
B = 39.5N
By substituting the value of B in equation (1)
A = 78.9 - 0.6668× 39.5
A = 52.5N
Hence, the magnitude of the tension in the string marked A is 52.5N
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M1 is a spherical mass (25.0 kg) at the origin. M2 is also a spherical mass (10.6 kg) and is located on the x-axis at x = 94.8 m. At what value of x would a third mass with a 19.0 kg mass experience no net gravitational force due to M1 and M2?
Answers
The point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.
Position of the third mass
m1<------(x)------> m3 <-----------(94.8 m - x)-------->m2
a point, x, where m3 experiences a zero net gravitational force due to M1 and m2;
Force on m3 due to m1 = Force on m3 due to m2
Gm1m3/d² = Gm2m3/r²
m1/d² = m2/r²
where;
d is the distance between m1 and m3 = xr is the distance between m3 and m2 = 94.8 - xm1/(x²) = m2/(94.8 - x)²
m1(94.8 - x)² = m2x²
(94.8 - x)² = (m2/m1)x²
(94.8 - x)² = (10.6/25)x²
(94.8 - x)² = 0.424x²
(94.8 - x)² = (0.651)²x²
94.8 - x = 0.651x
94.8 = 1.651x
x = 94.8/1.651
x = 57.42 m
Thus, the point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.
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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
Answers
The vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
Tension in the cable
Apply the principle of moment and calculate the tension in the cable;
Clockwise torque = TL sinθ
Anticlockwise torque = ¹/₂WL
TL sinθ = ¹/₂WL
T sinθ = ¹/₂W
T = (W)/(2 sinθ)
T = (29 x 9.8)/(2 x sin57)
T = 169.43 N
Vertical component of the forceT + F = W
F = W - T
F = (9.8 x 29) - 169.43
F = 114.77 N
Thus, the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
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100gm o2 gas is pressurized to 20 degree Celsius. done Also, how much heat energy will be converted into mechanical energy?
Answers
The heat energy that will be converted into mechanical energy is 1.83 kJ.
Heat capacity of the O2 gas
The heat energy that will be converted into mechanical energy is calculated as follows;
Q = mcΔθ
where;
m is mass = 100 g = 0.1 kgΔθ is change in temperaturec is specific heat capacityat 20 ⁰C = 293 K, C = 0.915 kJ/kg K
Q = (0.1 kg)(0.915)(20 )
Q = 1.83 kJ
Thus, the heat energy that will be converted into mechanical energy is 1.83 kJ.
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A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in Figure 4.29(b). The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
Figure 4.29
(a) Calculate the minimum force F he must exert to get the block moving.
40.873
(b) What is its acceleration once it starts to move, if that force is maintained?
m/s2
Answers
(a) The minimum force F he must exert to get the block moving is 38.9 N.
(b) The acceleration of the block is 0.79 m/s².
Minimum force to be applied
The minimum force F he must exert to get the block moving is calculated as follows;
Fcosθ = μ(s)Fₙ
Fcosθ = μ(s)mg
where;
μ(s) is coefficient of static frictionm is mass of the blockg is acceleration due to gravityF = [0.1(36)(9.8)] / [(cos(25)]
F = 38.9 N
Acceleration of the blockF(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32
a = F(net)/m
a = 28.32/36
a = 0.79 m/s²
Thus, the minimum force F he must exert to get the block moving is 38.9 N.
The acceleration of the block is 0.79 m/s².
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For the circuit shown below with a 12.0 V battery, find the total current through the battery and the current through each resistor with the following resistances: R1=5.0 Ohms, R2=10.0 Ohms, R3=15.0 Ohms, R4=33.0 Ohms. Show all your work!
Answers
The total current through the battery is 4.77 A.
The current through resistor R1 is 2.4 A.
The current through resistor R2 is 1.2 A
The current through resistor R3 is 0.8 A
The current through resistor R4 is 0.36 A.
Total resistance of the circuitThe total resistance of the circuit is calculated as follows;
1/Rt = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄
1/Rt = 1/5 + 1/10 + 1/15 + 1/33
1/Rt = 0.397
Rt = 1/0.397
Rt = 2.518 ohms
Total current flowing in the circuitI = V/Rt
I = 12/2.518
I = 4.77 A
Current in resistor R₁I₁ = V/R₁
I₁ = 12/5
I₁ = 2.4 A
Current in resistor R₂I₂ = 12/10
I₂ = 1.2 A
Current in resistor R₃I₃ = 12/15
I₃ = 0.8 A
Current in resistor R₄I₄ = 12/33
I₄ = 0.36 A
Thus, the total current through the battery is 4.77 A.
The current through resistor R1 is 2.4 A.
The current through resistor R2 is 1.2 A
The current through resistor R3 is 0.8 A
The current through resistor R4 is 0.36 A.
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6. Determine the number of significant figure of:
a. 0.2001 b. 2.0000 c. 243 d. 0.00010300
Answers
Answer:
b
Explanation:
the answer is b because if you do the math it equals to b
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.17 times a second. A tack is stuck in the tire at a distance of 0.351 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed
Answers
The tangential speed of the wheel is determined as 4.786 m/s.
Tangential speed of the wheel
The tangential speed of the wheel is calculated as follows;
v = ωr
where;
ω is angular speed in rad/sr is radius of the circular pathv = (2.17 x 2π rad)/s x 0.351 m
v = 4.786 m/s
Thus, the tangential speed of the wheel is determined as 4.786 m/s.
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the temperature at which the velocity of sound in air is twice its velocity at 15°C
Answers
With the use of below formula, at 879 °C, velocity will be double the velocity at 15 °C.
What is the relationship between Velocity and sound ?The velocity of sound waves in air is proportional to the square root of Thermodynamic temperature. That is, V = K[tex]\sqrt{T}[/tex]
Given that the temperature at which the velocity of sound in air is twice its velocity at 15°C, Let us make use of the formula;
(v2/v1) = √(T2 / T1)
Where
v2 = final velocityv1 = initial velocityT2 = final absolute temperatureT1 = initial temperature.Recall that absolute temperature = °C + 273.
If v2 = 2 × v1 and temperature in degree Celsius = 15°C, then,
Temperature in Kelvin K = 15 + 273 = 288
Substitute all the parameters into the formula
(2 × v1)/v1 = √(T2/288)
2 = √ (T2 /288)
Square both sides
4 = (T2/288)
T2 = 4 × 288
T2 = 1152K
Temperature in degrees Celsius = 1152 - 273 = 879 °C.
Therefore, at 879 °C, velocity will be double the velocity at 15 °C.
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Thanks to the direction finding feature in gmaps application, which most of us use, we can find our way. Here, the maps application offers us alternative routes. Among these suggestions, I want to choose the path that will have the least fuel and do this based on calculations. For example, one of the two directions may be short, but if that short route is also uphill, it will not be an economical route. In my opinion the most important factor is elevation. If we take elevation into account other factors such as friction, where assuming the same asphalt type is often used in the same area for friction, I think the correct result will be achieved. In your opinion, what are the input data required to find the least energy path, what assumptions can be made and what are the necessary formulations and calculations?
Answers
In my opinion, I think that the input data that are required to find the least energy path are:
ElevationDistanceWhat is a Map?This refers to the use of a diagram to represent the features of a place that shows its physical landforms to help in navigation.
Hence, we can see that when using maps like gmaps, it is important to consider both elevation and distance to be able to find the path that uses the least energy.
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How many bytes of memory space are there in an 80 GB hard disk and 256 GB card?
Answers
Answer:
80GB= 80000000000 bytes
256GB= 274877906944 bytes
Explanation:80GB= 80000000000 bytes
256GB= 274877906944 bytes