Answer:
I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43
It could have any direction of
θ = (225 - 180) ± arcsin(13/30)
θ = 45 ± 25.679...
70.679 ≤ θ ≤ 19.321
components of vector B would be
Bx = |B|cosθ
By = |B|sinθ
My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
What is magnitude of the resultant?IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43
It could have any direction of
θ = (225 - 180) ± arcsin(13/30)
θ = 45 ± 25.679...
70.679 ≤ θ ≤ 19.321
components of vector B would be
Bx = |B|cosθ
By = |B|sinθ
My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird. A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.
Therefore, My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.
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WILL MARK BRAINLIEST
1. What is the difference between distance and displacement?
A. Distance is measured in feet. Displacement is measured in meters
B. Distance is the difference between how quickly you move between two points. Displacement is the average of that motion
C. Distance is a measure of length of travel. Displacement is the measure of the time it takes to move that length
D. Distance is the length of travel. Displacement is the straight line distance between the starting point and the ending point
2. When forces are in opposite directions, they subtract from one another
A. True
B. False
3. When two forces are in opposite directions, and they are the exact same magnitude, the forces will
A. Subtract from each other
B. Cancel Out
C. Go on Infinitely
D. eventually reach equilibrium
4. Acceleration is a change in motion over time
True or False
Answer:
1. C
2. True
3. D
4. True
Explanation:
While distance refers to “how much land an object has traversed throughout its journey,” displacement measures “how far an object is out of place.” In this post, let's look at the difference between distance and displacement. Thus, option C is correct.
What are the distance and displacement?Displacement and distance are two distinct ideas. The total distance travelled is larger than the displacement between those two points if an object changes direction while travelling.
Distance is the length of any path connecting any two places. Displacement is the straight-line distance between any two points when calculated along the shortest route.
Therefore, The direction is ignored when calculating distance. The direction is accounted for in the displacement calculation.
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Describe all the ways a bicyclist can accelerate
Which processes result in the release of carbon? Select three options.
Animals break down food molecules to obtain energy.
The remains of producers are broken down by decomposers.
The remains of consumers are broken down by soil decomposers.
Producers take in carbon dioxide.
Producers make sugars and starches.
Answer:
The remains of producers are broken down by decomposers.
The remains of consumers are broken down by soil decomposers.
Animals break down food molecules to obtain energy.
Explanation:
Answer:
It is A. B. and C.
Explanation:
Hope this helps
A 12.0 kg box resting on a horizontal, frictionless surface is attached to a 5.00 kg weight by a thin, light wire that passes over a frictionless pulley (Figure 1). The pulley has the shape of a uniform solid disk of mass 2.60 kg and diameter 0.500 m. A) After the system is released, find the tension in the horizontal segment of the wire. B) After the system is released, find the tension in the vertical segment of the wire. C) After the system is released, find the x-component of the acceleration of the box. D) After the system is released, find the x- and y-components of the force that the axle exerts on the pulley.
Answer:
Explanation:
Assuming x direction is horizontal any y direction is vertical
As the moment of inertia of the pulley is I = (½)mR²
The pulley mass will appear as ½ its actual mass in the system.
F = ma
5.00(9.81) = (5.00 + 2.60/2 + 12.0)a
a = 49.05 / 18.3
a = 2.6803278...
a = 2.68 m/s²
A) T = 12.0(2.68) = 32.16... = 32.2 N
B) T = 5.00(9.81 - 2.68) = 35.65 = 35.7 N
C) aₓ = 2.68 m/s²
D) Fx = 32.2 N
Fy = 35.7 + 2.60(9.81) = 61.2 N
A)The tension in the horizontal segment of the wire will be 32.2 N
B)The tension in the vertical segment of the wire will be 35.7 N
What is acceleration?Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².It is a vector quantity. It requires both magnitudes as well as direction to define.
The given data in the problem is;
m is the mass box resting on a horizontal =12.0 kg
the uniform solid disk of mass of 2.60 kg
d is the diameter = 0.500 m.
A)The tension in the horizontal segment of the wire will be 32.2 N
The acceleration is found as;
[tex]\rm F = ma \\\\ a= \frac{49.05}{18.3} \\\\ \rm a= 2.68 \ m/sec^2[/tex]
The tension in the horizontal segment of the wire is found as;
T=ma
T= 12.0 ×2.68
T=32.2 N
Hence the tension in the horizontal segment of the wire will be 32.2 N
B)The tension in the vertical segment of the wire will be 35.7 N
T=m(g-a )
T= 5.00 × (9.8-2.68)
T=35.7 N
Hence the tension in the vertical segment of the wire will be 35.7 N
C) the x-component of the acceleration of the box. will be 2.68 m/s².
aₓ = 2.68 m/s²
D)The x- and y-components of the force that the axle exerts on the pulley will be 32.2 and 61.2 N respectively.
Fx = 32.2 N
Fy = 35.7 + 2.60(9.81) = 61.2 N
Hence the x- and y-components of the force that the axle exerts on the pulley will be 32.2 and 61.2 N respectively.
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You have a 25 W and a 100 W bulb at home. When you connect one or the other, which bulb carries the greater current
Answer:
100 w bulb has more current.
Explanation:
P=V^2/R.
when velocity is constant power is inversly proportional to resistence, so 25 W will have an hogher resistance tjan a 100 w bulb.
P=VI
when v is constant, power is directly proportional to I. hence, 100 w bulbs will carry more current.
When you connect one or the other, the bulb that's connected carries more current than the one that's not connected.
When you connect BOTH of them, the 100W bulb carries more current than the 25W one.
A body at higher temperature contains more heat? Is this true ?
What will be the speed of these waves (in terms of V) if we increase M by a factor of 18.0, which stretches the wire to double its original length
The speed of the chord wave allows finding a new speed when the mass is increased and the chord length is:
The velocity is: v = [tex]\frac{v_o}{3}[/tex]
A wave is a periodic movement of the particles that carries energy, but not matter, the speed of the waves is related to the properties of the medium by the relationship.
[tex]v = \sqrt{\frac{T}{\mu } }[/tex]
Where v is the speed of the wave, T the force and μ is the linear density.
Indicates that the applied mass increases by a factor of 18.0 and the length of the head is increased to twice the original.
The linear density of the cable is
[tex]\mu = \frac{m}{l}[/tex]
Where m is the mass of the cable and l is the length.
Let's use the subscript "o" for the initial conditions.
M = 18.0 m₀
l = 2 l₀
We look for the density.
[tex]\mu = \frac{18.0 m_o}{2.0 l_o}[/tex]mu = 18.0 mo / 2 lo
[tex]\mu = 9.0 \mu_o[/tex]
We substitute in the expression for the velocity assuming that the tension is kept constant.
[tex]v= \sqrt{\frac{T}{9.0 \mu} }[/tex]
[tex]v= \frac{v_o}{3}[/tex]
In conclusion, using the speed of the chord wave we can find a new speed when the mass is increased and the chord length is:
The velocity is: v = [tex]\frac{v_o}{3}[/tex]
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Another box of samples is hoisted up by the same rope. If the rope is shaken with the same frequency as before, and the wavelength is found to be 7.9 m , what is the mass of this box of samples
The motion of the rope which is perpendicular to the direction of the
propagation of the wave is a transverse wave motion.
The mass of the box is approximately 9.93 kgReasons:
The given function for the wave speed is presented as follows;
[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]
Where;
[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]
Taking the mass of the rope as, m = 2.00 kg
The length of the rope, L = 80.0 m
The mass hanging on the rope, M = 20.0 kg
We have;
T = 20.0 kg × 9.81 m/s² = 196.2 N
[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]
Therefore;
Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;
v = f × λ
Therefore;
v = 7.9 Hz × 7.9 m = 62.41 m/s
Which gives;
[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]
T = 62.41² × 0.025 = 97.3752025
[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]
Where;
g = The acceleration due to gravity which is approximately 9.81 m/s²
[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]
Therefore;
The mass of the box, m ≈ 9.93 kg
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The parameters obtained from a similar question online are;
[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]
Length of the rope, L = 80.0 m
Mass of the rope, m = 2.0 kg
Frequency of a point on the rope, f = 20 Hz
How did the study of the atom contribute to our understanding of the periodic table of the elements? (1 point) Atoms are representative of elements, so scientists scaled up O atomic char- derstand elemental characteristics, allowing sc Highlight e elements in a periodic table. The determination of electron charge led to an understanding of O how atoms interact with one another, which facilitated the organization of the periodic table. Elements are made of atoms, so understanding atoms provided O information about elements, which led to the organization of the periodic table. Experiments that identified characteristics of atoms provided O scientists with atomic weights and atomic numbers, which were used to organize the periodic table.
The study of the atom contribute d to our understanding of the periodic table of the elements by virtue of the following;
Elements are made of atoms, so understanding atoms provided information about elements, which led to the organization of the periodic tableExperiments that identified characteristics of atoms provided scientists with atomic weights and atomic numbers, which were used to organize the periodic table.As we know; the periodic table is an array of elements in order of their atomic number.
In which case, the periodic table is made up of 7 rows otherwise called Periods and 8 columns otherwise called Groups.
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Easiest way to Find fahrenhiet to celsius please i need necessary 20 for the fastest correct answer
Answer:
thx for the points
Explanation:
no need brainliest
A tow truck pulls a car 5.00 km along a horizontal roadway using a cable having a tension of 850 N. (a.) How much work does the cable do on the car if it pulls horizontally? If it pulls at 35.0° above the horizontal? (b.) How much work does the cable do on the tow truck in both cases of part (a)? (c.) How much work does gravity do on the car in part (a)?
I think 1980is the answer because you add???
Identify:
In each case the forces are constant and the displacement is along a straight line, so
[tex]$$W=F s \cos \phi \text {. }$$[/tex]
Set-Up:
In part (a), when the cable pulls horizontally [tex]$\phi=0^{\circ}$[/tex] and when it pulls [tex]$35.0^{\circ}$[/tex] above the horizontal [tex]$\phi=35.0^{\circ}$[/tex]. In part (b), if the cable pulls horizontally[tex]$\phi=180^{\circ}$[/tex]. If the cable pulls on the car [tex]$35.0^{\circ}$[/tex] above the horizontal it pulls on the truck at below the horizontal and [tex]$\phi=145.0^{\circ}$[/tex]. For the gravity force [tex]$\phi=90^{\circ}$[/tex], since the force is vertical and the displacement is horizontal.
Execute:
(a) When the cable is horizontal, [tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 0^{\circ}=4.25 \times 10^{6} \mathrm{~J}$[/tex].
When the cable is[tex]$35.0^{\circ}$[/tex] above the horizontal,[tex]$W=(850 \mathrm{~N})\left(5.00 \times 10^{3} \mathrm{~m}\right) \cos 35.0^{\circ}=3.48 \times 10^{6} \mathrm{~J}$[/tex].
(b)[tex]$\cos 180^{\circ}=-\cos 0^{\circ}$[/tex] and [tex]$\cos 145.0^{\circ}=-\cos 35.0^{\circ}$[/tex],
So the answers are [tex]$-4.26 \times 10^{6} \mathrm{~J}$[/tex] and [tex]$-3.48 \times 10^{6} \mathrm{~J}$[/tex].
(c) Since [tex]$\cos \phi=\cos 90^{\circ}=0, W=0$[/tex] in both cases.
Evaluate: If the car and truck are taken together as the system, the tension in the cable does no net wnetwork
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A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it. (a) What is the force constant of the spring
Answer:
Explanation:
The change in force = 9.81(1.95 - 0.300) = 16.2 N
The change in length is 0.750 - 0.200 = 0.550 m
K = ΔF/Δx = 16.2/0.550 = 29.4 N/m
How much force must be applied to push a 1.35 kg book across the desk at constant speed if the coefficient of sliding friction is 0.30?
The magnitude of the force that must be applied to push the book across the desk is 3.97 N.
The given parameters;
mass, m = 1.35 kgcoefficient of friction, μ = 0.3The acceleration of the book across the desk is calculated as follows;
a = μg
where;
g is acceleration due to gravitya = 0.3 x 9.8
a = 2.94 m/s²
The magnitude of the force that must be applied is calculated as follows;
F = ma
F = 1.35 x 2.94
F = 3.97 N
Thus, the magnitude of the force that must be applied to push the book across the desk is 3.97 N.
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Two parallel plates, separated by 0.20 m, are connected to a 12-V battery. An electron released from rest at a location 0.10 m from the negative plate. When the electron arrives at a distance 0.050 m from the positive plate, what is the potential difference between the initial and final points
The potential difference between the initial and final point is 3.0 V.
The given parameters:
distance between the plates, d = 0.2 mvoltage across the plates, V = 12 Vposition of the electron from negative plate, x₁ = 0.1 mposition of the electron from the positive plate, x₂ = 0.0 5mThe potential difference between the initial and final point is calculated as follows;
[tex]E = \frac{V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2}[/tex]
where:
[tex]d_2[/tex] is the distance of the electron between the positive and negative plate
[tex]0.1 + d_2 + 0.05 = 0.2\\\\d_2 + 0.15 = 0.2\\\\d_2 = 0.2 - 0.15\\\\d_2 = 0.05 \ m[/tex]
[tex]V_2 = \frac{V_1d_2}{d_1} \\\\V_2 = \frac{12 \times 0.05}{0.2} \\\\V_2 = 3.0 \ V[/tex]
Thus, the potential difference between the initial and final point is 3.0 V.
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what is translator disadvantage?
Machine translators:
Level of accuracy can be very low.
Accuracy is also very inconsistent across different languages.
Machines can't translate context.
Mistakes are sometimes costly.
Sometimes translation simply doesn't work.
Human translators:
Turnaround time is longer.
Translators rarely work for free.
Unless you use a translation agency, with access to thousands of translators, you're limited to the languages any one translator can work with.
A power plant running at 31 % efficiency generates 270 MW of electric power. Part A At what rate (in MW) is heat energy exhausted to the river that cools the plant
The rate of heat energy exhausted to the river is 600.96 MW.
What is efficiency?The ratio of usable output to total input can be used to objectively measure efficiency. The efficiency of the device is defined as the ratio of energy converted to a useable form to the original amount of energy supplied.
Given parameters:
Efficiency of the power plant; η = 31 %
Output electric power; O = 270 MW.
We know that, Efficiency of the power plant;
η = (Output electric power/ input power)× 100%
⇒ input power = (Output electric power × 100)/η
⇒ input power = (270 × 100)/31 MW
= 870.96 MW.
So, the rate of heat energy exhausted to the river that cools the plant = Input power- output power
= (870.96 - 270) MW
= 600.96 MW.
Hence, heat energy exhausted to the river that cools the plant is 600.96 MW.
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What are the two factors that affect the frictional force between objects
For a certain transverse standing wave on a long string, an antinode is at x = 0 and an adjacent node is at x = 0.20 m. The displacement y(t) of the string particle at x = 0 is shown in the figure, where the scale of the y axis is set by ys = 4.3 cm. When t = 0.90 s, what is the displacement of the string particle at (a) x = 0.30 m and (b) x = 0.40 m ? What is the transverse velocity of the string particle at x = 0.30 m at (c) t = 0.90 s and (d) t = 1.3 s?
The expressions for the traveling and standing wave to find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
The traveling wave is a disturbance in the medium that moves at constant speed, in the case of a transverse wave the expression for the perpendicular oscillation is:
y = A sin (kx - wt)
Where y is the oscillation perpendicular to the direction of the displacement, A the amplitude, k in wave number and w the angular velocity.
Standing waves are formed when a traveling wave collides with an obstacle and is reflected, in this case the sum of the two waves gives a wave that does not shift in time and fulfills the relationship
[tex]\frac{\lambda}{2} = \frac{L}{n}[/tex]
Where λ is the wavelength, L the distance between the reflection points and n the number of nodes.
Indicates that for the standing wave the distance between an antinode and the node is x = 0.20 m, therefore
[tex]\frac{\lambda}{4} = \frac{L}{1}[/tex]
λ = 4L
λ = 4 0.20
λ = 0.80 m
The wave number.
k = [tex]\frac{2\pi }{\lambda }[/tex]
k = [tex]\frac{2 \pi }{0.80 }[/tex]
k = 2.5π i m⁻¹
In the associated traveling wave, from the graph we can see that the period of the wave is:
T = 2.8 s
the angular velocity is related to the period.
[tex]w=\frac{2\pi}{T} \\w = \frac{2\pi }{2.8}[/tex]
w = 0.714π rad/s
indicate the maximum displacement that is the amplitude of the wave.
A = [tex]y_s[/tex]
A = 4.3 cm
Let's write the equation of the traveling wave.
y = 4.3 sin [π (2.5 x - 0.714 t)]
with this expression we can answer the questions.
a) the displacement of the particle for x = 0.30 m
y = 4.3 sin (π (2.5 0.30 - 0.714 t))
y = 4.3 sin π( 0.75 - 0.714 t(
Remember that the angles must be in radians. For time t = 0 the displacement is
y = 4.3 0.707
y = 3.04 cm
b) The displacement for x = 0.4m
y = 4.3 sin (π 2.5 0.4)
y = 0 cm
c) the transverse velocity of the wave at x = 0.30 m for the time of t = 0.90s
the speed of the wave is
[tex]v= \frac{dy}{dt} \\v= A w cos ( kx - wt)[/tex]
v = 4.3 0.714π cos π(2.5 0.3 - 0.714 t)
v = 9.65 cos π(0.75 - 0.714 t)
For time t = 0.90 s the velocity is:
v = 9.65 cos π(0.75 - 0.714 0.9)
v = 9.65 0.9436
v = 9.11 cm / s
d) The velocity for time t = 1.3 s
v = 9.65 cos π(0.75 - 0.714 1.3)
v = 9.65 0.9999
v = 9.65 cm / s
In conclusion, using the expressions for the traveling and standing wave, we can find the results for the questions about the displacement and speed of the particle are:
a) For time zero, the displacement at position x = 0.30 m is y = 3.04 cm
b) For time zero, the displacement at position x = 0.40 m is: y = 0
c) For the point x = 0.30 and time t = 0.9s, the velocity of the particle is:
v = 9.11 cm / s
d) For the point x = 0.30 and time t = 1.3s, the velocity of the particle is:
v = 9.65 cm / s
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A racecar can be slowed with a constant acceleration of -14 m/s2. If the car is going 75 m/s, how many meters will take to stop?
Answer:
Explanation:
v² = u² + 2as
s = (v² - u²)/2a
s = (0² - 75²) / (2(-14))
s = 200.8928
s = 200 m
Answer:
Identification
Explanation:
:-;..........
A less than youthful 82.6 kg physics professor decides to run the 26.2 mile (42.195 km) Los Angeles Marathon. During his months of training, he realizes that one important component in running a successful marathon is carbo-loading, the consumption of a sufficient quantity of carbohydrates prior to the race that the body can store as glycogen to burn during the race. The typical energy requirement for runners is 1 kcal/km per kilogram of body weight, and each mole of oxygen intake allows for the release of 120 kcal of energy by oxidizing (burning) glycogen.
(a) If the professor finishes the marathon in 4:45:00 h, what is the professor's oxygen intake rate, in liters per minute, during the race if he metabolizes all of the carbo-loaded glycogen during the race and the ambient temperature is 21.5°C? 2.28 Read the problem statement again carefully. Is the air at standard temperature and pressure during the marathon? How would this affect the volume of 1 mol of oxygen? L/min
(b) The human body has an efficiency of 25.0%. Only 25.0% of the energy released from oxidizing glycogen is used as macroscopic mechanical energy, and the remaining 75.0% is used for body processes such as pumping blood and respiration, and then leaves the body through the skin via radiation, evaporative cooling, and other processes. What is the average mechanical power (in W) generated by the professor during the run? 197.561 What is the total energy required by the professor during the run? How efficient is the human body, and how long did the race last? W
(c) What is the change in entropy (in J/K) of the professor's body if his core temperature has risen to 38.3°C during the run and his skin temperature is at 36.0°C during the marathon? J/K
(d) What is the change of the entropy (in J/K) of the air surrounding the professor during the race if the ambient temperature remains constant at 21.5°C? J/K
A merry-go-round rotates from rest with an angular acceleration of 1.08 rad/s2. How long does it take to rotate through (a) the first 3.74 rev and (b) the next 3.74 rev
Answer:
Let ω1 be the initial angular speed and ω2 the final angular speed:
α = (ω2- α1) / t
corresponding to a = (v2 - v1) / t
S (distance corresponds to theta)
1 rev = 2 pi and 3.74 rev = 7.48 pi = 23.5 radians
S = 1/2 a t^2 linear or S = 1/2 α t^2 angular acceleration
23.5 = 1/2 * 1.08 t^2 and t = 6.60 sec for first 3.84 rev
b) ω1 = 1.08 * 6.6 = 7.13 rad/sec initial speed for second 3.74
23.5 = 7.13 t + .54 t^2 compare to S = v1 t + 1/2 a t^2
.54 t^2 + 7.13 t - 23.5 = 0
t^2 + 13.2 t - 43.5 = 0
t = 2.7 sec for next 3.74
Check:
7.13 * 2.7 + .54 * 2.7^2 = 23.2 rad = 3.7 rad
what is a non-economic benefit to international trade
Answer:
non economic benefits to international trade means not being too much profit not selling too much things /items in international trade
2) A ray of light in air is approaching the boundary with water at an
angle of 52 degrees. Determine the angle of refraction of the light
ray. (Refractive index of air = 1, water = 1.33)
=
Answer:
Explanation:
ASSUMING the 52° is the angle of incidence measured from the perpendicular to the surface
n₁sinθ₁ = n₂sinθ₂
1 sin52 = 1.33sinθ₂
θ₂ = arcsin(sin52 / 1.33)
θ₂ = 36°
as measured from the perpendicular to the surface
An airplane accelerates from a speed of 33 m/s at the constant rate of 3.0 m/s2 over a distance of 500 m. What the final velocity?
Answer:
Explanation:
v² = u² + 2as
v² = 33² + 2(3.0)(500)
v² = 4089
v = 63.9452...
v = 64 m/s
8.92 A 45.0 kg woman stands up in a 60.0 kg canoe 5.00 m long. She walks from a point 1.00 m from one end to a point I .00 m from the other end (Fig. P8.92). If you ignore resistance to motion Of the canoe in the water, how far does the canoe move during fis process
The distance that the canoe moves in this process is 1.29 meters.
We first have to find the center of mass
[tex]X = \frac{MsXs+McXc}{Mw+Mc} \\\\[/tex]
Where
Ms = Woman's mass = 45
Mc = Canoe's mass = 60kg
Xs = position from left= 1 cm
Xc = position from left end of canoe's mass = 2.5cm
When we put these values into the equation we have:
[tex]X=\frac{45*1+60*2.5}{45+60} \\\\= 1.857\\[/tex]
The center of gravity lies at the center of this boat. Therefore,
[tex]Xc = \frac{L}{2} \\\\L = 5 m long\\\\\frac{5}{2} =2.5[/tex]
5.00 - 1. 00 = 4 meters
[tex]\frac{45*4+60*2.5}{45+60} = 3.14meters\\\\[/tex]
To get the distance that is moved by this canoe
distance = 3.143-1.857
= 1.286
≈ 1.29 meters
The distance that the canoe moves in this process is 1.29 meters.
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Explain different layers of atmosphere and the pressure in each layer. Draw diagram
Answer:
Our atmosphere has five different layers. They are:
1. Troposphere: This is the most important layer of the atmosphere with an average height of 13 km from the earth. It is in this layer that we find the air that we breathe. Almost all the weather phenomena such as rainfall, fog and hailstorm occur here.
2. Stratosphere: This layer extends up to a height of 50 km. It presents the most ideal condition for flying airplanes. It contains a layer of ozone gas which protects us from the harmful effect of the sun rays.
3. Mesosphere: This layer extends up to a height of 80 km. Meteorites bum up in this layer on entering from the space.
4. Thermosphere: In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It extends between 80-400 km. This layer helps in radio transmission. Radio waves transmitted from the earth the reflected back to the earth by this layer.
5. Exosphere: It is the uppermost layer where there is very thin air. Light gases such as helium and hydrogen float into space from here.
Calculate the following
Earth's mass: Man's Mass
Earth's Radiusi man Radius
Earth's Density: mon Density
Farth's granty: man granty
Paano ka makaka tulong sa upang maiwasan ang suliranin ukol sa climate change
Answer:
Climate Change
Ito ang pagbabago sa normal na panahon sa isang lugar. Maaaring abnormal na pagbabago sa ulan na nattanggap ng isang lugar sa isang taon o ang abnormal na pagbabago sa init ng panahon.
Paano masusulusyunan ito?
Tanggalin sa pagkakasaksak ang mga gadgets kapag puno na
Huwag pumutol ng mga puno
Iwasan ang pagsakay sa transportasyon kung malapit lang ang pupuntahan
Explanation:
pa brainliest
What happened when a lobster release a claw? a. The lobster attacks with the claw. b. The lobster let the claw go. C. The lobster bites off the claw. D. The lobster eats with the claw.
Answer:b
Explanation:
Answer:
The lobster let the claw go. so, B.
Explanation:
i hope this helps.
Name two environmental factors, one natural and one human-made, that could account for the trend or pattern in bird (quail, wren) and rodent (mice, rabbits) populations before 1997. Help!
Factors such as more crops, fewer predators, invasive species, and changes in the environment explain a decrease or increase in the birds and rodent populations.
Before 1997, both rodents and birds populations increased due to different factors. Moreover, these factors can be classified as natural or human-made factors. Here are the factors that mainly affected these animals:
Bird population:
Increase:
Decrease in predators that increase birds chances to survive (natural factor)
Increase in crops that improve food access for birds (human factor)
Decrease:
Invasive species or increase of predators (natural factor)
Destruction of natural habitats due to an increase in industry and extraction (human factor)
Rodent population:
Increase:
More adaptability that increased rodents chances to survive (natural factor)Increase in crops that improve food access for rodents (human factor)Decrease:
Invasive species or increase of predators (natural factor)Increase in rabbit consumption and use of rodent control products (human factor)Learn more in: https://brainly.com/question/18123593