Answer:
238 Degree
Explanation:
Data given
F1=(3.3,-0.5) and F2=(-3.8,-0.3).
To determine the angle F1+F2 makes with the positive x-axis, we need to determine the magnitude of the force F1+F2.
Since force is a vector quantity, we add the vectors component by component
[tex]F_{1}+F_{2}=<3.3,-0.5> +<-3.8,-0.3>\\F_{1}+F_{2}=<3.3+(-3.8), -0.5+(-0.3)>\\ F_{1}+F_{2}=<-0.5,-0.8>\\F_{1}+F_{2}=(-0.5,-0.8)[/tex]
To determine the angle, we use
[tex]F=(x,y)\\\alpha=arctan(\frac{y}{x} )\\Hence for \\F_{1}+F_{2}=(-0.5,-0.8)\\\alpha=arctan(\frac{-0.8}{-0.5} )\\\alpha=58^{0}[/tex]
Since the component of the force F1+F2 is a negative y and negative x which are located in the 3rd quadrant, the angle can be calculated as
∝=58+180=238 degree
Hence The angle is measured counterclockwise from the positive x-axis and must be in the range from 0 to 360 degrees is 238 Degree
The angle measured counterclockwise from the positive x-axis is θ = 57.99°
Finding the direction of the force.
Here we know that:
F1 = (3.3, -0.5)F2 = (-3.8, -0.3)First, we need to add the forces, we will get:
F1 + F2 = (3.3, -0.5) + (-3.8, -0.3) = (3.3 - 3.8, -0.5 - 0.3))
F1 + F2 = (-0.5, -0.8)
Now, the angle measured counterclockwise from the positive x-axis of a vector
(a, b) is given by:
θ = Atan(b/a).
Where Atan(x) is the inverse tangent function.
So in this case the angle will be:
θ = Atan(-0.8/-0.5) = 57.99°
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(a) If a flea can jump straight up to a height of 0.440 m, what is its initial speed as it leaves the ground? (b) How long is it in the air?
Answer
given,
height of the jump = 0.44 m
acceleration due to gravity, g = 9.8 m/s²
velocity at the height point = 0 m/s
initial speed = ?
Using equation of motion for speed calculation
v² = u² + 2 g h
0 = u² - 2 x 9.8 x 0.44
u = √8.624
u = 2.94 m/s
time taken to reach the highest point
v = u + g t
0 = 2.94 - 9.8 x t
t = 0.3 s
total time of flight will be equal to double of the time taken to reach the maximum height.
Total time = 2 x 0.3 = 0.6 s
A flea jumps straight up to a height of 0.440 m with an initial speed of 2.94 m/s and is in the air for 0.600 s.
A flea jumps straight up to a height (h) of 0.440 m. At the maximum height, the final speed (v) is zero. Given that the gravity (g) is 9.81 m/s², we can calculate the initial speed (u) using the following kinematic expression (We will assume y+ as the positive direction)
[tex]v^{2} = u^{2} + 2 \times g \times h \\\\(0m/s)^{2} = u^{2} + 2 \times (-9.81m/s^{2} ) \times 0.440m\\\\u = 2.94 m/s[/tex]
We can calculate the time (t) required to reach the maximum height using the following kinematic expression.
[tex]v = u + g \times t\\\\0m/s = 2.94m/s - 9.81m/s^{2} \times t\\\\t = 0.300 s[/tex]
The time in the air will be double the time to reach the maximum height.
[tex]t_{total} = 2 \times 0.300 s = 0.600 s[/tex]
A flea jumps straight up to a height of 0.440 m with an initial speed of 2.94 m/s and is in the air for 0.600 s.
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The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker.
Approximately how much time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer? Select the best answer from the choices provided.
a) 1/6 seconds
b) 1/4 seconds
c) 1/2 seconds
d)1/3 seconds
e) 1/5 seconds
Answer:
1/3
Explanation:
Take the displacement and divide it by the speed of light. Meters will cancel leaving you with seconds. :)
110m / 343m/s = 0.32069971 seconds or 1/3 seconds
The time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer 1/3 second ( approx.). So, option (d) is correct.
What is velocity?The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.
Given The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker.
So, distance between the first dancer in the line and the last dancer in the line be = 120 m -10 m = 110 m
Velocity of sound in air be = 343 m/s.
Hence, The time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer = distance/velocity = 110/343 second = 1/3 second ( approx.).
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An amount of work W is done on an object of mass m initially at rest, and as result it winds up moving at speed v. Suppose instead it were already moving at speed v and the same amount of work W was done on it. What would be its final speed
Answer:
Explanation:
Given
W amount of work is done on the system such that it acquires v velocity after operation(initial velocity)
According to work energy theorem work done by all the forces is equal to change in kinetic energy of object
[tex]W=\frac{1}{2}mv^2---1[/tex]
where m=mass of object
v=velocity of object
When the object is already have velocity v then the final speed is given by work energy theorem
[tex]W=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2-----2[/tex]
From 1 and 2 we get
[tex]\frac{1}{2}mv^2=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2[/tex]
[tex]2\times \frac{1}{2}mv^2=\frac{1}{2}mv_f^2[/tex]
[tex]v_f^2=2v^2[/tex]
[tex]v_f=\sqrt{2}v[/tex]
"A boat that can travel at 4.0 km/h in still water crosses a river with a current of 2.0 km/h. At what angle must the boat be pointed upstream (that is, relative to its actual path) to go straight across the river?
Answer:
30 degrees
Explanation:
The boat to go straight across the river uptream will have to make angle of 30 degree with the resultant velocity vector of boat.
What is vector law of addition ?
Vector addition is governed by the triangle law which states that when two vectors are represented by two triangle sides with their order of magnitude and direction, the resultant vector's magnitude and direction will be represented by the third triangle side.
It is given that speed of boat in still water v₁ = 4 km/h
speed of current v₂ = 2 km/h
Let the relative speed of boat with respect to current be = v km/h
and the angle made by v₁ with v be = θ
No find the angle θ with which the boat to go straight across the river uptream by triangle law of vector addition as shown below :
[tex]\begin{aligned}\theta &=\text{sin}^{-1}\left ( \frac{2}{4} \right )\\&= 30^{0}\end{aligned}[/tex]
Therefore, the boat to go straight across the river uptream will have to make angle of 30 degree with the resultant velocity vector of boat.
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An isolated, parallel‑plate capacitor carries a charge Q . If the separation between the plates is doubled, the electrical energy stored in the capacitor will be:a. Unchanged.b. Halved.c. Doubled.d. Quartered.e. Quadrupled.
Answer:
option C
Explanation:
Given,
Change on the capacitor = Q
Separation is doubled
Energy stored in the capacitor,E = ?
we know,
[tex]E = \dfrac{Q^2}{2C}[/tex]
and [tex]C = \dfrac{\epsilon_0A}{d}[/tex]
now,
[tex]E = \dfrac{Q^2}{2\epsilon A}\ d[/tex].......(1)
where d is the separation between the two plates.
now, when the separation is doubled
[tex]E' = \dfrac{Q^2}{2\epsilon A}\ (2d)[/tex]
[tex]E' = 2(\dfrac{Q^2}{2\epsilon_0 A}\ d)[/tex]
From equation (1)
E' = 2 E
Hence, the energy stored in the capacitor is doubled if the separation is increased.
The correct answer is option C.
A solid sphere and a hollow sphere have the same mass and radius. The two spheres are spun with matching angular velocities. Which statement is true? I. The hollow has the greater angular momentum II. The solid sphere has the greater angular momentum. III. The angular momentum is the same for both spheres. IV. The moment of inertia is the same for both spheres.
Answer:
I. The hollow sphere has the greater angular momentum.
Explanation:
Given that the mass and radius of hollow sphere and solid sphere are same. Let the mass and radius of two spheres be m and r respectively. The two spheres are rotating having same angular velocity ω .
Moment of inertia of solid sphere, I₁ = [tex]\frac{2}{5}\times{m}r^{2}[/tex]
Moment of inertia of hollow sphere, I₂ = [tex]\frac{2}{3}\times{m}r^{2}[/tex]
Since, I₁ and I₂ are not equal. Therefore, the statement iv is wrong.
The relation between angular momentum, moment of inertia and angular velocity is :
L = Iω
Let L₁ and L₂ be the angular momentum of solid sphere and hollow sphere respectively.
L₁ = I₁ω and L₂ = I₂ω
As ω is same for both spheres but I₂ is greater than I₁, hence L₂ is greater than L₁.
Therefore, statement I is correct that the hollow sphere has the greater angular momentum.
The angular momentum of the hollow sphere is greater than that of solid sphere.
The moment of inertia of solid sphere is given as follows;
[tex]I_{ss} = \frac{2}{5} mr^2 = 0.4mr^2[/tex]
The moment of inertia of hollow sphere is given as follows;
[tex]I_{hs} = \frac{2}{3} mr^2 = 0.67 mr^2[/tex]
The angular momentum of each sphere is calculated as follows;
[tex]L = I\omega \\\\L_{ss} = 0.4mr^2 \omega \\\\L_{hs} = 0.67 mr^2 \omega[/tex]
Thus, we can conclude that the angular momentum of the hollow sphere is greater than that of solid sphere.
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A vehicle of mass 1,500 kg is traveling at a speed of 50 km/hr. What is the kinetic energy stored in its mass? Calculate the energy that can be recovered by slowing the vehicle to a speed of 10 km/hr.
Answer:
Explanation:
Given
mass of vehicle [tex]m=1500\ kg[/tex]
Speed of vehicle [tex]u=50\ km/hr\approx 13.89\ m/s[/tex]
Kinetic Energy Possessed by mass
[tex]K_i=\frac{1}{2}mu^2[/tex]
[tex]K_i=\frac{1}{2}\times 1500\times (13.89)^2[/tex]
[tex]K_i=144.69\ kJ[/tex]
when vehicle is slowed down to speed of [tex]v=10\ km/hr\approx 2.78\ m/s[/tex]
Kinetic Energy at this speed
[tex]K_f=\frac{1}{2}mv^2[/tex]
[tex]K_f=\frac{1}{2}\times 1500\times (2.78)^2[/tex]
[tex]K_f=5.78\ kJ[/tex]
Energy Recovered [tex]=K_i-K_f[/tex]
Energy Recovered[tex]=144.69-5.78=138.9\ kJ[/tex]
A 2 kg block rests on a 34o incline. If the coefficient of static friction is 0.2, how much additional force, F, must be applied to keep the block from sliding down the incline?
Answer:
Explanation:
Reaction force of inclined surface = mg cosθ
= Friction force acting in upward direction = μ x mg cosθ
If F be force required in upward direction to keep the block at rest on the plane
F + μ x mg cosθ = mg sinθ
F = mg sinθ - μ x mg cosθ
F = mg( sinθ - μ cosθ)
= 2 x 9.8 ( sin34 - 0.2 cos34 )
= 19.6 ( .559 - .1658)
= 7.7 N
This is the minimum force required .
The additional force that needs to be applied to keep a 2 kg block from sliding down a 34 degree incline, given a coefficient of static friction of 0.2, is 7.89 N.
Explanation:To find out how much additional force, F, needs to be applied to keep the 2 kg block from sliding down the 34° incline, we first need to calculate the gravitational force pulling the block down the incline. This force is given by F_gravity = m * g * sin(theta), where m = 2 kg, g = 9.8 m/s² (roughly the acceleration due to gravity on Earth), and theta = 34 degrees. Thus, F_gravity = 2 kg * 9.8 m/s² * sin(34°) = 11.10 N.
Next, we calculate the static frictional force that is preventing the block from sliding down the incline. This force is given by F_friction = m * g * cos(theta) * mu_s, where mu_s = 0.2 is the coefficient of static friction. Thus, F_friction = 2 kg * 9.8 m/s² * cos(34°) * 0.2 = 3.21 N.
The additional force, F, that must be applied to keep the block from sliding down the incline is the difference between the gravitational force and the frictional force. That's F = F_gravity - F_friction = 11.10 N - 3.21 N = 7.89 N.
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A closed system consists of 0.5 kmol of ammonia occupying a volume of 6 m3. Determine (a) the weight of the system, in N, and (b) the specific volume, in m3/kmol and m3/kg. Let g 5 9.81 m/s2.
Explanation:
It is known that the molecular weight of ammonia ([tex]NH_{3}[/tex]) is as follows.
Molecular weight ([tex]NH_{3}[/tex]) = [tex]14 + 3 \times 1[/tex] = 17
(a) Therefore, we will calculate the mass as follows.
[tex]0.5 kmol \times (\frac{1000 mol}{1 kmol}) \times (\frac{17 g}{1 mol})[/tex]
= 8500 g
Now, formula to calculate weight of the system in N is as follows.
Weight = mass × g
= [tex]8500 g \times (\frac{1 kg}{1000 g}) \times (9.8 m/s^{2})[/tex]
= 83.3 N (1 [tex]kg m/s^{2}[/tex] = 1 N)
Hence, the weight of the system is 83.3 N.
(b) Relation between specific volume and number of moles is as follows.
[tex]v (m^{3}/kmol) = \frac{V}{n}[/tex]
Therefore, calculate the specific volume as follows.
[tex]V_m = \frac{6 m^{3}}{0.5 k mol}[/tex]
= 12 [tex]m^{3}/k mol[/tex]
Also,
[tex]v (m^{3}/kmol) = \frac{V}{m}[/tex]
v = [tex]\frac{6 m^{3}}{8.5 kg}[/tex]
= 0.705882 [tex]m^{3}/kg[/tex]
Therefore, we can conclude that the value of specific volume is 12 [tex]m^{3}/k mol[/tex] and 0.705882 [tex]m^{3}/kg[/tex].
Answer:
a) [tex]w=83.385\ N[/tex]
b) [tex]\bar V=12\ m^3.kmol^{-1}[/tex]
[tex]\b V=0.7059\ m^3.kg^{-1}[/tex]
Explanation:
Given:
no. of moles of ammonia in a closed system, [tex]n=0.5\ kmol=500\ mol[/tex]
volume of ammonia, [tex]V=6\ m^3[/tex]
We know the molecular formula of ammonia: [tex]NH_3[/tex]
The molecular mass of ammonia:
[tex]M=14+3\times 1=18\ g.mol^{-1}[/tex]
Now the mass of given ammonia:
[tex]m=n.M[/tex]
[tex]m=500\times 17[/tex]
[tex]m=8500\ g=8.5\ kg[/tex]
a)
Now weight:
[tex]w=m.g[/tex]
[tex]w=8.5\times 9.81[/tex]
[tex]w=83.385\ N[/tex]
b)
Specific volume:
[tex]\bar V=\frac{6}{0.5}[/tex]
[tex]\bar V=12\ m^3.kmol^{-1}[/tex]
also
[tex]\b V=\frac{V}{m}[/tex]
[tex]\b V=\frac{6}{8.5}[/tex]
[tex]\b V=0.7059\ m^3.kg^{-1}[/tex]
the position of a crate sliding down a ramp is given by x=0.25t^3 m, z = 6-0.75t^5/2 m, where t is in seconds. determine the magnitude of the crates velocity and acceleration when t = 2 seconds
Answer:
v = 30.15 m/s
a = 60.07 m/s2
Explanation:
Velocity is derivative of position with respect to time
[tex]v_x = x' = 3*0.25t^2 = 0.75t^2[/tex]
[tex]v_z = z' = 5*0.75t^4/2 = 1.875t^4[/tex]
Acceleration is derivative of velocity with respect to time
[tex]a_x = v_x' = 2*0.75 t = 1.5t[/tex]
[tex]a_z = v_z' = 4*1.875t^3 = 7.5t^3[/tex]
At t = 2 seconds
[tex]v_x = 0.75*2^2 = 3m/s[/tex]
[tex]v_z = 1.875*2^4 = 30m/s[/tex]
[tex]v = \sqrt{v_x^2 + v_z^2} = \sqrt{3^2 + 30^2} = \sqrt{909} = 30.15 m/s[/tex]
[tex]a_x = 1.5*2 = 3 m/s^2[/tex]
[tex]a_z = 7.5*2^3 = 60 m/s^2[/tex]
[tex]a = \sqrt{a_x^2 + a_z^2} = \sqrt{3^2 + 60^2} = \sqrt{3609} = 60.07 m/s[/tex]
The destination airport has one runway, 08-26, and the wind is calm. The normal approach in calm wind is a left hand pattern to runway 08. There is no other traffic at the airport. A thunderstorm about 6 miles west is beginning to develop to its mature stage, and rain is starting to reach the ground. The pilot decides to a. Fly normal approach to runway 8b. Fly a right hand approach to runway 8c. Fly the approach to runway 26d. Fly to the west for a fun ride
Answer:
C
Explanation:
Correct answer: C. Fly the approach to runway 26. The winds from the storm could suddenly gust up and you don’t want to be in a short final or even in the flare when a tailwind from a storm gusts up. No other traffic and calm winds currently means that you can land on any runway you want. Go for the safe bet and land on 26.
In this scenario, the pilot should fly a left hand approach to runway 08.
Explanation:In this scenario, with a calm wind, the pilot should fly a left hand approach to runway 08.
Since the wind is calm, there is no need for the pilot to adjust the approach. Flying in a clockwise pattern to runway 08 is the normal procedure.
Choosing any other option would not be appropriate or necessary.
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A mass is attached to a spring with spring constant k = 20 N m . The spring is stretched to 10 cm past its resting position. How much work (in J) does the spring do when the object is released and the mass travels back to its initial position?
The spring does 0.1 Joules of work when the object is released and travels back to its initial position.
We have,
The work done by the spring can be calculated using the formula for the potential energy stored in a spring:
Potential Energy (PE) = 0.5 * k * x²,
where k is the spring constant and x is the displacement from the equilibrium position.
Given:
Spring constant (k) = 20 N/m,
Displacement (x) = 0.10 m (10 cm).
Calculate the potential energy when the spring is stretched to 10 cm:
PE = 0.5 * k * x²
= 0.5 * 20 N/m * (0.10 m)²
= 0.5 * 20 * 0.01 J
= 0.1 J.
The spring stores 0.1 Joules of potential energy when stretched to 10 cm.
When the mass is released and returns to its initial position, this potential energy is converted back into kinetic energy as the mass accelerates.
Therefore,
The spring does 0.1 Joules of work when the object is released and travels back to its initial position.
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Final answer:
The spring does 0.1 Joules of work when the mass travels back to its initial position.
Explanation:
To find the work done by a spring when an object is released and travels back to its initial position, we can use the formula:
Work = (1/2) * k * x^2
Where k is the spring constant and x is the displacement from the equilibrium position. In this case, the spring constant is 20 N/m and the displacement is 10 cm (which is 0.1 m). Plugging these values into the formula, we get:
Work = (1/2) * 20 N/m * (0.1 m)^2 = 0.1 J
So, the spring does 0.1 Joules of work when the mass travels back to its initial position.
10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when it crosses over onto a floor that has a coefficient of kinetic friction = 0.35. How far will the puck travel across the floor before it stops?
Answer:
[tex] d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m [/tex]
Explanation:
For this case we can use the second law of Newton given by:
[tex] \sum F = ma[/tex]
The friction force on this case is defined as :
[tex] F_f = \mu_k N = \mu_k mg [/tex]
Where N represent the normal force, [tex]\mu_k [/tex] the kinetic friction coeffient and a the acceleration.
For this case we can assume that the only force is the friction force and we have:
[tex] F_f = ma[/tex]
Replacing the friction force we got:
[tex] \mu_k mg = ma[/tex]
We can cancel the mass and we have:
[tex] a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}[/tex]
And now we can use the following kinematic formula in order to find the distance travelled:
[tex] v^2_f = v^2_i - 2ad[/tex]
Assuming the final velocity is 0 we can find the distance like this:
[tex] d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m [/tex]
The hockey puck will travel approximately 58.5 meters across the floor before it stops.
Explanation:To solve this problem, we will use the concepts of kinetic energy and the work-energy theorem.
Initially, the hockey puck has a kinetic energy of KE1 = 0.5*m*v^2 = 0.5*0.3 kg*(20 m/s)^2 =60 J
As it travels across the floor, the friction does work on it until it stops. The work done by the friction force is W = μ*m*g*d, where: μ is the coefficient of kinetic friction (0.35), m is the mass (0.3 kg), g is the gravitational constant (9.8 m/s^2), and d is the distance traveled.
The work done by the friction is equal to the initial kinetic energy (work-energy theorem), so we can solve for d: 60 J = 0.35 * 0.3 kg * 9.8 m/s^2 * d. Simplifying this equation gives d = 60 J / (0.35 * 0.3 kg * 9.8 m/s^2) = 58.5 m
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3 An empty hot tub has a mass of 320 kg. When filled, the tub holds 600 gallons of water (rho = 62.4 lbm/ft3). The local acceleration due to gravity is 32 ft/s2. Determine the total weight of the hot tub and water in pounds-force (lbf)
Answer:
Total weight of the hot tub and water is 5676.6 pounds-force
Explanation:
rho = 62.4lbm/ft^3 × 1ft^3/7.481gal = 8.34lbm/gal
Mass of water = rho × volume = 8.34lbm/gal × 600 gallons = 5004lbm = 5004×0.45359kg = 2269.8kg
Total mass of hot tub and water = 320kg + 2269.8kg = 2589.8kg
Local acceleration due to gravity = 32ft/s^2 = 32ft/s^2 × 1m/3.2808ft = 9.75m/s^2
Total weight of hot tub and water = 2589.8kg × 9.75m/s^2 = 25250.55N = 25250.55/4.4482 lbf = 5676.6 lbf
What is the voltage across an 8.00 nm–thick membrane if the electric field strength across it is 5.50 MV/m?
Answer:
0.044 V
Explanation:
E = Electric field = [tex]5.5\times 10^6\ V/m[/tex]
d = Thickness of membrane = 8 nm
When the electric field strength is multiplied by the membrane thickness we get the voltage
Voltage across a gap is given by
[tex]V=Ed\\\Rightarrow V=5.5\times 10^6\times 8\times 10^{-9}\\\Rightarrow V=0.044\ V[/tex]
The voltage across the membrane is 0.044 V
Final answer:
The voltage across an 8.00 nm-thick membrane with an electric field strength of 5.50 MV/m is calculated using the formula V = Ed, resulting in a voltage of 44.0 mV.
Explanation:
The voltage across a membrane can be determined by using the relationship between electric field strength (E), voltage (V), and the distance (d) the electric field spans. The electric field is uniform and the formula to use is V = Ed. In this case, the electric field (E) is given as 5.50 MV/m or 5.50 x 10⁶V/m, and the thickness of the membrane (d) is 8.00 nm or 8.00 x 10⁻⁹ m.
To find the voltage across the membrane, we simply multiply the electric field by the thickness of the membrane:
V = (5.50 x 10⁶ V/m) x (8.00 x 10⁻⁹ m)
Therefore:
V = 44 x 10⁻³ V
V = 44.0 mV
So, the voltage across an 8.00 nm-thick membrane with the given electric field strength is 44.0 mV.
charge densities such that are equal in magnitude but opposite in sign. The difference in potential between the plates is 490 V.
(a) Is the positive or the negative plate at the higher potential?
select
the positive plate
the negative plate
(b) What is the magnitude of the electric field between the plates?
____ kV/m
Answer:
a) The positive plate is at a higher potential than the negative plate.
b) Electric field between the plates = 5.76 KV/m
Explanation:
a) The positive plate is at a higher potential than the negative plate because of convention. We could define the negative plate to be "high potential." However, that convention has to be consistent. The positive plate is at the high potential, and that potential causes positive charges to flow from high potential to low potential. If we had defined the convention in the other direction, then the negative terminal would be the high potential, and that potential would cause negative charges to flow from high potential to low potential.
b) Electric field, E = potential difference across plate/distance or separation between the plates
E = V/d
V = 490V, d = 8.5cm = 0.085m
E = 490/0.085 = 5,764.706 V/m = 5.76 KV/m
Hope this helps!!!
An advertisement claims that a particular automobile can "stop on a dime". What net force would actually be necessary to stop an automobile of mass 970 kgkg traveling initially at a speed of 52.0 km/hkm/h in a distance equal to the diameter of a dime, which is 1.8 cmcm ?
Answer:
F=5618278.8 N
Explanation:
Given that
m = 970 kg
Initial speed ,u = 52 km/h
u = 14.44 m/s ( 1 km/h = 0.277 m/s)
Distance d= 1.8 cm = 0.018 m
The final speed ,v = 0 m/s
We know that
v² = u² + 2 a d
a=Acceleration
0² = 14.44² + 2 x a x 0.018
[tex]a=-\dfrac{14.44^2}{2\times 0.018}\ m/s^2[/tex]
a=5792.04 m/s²
We know that
Force = Mass x acceleration
F= m a
F= - 5792.04 x 970 N
F= - 5618278.8 N
Therefore the magnitude of the force will be 5618278.8 N.
F=5618278.8 N
The sound intensity from a jack hammer breaking concrete is 2.0W/m2 at a distance of 2.0 m from the point of impact. This is sufficiently loud to cause permanent hearing damage if the operator doesn't wear ear protection. What are (a) the sound intensity and (b) the sound intensity level for a person watching from 50 m away?
Answer:
(a) [tex]I_{1}=3.2*10^{-3}W/m^{2}[/tex]
(b) [tex]\beta =95dB[/tex]
Explanation:
Given data
Distance r₁=50 m
Distance r₂=2 m
Intensity I₂=2.0 W/m²
To find
(a) The Sound Intensity I₁
(b) The Sound Intensity level β
Solution
For (a) the Sound Intensity I₁
[tex]\frac{I_{1} }{I_{2}}=\frac{(r_{2})^{2} }{(r_{1})^{2} }\\I_{1} =I_{2}(\frac{(r_{2})^{2} }{(r_{1})^{2} })\\I_{1}=(2.0W/m^{2} )(\frac{(2m)^{2} }{(50m)^{2} })\\I_{1}=3.2*10^{-3}W/m^{2}[/tex]
For (b) the Sound Intensity level β
The Sound Intensity level β is calculated as follow
[tex]\beta =(10dB)log_{10}(\frac{I}{I_{o} } )\\\beta =(10dB)log_{10}(\frac{3.2*10^{-3}W/m^{2} }{1.0*10^{-12} W/m^{2} } )\\\beta =95dB[/tex]
A box is sitting on a 2 m long board at one end. A worker picks up the board at the end with the box so it makes an angle with the ground of 35o. The coefficient of static friction between the box and the board is 0.5 and the coefficient of kinetic friction is 0.3 . Will the box slide down the ramp when it is at 35o?
Answer:
Yes it will slide down the ramp
Explanation:
Let m be the mass of the box and gravitational acceleration g = 9.81m/s2, we can calculate the gravity that affects the box
W = mg
When the box is at 35 degree incline, this gravity is split into 2 components, 1 parallel to the incline (Wsin35) and the other one perpendicular with the incline (Wcos35).
The one perpendicular with the incline has an equal and opposite normal force of Wcos35
This normal force would dictate the static friction force where coefficient = 0.5. So the static friction is 0.5mgcos35
The box would slide when the parallel component of gravity wins over static friction force
mgsin35 > 0.5mgcos35
Since mg is positive we can cancel them out on both sides
sin35 > 0.5cos35
0.57 > 0.5*0.82
0.57 > 0.41
This is true so we can conclude that the box slides down the ramp
Assume that a clay model of a lion has a mass of 0.225 kg and travels on the ice at a speed of 0.85 m/s. It hits another clay model, which is initially motionless and has a mass of 0.37 kg. Both being soft clay, they naturally stick together.
What is their final velocity?
Answer:
Final velocity will be equal to 0.321 m/sec
Explanation:
We have given mass of clay model of lion [tex]m_1=0.225kg[/tex]
Its speed is 0.85 m/sec, so [tex]v_1=0.85m/sec[/tex]
Mass of another clay model [tex]m_2=0.37kg[/tex]
It is given that second clay is motionless
So its velocity [tex]v_2=0m/sec[/tex]
Now according to conservation of momentum
Momentum before collision will be equal to momentum after collision
So [tex]m_1v_!+m_2v_2=(m_1+m_2)v[/tex], here v is velocity after collision
So [tex]0.225\times 0.85+0.37\times 0+(0.225+0.37)v[/tex]
[tex]0.595v=0.191[/tex]
v = 0.321 m/sec
So final velocity will be equal to 0.321 m/sec
The net electric charge of an amber rod which has been rubbed with fur is called negative Group of answer choices because amber is an insulator by arbitrary convention so that the proton charge will be positive because like charges repel None of the above
Answer:
The right option is option E. None of the answer choices given are totally correct.
Explanation:
All insulators normally have an equal amount of positive and negative charges distributed on their surface.
The amber rod (an insulator) is called negative because after the coming together with fur (another insulator), the amber rod rubs off electrons from the fur onto itself and has an overall more negatively charged particles than positively charged particles on its surface.
The fur in turn becomes positive because it has more positive charges than negative on its surface.
So, the convention allows the now rubbed off amber rod to be called negative.
So, it is evident that none of the answer choices are totally correct, the right answer is more of a mix of some of the answer choices and more!
Hope this helps!!
Calculate the energy of the quantum involved in the excitation of (i) an electronic oscillation of period 1.0 fs, (ii) a molecular vibration of period 10 fs, (iii) a pendulum of period 1.0 s. Express the results in joules and kilojoules per mole.
Answer:
a) E = 3.99 × 10⁵J/mol = 3.99 × 10² KJ/mol = 399KJ = 400KJ/mol
b) E = 3.99 × 10⁴ J/mol = 3.99 × 10¹ KJ/mol = 39.9KJ/mol = 40 KJ/mol
c) E = (3.99 × 10^-10) J/mol = 3.99 × 10⁻⁷ KJ/mol
Explanation:
E = hf where E = energy, H = Planck's constant = 6.62607004 × 10⁻³⁴ J.s and f = 1/time period
a) Period = 1 fs = 1 × 10⁻¹⁵ s
f = 1/(10⁻¹⁵ ) = 10¹⁵ Hz
E = hf = 6.62607004 × 10⁻³⁴ × 10¹⁵ = 6.63 × ⁻10¹⁹ J = 6.63 × 10⁻¹⁶ KJ
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = 3.99 × 10⁵J/mol = 3.99 × 10² KJ/mol = 399KJ = 400KJ/mol
b) Period = 10 fs = 10 × 10⁻¹⁵ = 10⁻¹⁴ s
f = 1/period = 1/10⁻¹⁴ = 10¹⁴ Hz
E = hf = 6.62607004 × 10⁻³⁴ × 10¹⁴ = 6.63 × 10^-20 J = 6.63 × 10⁻¹⁷ KJ
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = 3.99 × 10⁴ J/mol = 3.99 × 10¹ KJ/mol = 39.9KJ/mol = 40 KJ/mol
c) Period = 1s
f = 1/period = 1.0 Hz
E = 6.62607004 × 10⁻³⁴ × 1 = 6.63 × 10⁻³⁴ J = 6.63 × 10⁻³¹ KJ.
In energy per mol, we multiply the energy with the avogadro's constant, 6.022 × 10²³ atoms/mol
E = (3.99 × 10^-10) J/mol = 3.99 × 10⁻⁷ KJ/mol
Answer:
(i) E = 6.626 X 10⁻¹⁹ J = 400 KJ/mol
(ii) E = 6.626 X 10⁻²⁰ J = 40 KJ/mol
(iii) E = 6.626 X 10⁻³⁴ J = 4 X 10⁻¹³ KJ/mol
Explanation:
Energy associated with excitation of a quantum is given as;
E = hf
where;
E is the energy of excitation
h is Planck's constant = 6.626 X 10⁻³⁴Js⁻¹
f is the threshold frequency in s⁻¹
Thus, E = h/t
Part (i)
E = (6.626 X 10⁻³⁴)/(1 X 10⁻¹⁵)
E = 6.626 X 10⁻¹⁹ J
In (KJ/mol) = 6.626 X 10⁻²² KJ X 6.022 X10²³ = 400 KJ/mol
Part (ii)
E = (6.626 X 10⁻³⁴)/(1 X 10⁻¹⁴)
E = 6.626 X 10⁻²⁰ J
In (KJ/mol) = 6.626 X 10⁻²³ KJ X 6.022 X10²³ = 40 KJ/mol
Part (iii)
E = (6.626 X 10⁻³⁴)/(1)
E = 6.626 X 10⁻³⁴ J
In (KJ/mol) = 6.626 X 10⁻³⁷ KJ X 6.022 X10²³ = 4 X 10⁻¹³ KJ/mol
A new, previously unknown, planet Vulcan was discovered in our solar system. We measure an orbital period of 103 Earth days for the new planet. We send some astronauts to travel to the planet and they measure the planet's gravitational acceleration to be 8.2 m/s2 on its surface. They also determine that the planet's radius is half the radius of Earth.
Q1. How far away from the sun is the new planet?
(A) 8.3 x 100 m
(B) 6.4 x 100 m
(C) 1.7 x 100 m
(D) 9.5 x 1010 m
(E) 2.3 x 1010 m
Q2. What is the mass of planet Vulcan?
(A) 1.3 x 1024 kg
(B) 5.4 x 1024 kg
(C) 9.8 x 1024 kg
(D) 0.2 x 102 kg
(E) 4.8 x 1024 kg
Q3. The astronauts also discover a moon that orbits planet Vulcan with a period of 63 days. How far away is the moon from the planet Vulcan?
(A) 1 x 108 m
(B) 20 x 108 m
(C) 460 x 108 m
(D) 9 x108 m
(E) 4 x 10 m
Q4. The astronauts visit the newly discovered moon to study it. They measure the gravitational acceleration on the surface of the moon to be 2.7 m/s. When their mission is finished, they try to escape from the moon. They measure that the minimum velocity to escape the moon is 3,000 m/s. What is the mass of the moon?
(A) 1.1 x 1023 kg
(B) 5.0 x 1023 kg
(C) 7.8 x 1023 kg
(D) 29 x 104 kg
(E) 370 x 1023 kg
Answer:
Q1. corect is B, Q2. it is A, Q3. E and Q4. A
Explanation:
Q1 For this exercise we can use Newton's second law where acceleration is centripetal.
F = m a
a = v² / r.
G m M / r² = m v² / r
G M / r = v²
The velocity has a constant magnitude whereby we can divide the length of the circular orbit (2π r) between the period
G M / r = (2π r / T)²
r³ = G M T2 / 4π²
Let's calculate
T = 103 day (24 h / 1 day) (3600 s / 1h) = 8,899 10⁶ s
r³ = 6.67 10⁻¹¹ 1.99 10³⁰ (8,899 10⁶) 2 / 4π²
r = ∛ (266.25 10³⁰)
r = 6.4 10¹⁰ m
The distance matches the value in part B
Q2 Astronauts have measured the acceleration of gravity, so we can use the second law with a body on the planet's surface
F = m g
G m M_p / R_p² = m g
G M_p / R_p² = g
M_p = g R_p² / G
They indicate that the radius of the planet is half the radius of the Earth
R_p = ½ R_earth
R_p = ½ 6.37 10⁶
R_p = 3.185 10⁶ m
Let's calculate
M_p = 8.2 (3,185 10⁶)² / 6.67 10⁻¹¹
M_p = 1.25 10²⁴ kg
The correct answer is A
Q3 We use Newton's second law again, with part Q1, where M is the mass of the planet and m is the mass of the moon
r³ = G M T² / 4π²
T = 63 days (24h / 1day) (3600s / 1h) = 5.443 10⁶ s
r³ = 6.67 10⁻¹¹ 1.25 10²⁴ (5.443 10⁶)² / 4π²
r = ∛ (62.56807 10²⁴)
r = 3.97 10⁸ m
The correct answer is E
Q4 To calculate this part let's use the conservation of mechanical energy,
Starting point The surface of the moon
Em₀ = K + U = ½ m v2 - G m M / r
Final point. Infinity with zero speed
[tex]Em_{f}[/tex] = 0
Em₀ = Em_{f}
½ m v² - G m M / R = 0
v² = 2 G M / r
M = v2 r / 2G
r = 2 G M / v²
Since we don't know the radius of the moon, we will also use the equation in part 2
M = g r² / G
r = √ GM / g
Let's replace
2G M / v² = √ G M / g
4 G M / v⁴ = 1 / g
M = v⁴ / (g 4G)
M = 3000⁴ / (2.7 4 6.67 10-11)
M = 1.12 10²³ kg
corract is A
Consider a concentric tube heat exchanger. Assuming there is no fin attached to any of the surfaces and considering negligible fouling, what would be the appropriate equation for calculating the overall heat transfer coefficient?
Answer:
[tex]Q = T1 - T2 (KA)/L[/tex]
Explanation:
Where Q is the co-efficient of heat transfer.
T-1 is initial temperature of exchanger
T-2 is final temperature of sink where has to to be dissipated
k is the co-efficient of thermal conductivity
A is the total area of exchanger surface...
and L is the total length of exchanger...
A tire has a tread pattern with a crevice every 2.00 cm. Each crevice makes a single vibration as the tire moves. What is the frequency of these vibrations if the car moves at 30.0 m/s?
Answer:
Frequency of the vibration due to crevice is 1500 Hz.
Explanation:
Frequency is defined as rate of vibration per second. Its S.I. unit is s⁻¹ or Hz.
According to the problem, we have to find number of crevice car makes in 1 second.
Speed of car, u = 30 m/s
Distance covered by car in 1 second, d = 30 m
For every 0.02 m, one crevice occurred by the tire of car.
Number of crevice occurred in 1 second by the car = [tex]\frac{30}{0.02}[/tex]
= 1500
Since, each crevice makes a single vibration. Thus, the frequency of these vibrations is 1500 Hz.
The frequency of the vibrations generated by the tire treads in this case, given that each crevice is 2.00 cm apart and the car moves at 30.0 m/s, is 1500 Hz.
Explanation:In this scenario, it is important to understand that the frequency of the vibrations is determined by the speed of the tire and the tread pattern. Here, the crevice, which causes the vibration, occurs every 2.00 cm (or 0.02 m). This means for every meter the car moves, 0.02 m into 1 m gives a total of 50 vibrations. Therefore, at a speed of 30.0 m/s, we will have 50 vibrations/m multiplied by 30 m/s, giving a frequency of 1500 Hz or vibrations per second.
Frequency in this question refers to the number of times the vibration from the crevices occur in a unit of time (in this case, per second). This concept is crucial in understanding wave motions and vibration patterns, and is often applied in physics-related disciplines.
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What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0∘ downward from the horizontal?
Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here
A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.
What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees downward from the horizontal?
Answer:
[tex]W_{work}=2.67*10^{-3}J[/tex]
Explanation:
Given data
Charge q=28 nC
Electric field E=5.00×10⁴ V/m.
Distance d=2.70 m
Angle α=45°
To find
Work done by electric force
Solution
[tex]W_{work}=F_{force}*D_{distance}Cos\alpha \\where\\F_{force}=q_{charge}*E_{Electric-Field}\\So\\W_{work}=qE*D*Cos\alpha \\W_{work}=(28*10^{-9}C )(5.00*10^{4}V/m )(2.70m)Cos(45)\\W_{work}=2.67*10^{-3}J[/tex]
A 67 Vrms source is placed across a load that consists of a 12 ohm resistor in series with an capacitor whose reactance is 5 ohms. Compute the following: a) The average power of the load b) The reactive power of the load c) The apparent power of the load d) The power factor of the load
a) The average true power is 318.3 W
b) The reactive power is 132.6 W
c) The apparent power is 344.8 W
d) The power factor is 0.92
Explanation:
a)
For a circuit made of a resistor and a capacitor, the average (true) power is given by the resistive part of the circuit only.
Therefore, the average true power is given by:
[tex]P=I^2R[/tex]
where
I is the current
R is the resistance
In this problem, we have
V = 67 V (rms voltage)
[tex]R=12 \Omega[/tex] is the resistance of the load
[tex]X=5\Omega[/tex] is the reactance of the circuit
First we have to find the impedance of the circuit:
[tex]Z=\sqrt{R^2+X^2}=\sqrt{12^2+5^2}=13 \Omega[/tex]
Then we can find the current in the circuit by using Ohm's law:
[tex]I=\frac{V}{Z}=\frac{67}{13}=5.15 A[/tex]
Therefore, the average true power is
[tex]P=I^2R=(5.15)^2(12)=318.3 W[/tex]
b)
The reactive power of a circuit consisting of a resistor and a capacitor is the power given by the capacitive part of the circuit.
Therefore, it is given by
[tex]Q=I^2X[/tex]
where
I is the current
X is the reactance of the circuit
In this circuit, we have
[tex]I=5.15 A[/tex] (current)
[tex]X=5 \Omega[/tex] (reactance)
Therefore, the reactive power is
[tex]Q=(5.15)^2(5)=132.6W[/tex]
c)
In a circuit with a resistor and a capacitor, the apparent power is given by both the resistive and capacitive part of the circuit.
Therefore, it is given by
[tex]S=I^2Z[/tex]
where
I is the current
Z is the impedance of the circuit
Here we have
I = 5.15 A (current)
[tex]Z=13 \Omega[/tex] (impedance)
Therefore, the apparent power is
[tex]S=I^2 Z=(5.15)^2(13)=344.8 W[/tex]
d)
For a circuit with a resistor and a capacitor, the power factor is the ratio between the true power and the apparent power. Mathematically:
[tex]PF=\frac{P}{S}[/tex]
where
P is the true power
S is the apparent power
For this circuit, we have
P = 318.3 W (true power)
S = 344.8 W (apparent power)
So, the power factor is
[tex]PF=\frac{318.3}{344.8}=0.92[/tex]
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A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.
a) What is the magnitude of the force of the wall on ladder?b) What is the magnitude of the normal force of the ground on ladder?c) What is the minimum coefficient of friction so the ladder does not slip?d) What is the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder?
A) Force of the wall on the ladder: 186.3 N
B) Normal force of the ground on the ladder: 725.2 N
C) Minimum value of the coefficient of friction: 0.257
D) Minimum absolute value of the coefficient of friction: 0.332
Explanation:
a)
The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:
[tex]W=mg[/tex]: weight of the ladder, with m = 20 kg (mass) and [tex]g=9.8 m/s^2[/tex] (acceleration of gravity)
[tex]W_M=Mg[/tex]: weight of the person, with M = 54 kg (mass)
[tex]N_1[/tex]: normal reaction exerted by the wall on the ladder
[tex]N_2[/tex]: normal reaction exerted by the floor on the ladder
[tex]F_f = \mu N_2[/tex]: force of friction between the floor and the ladder, with [tex]\mu[/tex] (coefficient of friction)
Also we have:
L = 4.1 m (length of the ladder)
d = 3.0 m (distance of the man from point A)
Taking the equilibrium of moments about point A:
[tex]W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}[/tex]
where
[tex]Wsin 21^{\circ}[/tex] is the component of the weight of the ladder perpendicular to the ladder
[tex]W_M sin 21^{\circ}[/tex] is the component of the weight of the man perpendicular to the ladder
[tex]N_1 sin 69^{\circ}[/tex] is the component of the normal force perpendicular to the ladder
And solving for [tex]N_1[/tex], we find the force exerted by the wall on the ladder:
[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N[/tex]
B)
Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of [tex]N_2[/tex].
We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.
Therefore, we have:
[tex]\sum F_y = 0\\N_2 - W - W_M =0[/tex]
And substituting and solving for N2, we find:
[tex]N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N[/tex]
C)
Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.
The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.
Therefore, we can write:
[tex]\sum F_x = 0\\F_f - N_1 = 0[/tex]
And re-writing the equation,
[tex]\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257[/tex]
So, the minimum value of the coefficient of friction is 0.257.
D)
Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.
From part C), we saw that the coefficient of friction can be written as
[tex]\mu = \frac{N_1}{N_2}[/tex]
This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was
[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}[/tex]
We see that this quantity is maximum when d is maximum, so when
d = L
Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:
[tex]N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)[/tex]
And substituting, we get
[tex]N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N[/tex]
And therefore, the minimum coefficient of friction in order for the ladder not to slip is
[tex]\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332[/tex]
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A friend is writing a science fiction screenplay about an asteroid on a collision course with Earth. She asks you to calculate some numbers so her scenario will be correct. Astronauts will attach a rocket engine to the asteroid in an attempt to divert it. The asteroid is moving at 21 km/s. The rocket will provide an acceleration of 0.035 km/s2 at a right angle to the original motion. The rocket only has enough fuel to provide this acceleration for 40 seconds. Will this change the direction of the asteroid’s motion by at least 22°, enough to miss Earth and save civilization? (15 pts, according to Grading for problem solving, see reverse side. Your group should submit one analysis, with all group member’s names, either on the Problem solving framework or the plain white paper provided.)
Answer:
No, the deviation in the path of asteroid is not by 22°
Explanation:
Given:
velocity of asteroid, [tex]v_a=21\ km.s^{-1}[/tex]
acceleration of the rocket, [tex]a=0.035\ km.s^{-2}[/tex]
time of acceleration, [tex]t=40\ s[/tex]
Now, the final velocity of the asteroid:
using the equation of motion,
[tex]v=u+a.t[/tex]
where:
[tex]v=[/tex] final velocity
[tex]u=[/tex] initial velocity in the direction
[tex]v=0+0.035\times 40[/tex]
[tex]v=1.4\ km.s^{-1}[/tex]
Now direction of the resultant velocity:
[tex]\tan\beta=\frac{v_a}{v}[/tex]
[tex]\tan\beta=\frac{21}{1.4}[/tex]
[tex]\beta=86.186^{\circ}[/tex]
So, the deviation in the asteroid:
[tex]\theta=90-\beta[/tex]
[tex]\theta=90-86.186[/tex]
[tex]\theta=3.814^{\circ}[/tex]
How do astronomers set about looking for extrasolar planets?
Explanation:
Implicit techniques for the discovery of extra-solar planets are used by astronomers. Evidence from the radial velocity of a change in the rotation of the star indicates the presence of a planet. If, with reference to Earth, the inclination of the planet's orbit is viewed as an edge-on, the detection of light originating from the star throughout its planetary transit would then be a verification.