Complete Question
Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of light to get from the Sun to the Earth (in s) is written in scientific notation. Note: the speed of light is approximately 3.0×108 m/s.
Answer:
The values of m and n is 5 and 2
Explanation:
From the question we are told that
The speed of light is [tex]v = 3.0 *10^{8} \ m/s[/tex]
Generally the distance between the sun and the earth has a constant value
[tex]d = 1.5*10^{11} \ m[/tex]
So the time taken is mathematically evaluated as
[tex]t = \frac{1.5*10^{11}}{3.0*10^{8}}[/tex]
[tex]t = 5.0 *10^{2}[/tex]
Generally scientific notation is represented as
[tex]m* 10^n[/tex]
So comparing the standard representation of scientific notation and the value obtained we see that
m = 5 and n= 2
14. A 5,000 kg train is traveling at a velocity of 100 m/s and hits another train. The two trains stick together, and the new velocity is 50 m/s. What is the mass of the second
train?
A. 15.000 kg
B. 8.000 kg
C. 5,000 kg
D. 10.000 kg
Please help ASAP
Answer:
C
Explanation:
Momentum of first train is = m * v= (5000 kg)·(100 m/s)
= 500,000 kg·m/s.
Momentum after collision is = (5000+M) kg)·(50 m/s) And using conservation of momentum5000+M) kg)·(50 m/s) = 500,000 kg·m/s
Dividing by 50 m/s and subtracting 5000 kg, we have ...
(5000 +M) kg = 10,000 kg
M kg = 5000 kg
Why is the trajectory of the positively charged particle curving down, to the bottom of the screen, when moving in the magnetic field?
Answer:
Because the force F is always at 90 degrees to the velocity of the particle.
Explanation:
The factors that determine the path of a particle in a uniform magnetic field depend on
The magnetic flux density
The charge on the particle
The speed of the particle
Why is the trajectory of the positively charged particle curving down, to the bottom of the screen, when moving in the magnetic field?
Because the force F is always at 90 degrees to the velocity of the particle.
This can be illustrated by the equation
F = BQVSINØ
The magnetic force is always acting perpendicular to the particle velocity and its direction can be found using the left hand rule.
7. When initially-unpolarized light passes through three polarizing filters, each oriented at 45 degree angles from the preceding filter, _____ of the initial light passes through the last filter.
Answer:
Explanation:
Let the intensity of unpolarised light be I₀ . After passing through the first polarising filter , the intensity is I₀ / 2 .
After second filter , the intensity will be I₀ / 2 x cos²45 = I₀ / 4
After third filter , the intensity will be I₀ / 4 x cos²45 = I₀ / 8 .
So,
1 / 8 the of initial light passes through the last filter .
Scientist start by asking a question about something what is the initial explanation to that question called
A hypothesis. It is the first step to the scientific method.
Two identical guitar strings are prepared such that they have the same length ( 0.62 m ) and are under the same amount of tension. The first string is plucked at one location, primarily exciting the second harmonic. The other string is plucked in a different location, primarily exciting the third harmonic. The resulting sounds give rise to a beat frequency of 324 Hz . What is the wave propagation speed on the guitar strings?
Answer:
The speed is 401.76m/s
Explanation:
To find the fundamental frequency we use
3rd.harmonic - 2nd.harmonic = 324Hz
And we know that
f=v/2L
So v=f2L
v= 324 x 2 x 0.62= 401.76m/s
A car accelerates from 14.0m/s to 21m/s in 6.0s. What was its acceleration?
How far did it travel in this time? Assume constant acceleration.
Answer:
Acceleration (a) = 1.167 m/s²
Distance travel = 105 m (Approx)
Explanation:
Given:
Initial speed (u) = 14 m/s
Final speed (v) = 21 m/s
Time taken (t) = 6 sec
Find:
Acceleration
Distance travel
Computation:
v = u + at
21 = 14 + a(6)
7 = 6a
Acceleration (a) = 1.167 m/s²
S = ut + (1/2)(a)(t²)
S = (14)(6) + (1/2)(1.167)(6²)
S = 84 + 21
S = 105 m (Approx)
Distance travel = 105 m (Approx)
Discuss the characteristics of the waves in the visible portion of the electromagnetic spectrum
Explanation:
The electromagnetic spectrum consists of series of waves that are arranged from longest wavelength to shortest.
In the electromagnetic spectrum, between infrared and ultraviolet, visible light falls.The frequency of visible light is of the order of 10¹⁴ Hz.Its wavelength is about 740 nm.Visible light spectrum is visible to humans. It consists of colors like violet, blue, green, yellow etc.A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water column on the vertical sides of the U is more than 10 cm deep. Then, oil with a density of 950 kg/m3 is poured into one side, until the column of oil is 6.0 cm tall. How much higher is the top surface of the oil on that side of the tube compared with the surface of the water on the other side of the tube?
Answer:
The value is [tex]\Delta h = 0.003 \ m[/tex]
Explanation:
From the question we are told that
The height of the water is [tex]h_1 = 10 \ cm = 0.10 \ m[/tex]
The density of oil is [tex]\rho_o = 950 \ kg/m^3[/tex]
The height of oil is [tex]h_2 = 6 \ cm = 0.06 \ m[/tex]
Given that both arms of the tube are open then the pressure on both side is the same
So
[tex]P_a = P_b[/tex]
=> Here
[tex]P_a = P_z + \rho_w * g * h[/tex]
where [tex]\rho_w[/tex] is the density of water with value [tex]\rho_w = 1000 \ kg/m^3[/tex]
and [tex]P_z[/tex] is the atmospheric pressure
and
[tex]P_b = P_z + \rho_o * g * h_2[/tex]
=> [tex]P_z + \rho_w * g * h = P_z + \rho_o * g * h_2[/tex]
=> [tex]\rho_w * h = \rho_o * h_2[/tex]
=> [tex]h = \frac{950 * 0.06 }{1000}[/tex]
=> [tex]h = 0.057 \ m[/tex]
The difference in height is evaluated as
[tex]\Delta h = 0.06 - 0.057[/tex]
[tex]\Delta h = 0.003 \ m[/tex]
When discharging a fire extinguisher, it is important to aim: a. On top of the flames b. At the top of the flames c. Anywhere d. At the base of the fire
Answer:
d. At the base of the fire
Explanation:
Most fire extinguisher utilize the non-combustible nature of carbon-dioxide . carbon-dioxide is what is used in fire extinguishers, because it does not support burning. Also, carbon-dioxide is denser than air, allowing it to sink down. Aiming at the bottom of the flame allows the carbon-dioxide to sink down on the base of the flame, covering the base of the fire, and cutting off its supply of oxygen which is vital for combustion.
At what distance will a 80 W lightbulb have the same apparent brightness as a 120 W bulb viewed from a distance of 40 m
Answer:
32.6mm
Explanation:
Using area of a sphere(bulb) = 4πr²
So A is proportional to radius²
So the Energy will be proportional to r²
But 120/80 = 1.5 is the energy factor so
Using
1.5/d² = 1/r²
1.5/40²= 1/r^2
r = √( 40²/ 1.5)
r = 32.6m
hydraulic lift is to be used to lift a 2100-kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm. Determine the diameter of the piston on which the weight is to be placed.
Answer:
840 cm
Explanation:
Note: A hydraulic press operate based on pascal's principle.
From pascal's principle
W₁/d₁ = W₂/d₂...................... Equation 1
Where W₁ and W₂ are the first and second weight, and d₁ and d₂ are the first and second diameter of the piston.
make d₁ the subject of the equation
d₁ = W₁×d₂/W₂................ Equation 2
Given: W₁ = 2100 kg, W₂ = 25 kg, d₂ = 10 cm = 0.1 m.
Substitute these values into equation 2
d₁ = 2100(0.1)/25
d₁ = 8.4 m
d₁ = 840 cm
Suppose that an object in free fall were somehow equipped with a speedometer. By how much would its speed readings increase with each second of fall? Express your answer to two significant figures and include the appropriate units.
The period of a simple pendulum is 8 s. If the same pendulum experiment is repeated on a plant of g = 1/4 gE, the new period is
Answer:
16 secs.
Explanation:
Data obtained from the question include the following:
Period on earth (T) = 8 secs
Next period (Tn) =?
Next, we shall determine the length of the string. This can be obtained as follow:
Period on earth (T) = 8 secs
Acceleration due to gravity on earth (gE) = 9.8 m/s²
Pi (π) = 3.14
Length of string (L) =?
T = 2π√(L/gE)
8 = 2 × 3.14 ×√(L/9.8)
8 = 6.28 × √(L/9.8)
Divide both side by 6.28
8/6.28 = √(L/9.8)
Take the square of both side
(8/6.28)² = L/9.8
Cross multiply
L = (8/6.28)² × 9.8
L = 15.9 m
Therefore the length of string is 15.9 m
Next, we shall determine the new period of the pendulum as follow:
Length (L) = constant = 15.9 m
Pi (π) = 3.14
Acceleration due to gravity on earth (gE) = 9.8 m/s²
Acceleration due to gravity on the planet (g) = 1/4 gE = 1/4 × 9.8 m/s²
= 2.45 m/s²
New period (Tn) =?
Tn = 2π√(L/g)
Tn = 2 × 3.14 × √(15.9/2.45)
Tn = 6.28 × √(15.9/2.45)
Tn = 16 secs.
Therefore, the new period of the pendulum is 16 secs.
What is the primary reason that astronomers suspect that some jovian moons were captured into their current orbits
Answer:
The summary including its circumstances in question is outlined in the illustrating segment below.
Explanation:
An astronomer seems to be an astronomical specialist who performs his information about the research problem or area elsewhere in Earth's view.A few other moons include "backward" orbits (particularly in comparison to the movement or rotation of another planet), or are favorably committed to the equator of their continent.Name at least three fundamental differences between the harmonic oscillator dynamics and the simple pendulum dynamics
Answer: please find the answer in the explanation.
Explanation:
Harmonic can be experienced by any body that repeats itself. The pattern can be sinusoidal, square, tooth etc.
The fundamental differences between the harmonic oscillator dynamics and the simple pendulum dynamics are:
1.) The harmonic oscillator dynamics can be sinusoidal or square wave so far the motion is periodic while the simple pendulum dynamics is always sinusoidal.
2.) In simple pendulum dynamics, the period of oscillation is independent of the amplitude. While the period in harmonic oscillator dynamics depends on the amplitude.
3.) Differential equation is only one method to analyze the simple pendulum dynamics where there are several methods to analyze the harmonic oscillator dynamics.
A truck travels due east for a distance of 1.9 km, turns around and goes due west for 9.3 km, and finally turns around again and travels 3.4 km due east.
What is the total distance that the truck travels? What are the magnitude and direction of the truck's displacement?
Answer:
The total distance of the truck is 14.6 km
The magnitude and direction of truck's disaplacement is 4 km east
Explanation:
Distance is the total path covered by the trcuk regardless of the direction since distance is described by magnitude only.
The total distance traveled by the truck is given by;
Total Distance = 1.9 km + 9.3 km + 3.4 km
Total Distance = 14.6 km
Displacement is the change in postion at a time interval. In dispalcement direction is considered.
let eastward motion be positive
let westward motion be negative
Displacement = 1.9km east - 9.3 km west + 3.4 km east
Displacement = -4 km east
Therefore, the magnitude and direction of dispalcement = 4 km east
A 0.16 pF parallel - plate capacitor is charged to 10 V. Then the battery is disconnected from the capacitor. When 1.00 107 electrons are now placed on the negative plate of the capacitor, the voltage between the plates changes by
Answer:
The value is [tex]V_d = 5 \ V[/tex]
Explanation:
From the question we are told that
The capacitance of the capacitor is [tex]C = 0.16 \ pF = 0.16*10^{-12} \ F[/tex]
The voltage is [tex]V = 10\ V[/tex]
The number of electrons present on the negative plate is [tex]N_e = 1.00 *10^{7} \ electrons[/tex]
Generally the charge on the positive plate at 10 Volt is mathematically represented as
[tex]Q_a = C * V[/tex]
=> [tex]Q_a = 0.16*10^{-12} * 10[/tex]
=> [tex]Q_a = 0.16*10^{-11} \ C[/tex]
Now the charge on the plate when the electron where placed is evaluated as
[tex]Q_b = Q_a + ( N_e * e)[/tex]
Where e is the charge on each electron with a value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]Q_b = 0.16*10^{-11} + (1.0*10^7 * 1.60*10^{-19})[/tex]
=> [tex]Q_b = 3.2 *10^{-12 } \ C[/tex]
Generally the voltage between the two plate is evaluated as
[tex]V_d = \frac{ Q_b - Q_a }{ 2 * C }[/tex]
=> [tex]V_d = \frac{ 3.2*10^{-12} - 0.16*10^{-11}}{2 * 0.16 *10^{-12}}[/tex]
=> [tex]V_d = 5 \ V[/tex]
A sign is hung between two cables. If the sign weighs 350 N, what is the tension (in N) in each cable?
Answer:
Tension in each cable = 175 N
Explanation:
Given:
Number of cables = 2
Weight of sign board(T) = 350 N
Find:
Tension in each cable
Computation:
We know that,
T1 = T2
So,
T1 + T2 = T
T1 + T2 = 350
T1 + T1 = 350
T1 = 175
Tension in each cable = 175 N
Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s.A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat of frequency 6.00 Hz, which increases to 7.00 Hz when she tightens her violin string slightly.
(a) What was the frequency of the note played by her violin when she heard the 3-Hz beats?
(b) To get her violin perfectly tuned to concert A, should she tighten or loosen her string from what it was when she heard the 3-Hz beats?
Answer:
A) 443 Hz
B) She has to loosen the string
Explanation:
A) Given;
Beat frequency;f_beat = 3 Hz
Frequency of electronically generated tone; f_e = 440 Hz
We know that formula for beat frequency is given by;
f_beat = |f1 - f2|
Now, applying it to this question, we have;
f_beat = f_v - f_e
Where f_v is frequency of the note played by the violinist
Thus, plugging in the relevant values;
3 = f_v - 440
f_v = 3 + 440
f_v = 443 Hz
B) In the concept of wave travelling in a string, the frequency is directly proportional to the square root of the force acting on the string.
Now, for the violinist to get her violin perfectly tuned to concert A from what it was when she heard the 3-Hz beats, the beat frequency will have to be zero. Which means it has to decrease by 3 Hz. For it to decrease, it means that the force applied has to decrease as we have seen that frequency is directly proportional to the square root of the force acting on the string.
Thus, she would have to loosen the string.
A cosmic-ray proton in interstellar space has an energy of 10.0 MeV and executes a circular orbit having a radius equal to that of Mercury’s orbit around the Sun (5.80 × 1010m). What is the magnetic field in that region of space?
Answer:
The value is [tex]B = 7.7106 *10^{-12 } \ T[/tex]
Explanation:
From the question we are told that
The energy is [tex]E = 10.0 MeV = 10.0 *10^{6} \ eV = 10.0 *10^{6} * 1.60*10^{-19} = 1.6 *10^{-12} \ J[/tex]
The radius is [tex]R = 5.80 *10^{10} \ m[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \frac{m * v }{q * R }[/tex]
Where m is the mass of proton with value [tex]m = 1.6*10^{-27} \ kg[/tex]
v is the velocity of the proton which is mathematically deduced from the formula for kinetic energy as
[tex]v = \sqrt{ \frac{E }{ 0.5 * m } }[/tex]
Here E is also equivalent to kinetic energy of the proton so
[tex]v = \sqrt{ \frac{1.6 *10^{-12} }{ 0.5 * 1.6*10^{-27} } }[/tex]
[tex]v = 4.47214 *10^{7} \ m/s[/tex]
So
[tex]B = \frac{1.60 *10^{-27} * 4.47214 *10^{7}}{ 1.60 *10^{-19} * 5.80*10^{10}}[/tex]
[tex]B = 7.7106 *10^{-12 } \ T[/tex]
The magnetic field in the region has a magnitude of 7.7×[tex]10^{-12}[/tex] T.
Let's consider the magnetic field in the region be B. Now the proton in the given orbit is in equilibrium under the influence of centripetal force ([tex]F_{c}[/tex]) and magnetic force([tex]F_{b}[/tex]).
[tex]F_{c}[/tex] = [tex]F_{b}[/tex]
m[tex]v^{2}[/tex]/R = qvB
⇒ B = mv/qR
where, m is the mass of proton = 1.6 × [tex]10^{-27}[/tex] kg
q is the charge of proton = 1.6 × [tex]10^{-19}[/tex] C
R is the radius of the orbit = 5.8 × [tex]10^{10}[/tex] m
now we can calculate the velocity of the proton, v, from the energy of the proton given in the question that is E = 10 MeV.
1 MeV = 1.6 × [tex]10^{-19}[/tex] J
v = [tex]\sqrt{\frac{2E}{m} }[/tex]
=[tex]\sqrt{\frac{2*10*10^{6}* 1.6*10^{-19} }{1.6*10^{-27} } }[/tex]
v = 4.47 × [tex]10^{7}[/tex] m/s
now,
B = mv/qR
= [tex]\frac{1.6*10^{-27}*4.47*10^{7} }{1.6*10^{-19}*5.8*10^{10} }[/tex]
B = 7.7 ×[tex]10^{-12}[/tex] T
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A small uniform disk and a small uniform sphere are released simultaneously at the top of a high inclined plane, and they roll down without slipping. Which one will reach the bottom first?
Answer:
the sphere
Explanation:
From the given information,
A free flow body diagrammatic expression for the small uniform disk and a small uniform sphere which are released simultaneously at the top of a high inclined plane can be seen in the image attached below.
From the diagram;
The Normal force mgsinθ - Friction force F = mass m × acceleration a
Meanwhile; the frictional force
[tex]F = \dfrac{I \alpha }{R}[/tex]
where
[tex]\alpha = \dfrac{a}{R}[/tex] in a rolling motion
Then;
[tex]F = \dfrac{I a }{R^2}[/tex]
∴
The Normal force mgsinθ - F = m × a can be re-written as:
[tex]\mathtt{mg sin \ \theta- \dfrac{Ia}{R^2} = ma}[/tex]
making a the subject of the formula, we have:
[tex]a = (\dfrac{mg \ sin \theta}{m + \dfrac{I}{R^2}})[/tex]
Similarly;
I = mk² in which k is the radius of gyration
∴
replacing I = mk² into the above equation , we have:
[tex]a = (\dfrac{mg \ sin \theta}{m + \dfrac{mk^2}{R^2}})[/tex]
where;
the uniform disk [tex]\dfrac{k^2}{R^2 }= \dfrac{1}{2}[/tex]
the uniform sphere [tex]\dfrac{k^2}{R^2 }= \dfrac{2}{5}[/tex]
∴
[tex]a = \dfrac{2}{3} \ g sin \theta \ for \ the \ uniform \ disk[/tex]
[tex]a = \dfrac{5}{7} \ g sin \theta \ for \ the \ uniform \ sphere[/tex]
We can now see that the uniform sphere is greater than the disk as such the sphere will reach the bottom first.
Sound waves reaching the olfactory mucosa do not elicit olfactory perception due to the ____________ not matching the type of stimulus of the receptor.
Answer:
Sound waves reaching the olfactory mucosa do not elicit olfactory perception due to the ___MODALITY_________ not matching the type of stimulus of the receptos
Explanation:
Because modality of sensation refers to what is perceived after the stimulus is effected eg the pressure modality is effected when the pressure receptors of the skin are stimulated so in this case sound will not elite olfactory sensation because the modality here is sound rather than smell
A rectangular wooden block, 22 cm x 13.2 cm x 4.4 cm, has a mass of 1562.0 g. The density of the wood in kg/m 3 is:
Answer:
1222.45 Kg/m³
Explanation:
The following data were obtained from the question:
Dimension of rectangular block = 22 cm x 13.2 cm x 4.4 cm
Mass (m) = 1562.0 g
Density (D) =?
Next, we shall determine the volume of the rectangular block.
This is illustrated below:
Volume = length x breadth x height
V = 22 cm x 13.2 cm x 4.4 cm
V = 1277.76 cm³
Next, we shall convert 1277.76 cm³ to m³.
This can be obtained as follow:
1 cm³ = 1×10¯⁶ m³
Therefore,
1277.76 cm³ = 1277.76 × 1×10¯⁶
1277.76 cm³ = 1.27776×10¯³ m³
Therefore, 1277.76 cm³ is equivalent to 1.27776×10¯³ m³.
Next, we shall convert 1562.0 g to kg. This can be obtained as follow:
1000 g = 1 kg
Therefore,
1562.0 g = 1562/1000 = 1.562 kg
Therefore, 1562.0 g is equivalent to 1.562 Kg.
Finally, we shall determine the density of rectangular block as follow:
Volume = 1.27776×10¯³ m³.
Mass = 1.562 Kg.
Density =?
Density = mass /volume
Density = 1.562 / 1.27776×10¯³
Density = 1222.45 Kg/m³
Therefore, the density of rectangular block is 1222.45 Kg/m³
light of a wavelength 600 nm shines on a soap bubble film. For what soap film thickness will destructive interference occur
Answer:
The minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.
Explanation:
Given;
wavelength of light, λ = 600 nm
The minimum thickness of the soap bubble for destructive interference to occur is given by;
[tex]t = \frac{\lambda/n}{2}\\\\t = \frac{\lambda}{2n}[/tex]
where;
n is refractive index of soap film = 1.33
[tex]t = \frac{\lambda}{2n} \\\\t = \frac{600*10^{-9}}{2(1.33)}\\\\t = 2.2556 *10^{-7} \ m\\\\t = 225.56 *10^{-9} \ m\\\\t = 225.56 \ nm[/tex]
Therefore, the minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.
Is the sinusoidal pattern on a string longer or shorter when there is a greater propagation velocity?
Answer:
We know that
Velocity= wavelength x frequency
And f= 1/2π√T/u
So we also know that
The frequency of nth harmonic standing wave is
fn=nv/2L
So from the relationships above
We can see that as propagational velocity increases, the frequency increases and the wavelength decreases. So finally we can say that sinusoidal pattern on a string is shorter when there is a greater propagation velocity
Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.
Answer:
14.57 ohms
Explanation:
Here in the figure ,Rb & R₄are in series & also Rc & R₅ are in series. As they are in series , ( Rb + R₄ ) & (Rc & R₅) are in parallel . So the equivalent resistance in that branch = ( 2 + 18 ) ║ ( 3 + 12 )
= 20 ║ 15
= (20×15) / (20 + 15)
= 8.57 ohms
Also Ra ( 6 ohm ) is in series with that branch ,. So the equivalent resistance of the whole circuit = 8.57 + 6 = 14.57 ohms.
total resistance is 26 ohm and voltage across [tex]R_C[/tex] resister is 1.1554 Ampere.
According to the circuit,
[tex]R_A[/tex] = 6 Ω.
[tex]R_B[/tex] = 2 Ω.
[tex]R_C[/tex] = 3 Ω.
R₄ = 18 Ω.
R₅ = 12 Ω.
V = 10 V.
In the circuit, the series combination of [tex]R_C[/tex] and R₅ is parallel connected with series combination of [tex]R_B[/tex] and R₄. And this network is connected with [tex]R_A[/tex] in series combination.
So, Equivalent resistance of the circuit is; R = [tex]R_A[/tex] + ( [tex]R_C[/tex] + R₅) II ([tex]R_B[/tex] +R₄)
= 18 Ω + ( 3 Ω + 12 Ω)II(2 Ω+ 18 Ω)
= 18 Ω + 15ΩII20Ω
= 18 Ω + (15×20)/(15+20)Ω
= 18 Ω + 8.57 Ω
= 26 Ω
So, current flowing through the circuit: I = V/ R = 10V / 26Ω = 0.38 Amp.
Voltage drop across the parallel network; V₁ = V - I[tex]R_A[/tex]
= 10V - 0.38×6 V
= 7.72 V.
So, voltage across the resister [tex]R_C[/tex] = V₁[tex]R_C[/tex]/[tex](R_C[/tex] + R₅)
= 7.72 × 3/(3 + 12) Amp
= 1.1544 Amp.
Hence, total resistance is 26 ohm and voltage across specific resister is 1.1554 Ampere.
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Find the magnitude of u × v and the unit vector parrallel to u × v in the direction of u × v. u = 2i + 2j - k, v = -i + k
Answer:
magnitude = 3
unit vector = [tex]\frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}[/tex]
Explanation:
Given vectors:
u = 2i + 2j - k
v = -i + k = -i + 0j + k
(a) u x v is the cross product of u and v, and is given by;
[tex]u X v = \left[\begin{array}{ccc}i&j&k\\2&2&-1\\-1&0&1\end{array}\right][/tex]
u x v = i(2+0) - j(2 - 1) + k(0 - 2)
u x v = 2i - j - 2k
Now the magnitude of u x v is calculated as follows:
| u x v | = [tex]\sqrt{2^2 + (-1)^2 + (-2)^2}[/tex]
| u x v | = [tex]\sqrt{4 + 1 + 4}[/tex]
| u x v | = [tex]\sqrt{9}[/tex]
| u x v | = 3
Therefore, the magnitude of u x v is 3
(b) The unit vector û parallel to u x v in the direction of u x v is given by the ratio of u x v and the magnitude of u x v. i.e
û = [tex]\frac{u X v}{|u X v|}[/tex]
u x v = 2i - j - 2k [calculated in (a) above]
|u x v| = 3 [calculated in (a) above]
∴ û = [tex]\frac{2i - j - 2k}{3}[/tex]
∴ û = [tex]\frac{2i}{3} - \frac{j}{3} - \frac{2k}{3}[/tex]
A child of mass 22.0 kg is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she experience to stay on if she is 1.25 m from the center
Answer:
Centripetal Force = 483.3 N
Explanation:
A centripetal force is the force that tends to keep a mocing object along a curved path and it is directed towards the centre of the rotatio, while centrifugal force is an apparent force that tends to force a rotating object away from the center of the rotation.
The formula for centripetal force is given by:
[tex]F_c = \frac{mv^2}{r} \\where:\\F_C = centripetal\ force\\m = mass\ = 22kg\\\omega =angular\ velocity = 40.0\ rev/min[/tex]
Let us work on the angular velocity (ω), by converting to radians/ seconds
ω = 40 rev/min,
1 rev = 2π rad
∴ 40 rev = 2π × 40 rad = 80π rad
1 min = 60 seconds
[tex]\therefore\ 40\ rev \slash min = \frac{80\ \times\ \pi\ rad}{60\ seconds} \\40\ rev \slash min = 4.189\ rad \slash sec[/tex]
Next let us find the velocity (v) from the angular velocity. Velocity (v) and angulsr velocity (ω) are related by the equation:
v = ω × r (m/s)
v = 4.189 × 1.25
v = 5.24 m/s
Finally, the centripetal force is calculated thus:
[tex]F_c = \frac{mv^2}{r} \\\\F_c = \frac{22 \times (5.24)^2 }{1.25} \\\\F_c = \frac{604.07}{1.25}\\ F_c = 483.3N[/tex]
An object is continuously changing its velocity by the same rate. What is true
about this object?
a. The object has decreasing acceleration.
b. The object has increasing acceleration.
c. The object has zero acceleration.
d. The object has constant acceleration.
Answer:
d. The object has constant acceleration.
Explanation:
Acceleration is the rate of change of velocity. So if the rate of change is the same at all times, then the acceleration is constant.
What is the direction of the force? Answer with a positive angle measured counterclockwise from the positive x axis.
Answer:
hello the diagram related to your question is missing attached below is the missing diagram
answer : 107.07⁰
Explanation:
The direction of the force can be calculated as below
angle = 180 - ∝ - ∅ ------ 1
∅ = [tex]tan^{-1} b/a[/tex] = tan^-1 ( 0.429 )
∅ = 23.2⁰
∝ can be determined by applying the parallelogram law
∝ = [tex]tan^{-1} ( \frac{R24sin56.4}{f13 + R24cos56.4} )[/tex]
= tan^-1 ( 21.173 / 17.935 )
∝ = 49.73⁰
back to equation 1 direction of force with positive angle
= 180 - 49.73 - 23.2 = 107.07⁰