The Aster's final position from her initial position is 64 m
The angle is 300° and She has to head in West north direction to return to her initial position
What is Displacement ?The displacement is the distance travelled in a specific direction. Displacement is a vector quantity.
Given that Aster walks first 70 m in the direction 37° north of east, and then walks 82 m in the direction 20° south of east, and finally walks 28 m in the direction 30° west of north.
This can be solved by using bearing method. Cosine formula will be the best to solve for the distance D.
[tex]D^{2}[/tex] = [tex]70^{2}[/tex] + [tex]82^{2}[/tex] - 70 x 82 x Cos (37 + 20)
[tex]D^{2}[/tex] = 4900 + 6724 - 5740Cos57
[tex]D^{2}[/tex] = 11624 - 3126.23
[tex]D^{2}[/tex] = 8497.8
D = [tex]\sqrt{8497.8}[/tex]
D = 92.2 m
a) The Aster's final position from her initial position is 92.2 - 28 = 64 m
The angle = 270° + 30° = 300°
b) She has to head in West north direction to return to her initial position
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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?
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]
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?
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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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.
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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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.)
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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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
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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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.
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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the temperature at which the velocity of sound in air is twice its velocity at 15°C
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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An object 10mm in height is located 20cm from a crown glass spherical surface whose power is +10.00DS. Locate the image
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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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?
Answer:
Explanation:
2.1 x 10^2 - 20J
100gm o2 gas is pressurized to 20 degree Celsius. done Also, how much heat energy will be converted into mechanical energy?
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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How many bytes of memory space are there in an 80 GB hard disk and 256 GB card?
Answer:
80GB= 80000000000 bytes
256GB= 274877906944 bytes
Explanation:80GB= 80000000000 bytes
256GB= 274877906944 bytes
Tuning an Instrument. A musician tunes the C-string of her instrumeut to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 m long and has a mass of 14.4 g. (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from C to D?
(a) The tension the musician must stretch it is 147.82 N.
(b) The percent increase in tension is needed to increase the frequency is 26%.
Tension in the stringv = √T/μ
where;
v is speed of the waveT is tensionμ is mass per unit length = 0.0144 kg / 0.6 m = 0.024 kg/mv = Fλ
in fundamental mode, v = F(2L)
v = 2FL
v = 2 x 65.4 x 0.6 = 78.48 m/s
v = √T/μ
v² = T/μ
T = μv²
T = 0.024 x (78.48)²
T = 147.82 N
When the frequency is 73.4 Hz;v = 2FL = 2 x 73.4 x 0.6 = 88.08 m/s
T = μv²
T = (0.02)(88.08)²
T = 186.19 N
Increase in the tension= (186.19 - 147.82)/(147.82)
= 0.26
= 0.26 x 100%
= 26 %
Thus, the tension the musician must stretch it is 147.82 N.
The percent increase in tension is needed to increase the frequency is 26%.
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A 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?
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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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?
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 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
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.
An electron gun shoots electrons at a metal plate that is 4.0 mm away in a vacuum. The plate is 5.0 V lower in potential than the gun. How fast must the electrons be moving as they leave the gun if they are to reach the plate?
The speed of the electron is 1.3 * 10^6 m/s
What is the velocity?We know that when the electron gun is shot, the potential energy of the electron is converted into kinetic energy. The mass of the electron is given as 9.11 * 10^-31 Kg.
The energy of the electron is;
eV = 1e * 5V = ev or 8 * 10^-19 J
Given that E = 1/2mv^2
8 * 10^-19 = 0.5 * 9.11 * 10^-31 * v^2
v = √ 8 * 10^-19/0.5 * 9.11 * 10^-31
v = √1.75 * 10^12
v = 1.3 * 10^6 m/s
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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?
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 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.
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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In our solar system, which celestial object is known as the dwarf planet?
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 now
light travel
3, 00,000 km/s. Is it velocity or speed?
What force causes a bike to move forward?
A. Air resistance
B. Thrust
C. Friction
D. Gravity
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.
6. Determine the number of significant figure of:
a. 0.2001 b. 2.0000 c. 243 d. 0.00010300
Answer:
b
Explanation:
the answer is b because if you do the math it equals to b
Calculate the amount of air in a room 6m long, 5m wide and 3mm high.
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 )
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!
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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The head of a hammer with a mass of 1.5 kg is allowed to fall onto a nail from a height of 0.70 m .
What is the maximum amount of work it could do on the nail?
Express your answer to two significant figures and include the appropriate units.
Why do people not just "let it fall" but add their own force to the hammer as it falls?
a)To make the trajectory of the fall more precise.
b)To increase the power of the hammer.
Maximum work done by the hammer on the is 10 J.
Force is added to the hammer in order to increase the power of the hammer; option B.
What is work done?Work done is defined as the product of force and the distance travelled by force.
Energy is used to do work.
The potential energy of the hammer is converted to work done.
Potential energy = mghThe maximum amount of work it could do on the nail is given below:
Maximum work = 1.5 * 9.81 * 0.7
Maximum work = 10 J
Force is added to the hammer in order to increase the power of the hammer.
In conclusion, the work done by the hammer is obtained from the potential energy of the hammer.
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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.)?
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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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?
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]
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.
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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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.
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])
A 1.0 mal sample of an ideal pas is kept at 0.0°c during an expansion from Bradba xa 3.0 litre to 10.0 litre a) How much work is done on the pas during the expansion? b) How much everly transfer by heat occurs with the surroundings in this Process?
The work done during expansion is 2.73×10³J and the heat transferred to the surrounding is -2.73×10³ J
What is isothermal process ?The thermodynamic process which takes place in constant temperature of the system is called isothermal process.
What is work done in isothermal process?Due to change in volume of the system at constant temperature, some work is done .As, the temperature of gas is fixed. This process is isothermal.Now, work done (W) = nRT ln ( V2/V1)where W = work done by the systemn= number of moles
R= universal gas constant
V2= volume of the system after expansion
V1= volume of the system before expansion
W= (1)(8.31)(273)ln ( 10/3)= 2.73×10³ J.
Also in isothermal process the amount of work done by the system = -( the amount of heat transferred to the surrounding)So the amount of heat transferred is -2.73×10³ JThus, we can conclude that a) the amount of work done is 2.73×10³J and b) the amount of heat transferred to the surrounding is -(2.73×10³J).
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