Answer:
Total distance covered (scalar quantity) = 23 km
Displacement (vector quantity) = 3 km north from the original starting point
Explanation:
Since he drove 13 km north and then 10 km south, the total distance he cover in his drive was: 13 km + 10 km = 23 km
On the other hand, his displacement was 3 km north from where he started.
light of wavelength 587.5 nm illuminates a slit of width 0.75 mm. at what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.85 mm from the central maximum?
1.085m
Explanation:
Using
a= lambda/sinစ
Sinစ= (587.5*10^-9) x 0.75*10^-3
= 0.000783
Sinစ=0.875*10^-3/d
0.000783= 0.875/d
d= 1.085m
What is the force on an object due to its mass and gravity called?
Answer:
The Gravitational force.
when a marble rolls down a slope which forces acts on it
Answer:
What forces act on a marble rolling down a ramp?
Answer: Gravity acts vertically downward. A normal force acts from the ruler toward the marble/ball in a direction that is perpendicular to the plane of the ruler. Friction acts in the direction opposite to which the marble/ball is moving. ... Friction slows down the marble/ball.
g A ball is thrown against the wall and bounces back with the same velocity. What type of collision is this
Answer:
Perfectly elastic collision
Explanation:
In a closed system, an elastic collision is a type of collision between two bodies, where the total momentum and kinetic energy are conserved.
We are told from the problem that the ball bounces back with its original velocity. For the ball to bounce back to the thrower in the first place, this is our first hint that the collision is elastic. If the collision was inelastic, the ball would most likely have stuck to the wall.
In addition to that, the velocity of the ball remains fairly unchanged even after the collision. This confirms that the kinetic energy it had before the collision is the same as the kinetic energy it has after the collision.
As a result of this, the collision is perfectly elastic
g The velocity (V) of a particle is given by 5t2 meters, where t is in s. Find the acceleration of the particle at t=3s.
Answer:
Acceleration, [tex]a=45\ m/s^2[/tex]
Explanation:
The velocity of a particle is given by :
[tex]V=5t^2[/tex]
t is time in seconds
The acceleration in terms of velocity is given by :
[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(5t^2)}{dt}\\\\a=5\times \dfrac{t^3}{3}[/tex]
We need to find the acceleration of the particle at t = 3 s. Put t = 3 s in the expression of a.
So,
[tex]a=\dfrac{5\times 3^3}{3}\\\\a=45\ m/s^2[/tex]
So, the acceleration of the particle at t = 3 s is [tex]45\ m/s^2[/tex].
(Application) You are asked to design a multiple lens system for the scenarios below. The lenses youhave available to you are:1. converging (f =7.5 cm)2. converging (f =15.0 cm)3. diverging (f = -15.0 cm)Think about the following scenarios:A) Lens 1 makes a real image i1, of which lens 2 makes a real image i2.B) Lens 1 makes a real image i1, of which diverging lens 2 makes a virtual image i2.C) Converging lens 1 makes a virtual image i1, of which converging lens 2 makes a virtual image i2.D) Converging lens 1 makes a virtual image i1, of which lens 2 makes a real image i2.For each scenario, decide:1) What type of lens is lens 1? Lens 2? Why? 2) Where should the object be placed with respect to the focal point to make the indicated type of image?3) Make a schematic drawing for Scenario D only. If you want extra practice, try to makeschematic drawings of the other three scenarios.
Answer:
1) A) Lens converging, B) lens 1 converging
2) A and B object is at a greater distance than its focal
3)
Explanation:
For this exercise we will give some more important characteristics of the lenses.
we use the equation of the constructor
1 / f = 1 / p + 1 / q
where f is the focal length and p and qlas the distance to the object and image, respectively.
We must also use how the lenses are combined
1 / = 1 / f₁ + 1 / f₂
where feq is the equivalent focal length and it is assumed that the two lenses are in contact.
By applying the equation of the constructor you get some characteristics of the image
Converging lens
* Object farther than focal length real and inverted image
* Oject after focal virtual image and right
Diverging lens
* for all virtual image distance and right
with this information let's analyze the different configurations
1) let's do an analysis for each scenario
A) since lens 1 and lens 2 create real image they must be converging lenses
B) Lens 1 must be convergent and lens 2 can be of both types
C) Lens 1 can be convergent with the object that is at a shorter distance than at focal point (p <f₁). The same analysis diverges for lens 2, the object that is the image of the other lens is shorter than its focal distance p₂ <f₂
D) the object is at a shorter distance than the focal length p₁ <f₁, the object for this lens is at a greater distance than its focal point p₂> f₂
2) let's analyze for each scenario
A) for each lens the object is at a greater distance than its focal length p₁> f₁ and p₂> f₂
B) for lens 1 p₁> f₁
for lens 2
if it is convergent p₂ <f₂
and if it is divergent any distance is possible
C) lens 1 p₁ <f₁ and for lens 2 p₂ <f₂
D) lens 1 p₁ <f₁ and for lens 2 p₂> f₂
3) see attached
The distance with subscript 1 is measured with respect to line1 and the distance with subscript 2 is measured with respect to the lens with subscript 2
After crossing the finish line, a race car slows down from 47 m/s to 32m/s in 3seconds. What is the car’s acceleration?
Answer: -5 m/s^2
Explanation: a = v - u/t
= 32 - 47/3
= -15/3
= -5 m/s^2
A bat emits a sound at a frequency of 30.0 kHz as it approaches a wall. The bat detects a beat frequency of 700 Hz. The speed of the bat is closest to
Answer:
3.948m/s
Explanation:
To solve this we need to apply Doppler effect theory
So
To find the frequency received by insect will be gotten when the Source and observer both are moving in same direction which is given by
f1 = f0 x (V - Vo)/(V - Vs)
f0 = 30.0 kHz
V = 344 m/s
Vs will now be the speed of the bat and
Vo will be the speed of the object which is = 0 m/s
So substituting we have
f1 = 30 x 10^3 x (344- 0)/(344- Vs)
Next to find the frequency reflected by wall we use
f2 = f1 x (V + Vs)/(V + Vo)
So substituting the value of f1 calculated above we have
f2 = 30 x 10^3 x (344 + Vs) x (344 - 0)/[(344 - Vs) x (344 + 0)]
f2 = 30 x 10^3 x (344 + Vs)/(344- Vs)
But the beat frequency detected by bat is 700 Hz,
So we say
f2 - f0 = 700 Hz
30 x 10^3 x (344+ Vs)/(344 - Vs) - 30x 10^3 = 700
(344 + Vs)/(344 - Vs) = 1 + 700/30000 = 1.023
344 + Vs = 344 x 1.023 - Vs x 1.0233
Vs = 344 x ( 1.023 - 1)/(1 + 1.023)
So finally
Vs = Speed of source that is the bat is = 3.949m/s
g Explain completely, but in the fewest words possible, why a ball thrown horizontally follows the path that it does
Answer:
Explanation:
A ball thrown horizontally is a type of projectile which moves in a trajectory path. It moves with respect to its momentum and gravitational pull of the earth.
The motion of the thrown ball can be described in two components, horizontal and vertical. In horizontal component of motion of the thrown ball, gravitational force is zero. While in the case of vertical component, the gravitational pull causes the path followed.
The force of gravity pulls the ball towards the surface of the earth gradually, thus forming a trajectory path.
When a circuit is arranged in parallel:________
a) There is only one path the electrons can take through the circuit, and they must pass through all circuit components
b) There are multiple paths the electrons can take through the circuit, and it is possible for the electron to pass through one circuit component but not another.
Answer:
Answer:
B .There are multiple paths the electrons can take through the circuit, and it is possible for the electron to pass through one circuit component but not another.
Explanation:
Because in a parallel circuit, all components are connected across each other, forming exactly two sets of electrically common points.
Why was miasma theory replaced?
The flight path of a jet aircraft as it takes off is defined by the parmetric equations x=1.25 t2 and y=0.03 t3, where t is the time after take-off, measured in seconds, and x and y are given in meters. If the plane starts to level off at t = 40 s, determine at this instant:a. The horizontal distance it is from the airport b. Its altitudec. Its speed.
Answer:
A. We are given t= 40s, x=1.25t²
So The horizontal distance it is from the airport will be = 1.25(40)²= 2000meters
B. y= 0.03t³
So y= 0.03(40)³= 1920m altitude
C. To find speed
We say
dx/dt=Vx= 2.5t= 100
dy/dy = 0.09t= 144
So speed= √100²+144²= 175.31m/s
When you attempt to swim, you brush your hands in a wide motion to push the water around you back behind you. This in turn, causes the water around you to push your forward. Which of Newton's Laws would explain this action? a. Newton's First Law b. Newton's Second Law c. Newton's Third Law d. None of Newton's Laws
Answer:
C. Newton's 3rd law
Explanation:
Swimmers must stroke downward in the water to stay afloat and propel forward. This movement is equal and opposite to the force the water exerts against the swimmer to stop them from moving.
Answer:
c. Newton's Third Law
Explanation:
I got it correct on the test
A laser with a power of 1.0 mW has a beam radius of 1.0 mm. What is the peak value of the electric field in that beam
Answer:
The peak value of the electric field is 489.64 V/m
Explanation:
Given;
power of the laser, P = 1.0 mW = 1 x 10⁻³ W
Radius of the beam, R = 1.0 mm = 1 x 10⁻³ m
Area of the beam = πr² = π(1 x 10⁻³ )² = 3.142 x 10⁻⁶ m²
The average intensity of the light = P / A
The average intensity of the light = ( 1 x 10⁻³) / (3.142 x 10⁻⁶)
The average intensity of the light = 318.27 W/m²
The peak value of the electric field is given by;
[tex]E_o = \sqrt{\frac{2I_{avg}}{c\epsilon_o}}\\\\E_o = \sqrt{\frac{2(318.27)}{(3*10^8)(8.85*10^{-12})}}\\\\E_o = 489.64 \ V/m[/tex]
Therefore, the peak value of the electric field is 489.64 V/m.
8. How did the measured angular magnification of the telescope compare with the theoretical prediction?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The theoretical angular magnification lies within the angular magnification range
Explanation:
From the question we are told that
The focal length of B is [tex]f_{objective } = 43.0 \ cm[/tex]
The focal length of A is [tex]f_{eye} = 10.4 \ cm[/tex]
The theoretical angular magnification is mathematically represented as
[tex]m = \frac{f_{objective }}{f_{eye}} = \frac{43.0}{10.4}[/tex]
[tex]m = \frac{f_{objective }}{f_{eye}} = 4.175[/tex]
Form the question the measured angular magnification ranges from 4 -5
So from the value calculated and the value given we can deduce that the theoretical angular magnification lies within the angular magnification range
A train which is traveling at 73mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased velocity is observed to be 56mi/hr as it passes a point 1/2mi beyond A. A car moving at 45mi/hr passes point B at the same instant that the train reaches point A. In an unwise effort to beat the train to the crossing, the driver "steps on the gas." Calculate the constant acceleration a that the car must have in order to beat the train to the crossing by 3.9sec and find the velocity v of the car as it reaches the crossing.
Answer:
First to find deceleration of the train we use
v²= u²+ 2as
56²= 73²+ 2(0.5)a
a= -2193mi/hr²
Then we find time in which the train does the intersection
Using
S= ut+ 1/2 at²
1= 73t-1/2(1293)t²
t =68.5s
But since the train is to intersect in 3.9s the time will be the difference which is
65.68s
So finding acceleration
S= ut + 1/2at²
1.3mi= 45/3600mi/s(65.58s)+ 1/2a(65.5)²
So a= 1.179ft/s²
To find velocity we use
V= u + at
= 45/3600mi/s + (-2.33E-4mis²)(65.58s)
V= 0.0271mi/s
= 97.6ft/s
A rough value of deceleration of a skidding automobile is about 7.0m\s^2.using this how long does it take for a car going at 30m\s to stop after the skid starts.How far dose the car go in this time??
Explanation:
Given:
v₀ = 30 m/s
v = 0 m/s
a = -7.0 m/s²
Find: t and Δx
v = at + v₀
0 m/s = (-7.0 m/s²) t + 30 m/s
t = 4.3 seconds
v² = v₀² + 2aΔx
(0 m/s)² = (30 m/s)² + 2 (-7.0 m/s²) Δx
Δx = 64 meters
Part B
Do the atoms in salt follow a pattern?
Answer: Yes.
Explanation: Because of the crystalline structure.
What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?
This question is incomplete, the complete question is;
A weightlifter holds a 1,300 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m?
What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?
Answer: Average force exerted by the weightlifter in the process = 1600N
Explanation:
To find Work done to lift a barbell and half of the hanging chain we say;
W₁ = ( 1300N + (1/2 × 400N)) × 1m
W₁ = (1300 + 200) Nm
W₁ = 1500J
now work done to lift the upper half of the chain we say:
W₂ = (1/2 × 400N) × (1/2 × 1m)
W₂ = 200N × 0.5m
W₂ = 100J
So total work done will be
W = W₁ + W₂
W = 1500J + 100J
W = 1600J
To find the average force exerted by the weight lifter, we say;
F = W/D
F = (1600 / 1m) N
F = 1600N
∴Average force = 1600N
A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 193 m/s and a frequency of 235 Hz . The amplitude of the standing wave at an antinode is 0.380 cm.
A) Calculate the amplitude at a point on the string a distance of 18.0 cm from the left-hand end of the string.
B) How much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point?
C) Calculate the maximum transverse velocity of the string at this point.
D) Calculate the maximum transverse acceleration of the string at this point.
Answer:
A. We know that amplitude at x is
Asin (kx)
But k= 2πf/v
k= 2*3.132*235/193= 7.65
So A = 0.35*sin( 7.65x 0.18)= 0.00841m
C
Vmax = Amplitude x angular velocity
= 0.0084 x 2πf
= 0.0084* 2*3.142* 235= 12.4m/s
D. Maximum acceleration = omega² x Amplitude
= (2πf)²* 0.00841= 183.40m/s²
According to Hookes Law, in order for a spring to apply a restorative force, what must be true about the displacement of the spring? a. The displacement must be proportional to the spring constant b. The displacement must be towards the spring, to compress it and cause it to force back c. The displacement must be measured away from a starting rest position in order to have a force from the spring d. The displacement must be away from the spring, to stretch it and cause it to pull back
Answer:
Its not A..
Explanation:
That was my choice and it was incorrect
The displacement must be towards the spring, to compress it and cause it to force back.
The displacement must be away from the spring, to stretch it and cause it to pull back.
What is the spring constant?The spring constant is used to define the stiffness of the spring, the greater the value of the spring constant stiffer the spring and it is more difficult to stretch the spring.
The spring's restoring force attempts to return it to its equilibrium position; as a result, the displacement must be away from the spring in order to stretch it and make it a drawback, and it must be toward the spring in order to compress it and force it back.
Thus, the correct answers are option B and option D.
Learn more about spring constant,
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Calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle at 16.2 m/s.
Answer:
The maximum height to which water could be squirted with the hose is 13.380 meters.
Explanation:
A line of current of a fluid can be explained sufficiently by Bernoulli's Theorem. In this case, the system can be simplified due to neglectance of changes in absolute pressure. Water is squirted with an initial speed and reaches its maximum height, where final speed is zero. That is to say:
[tex]\frac{v_{1}^{2}}{2\cdot g} +z_{1} = \frac{v_{2}^{2}}{2\cdot g} +z_{2}[/tex]
Where:
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final height of water, measured in meters.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of water, measured in meters per second.
If [tex]z_{1} = 0\,m[/tex], [tex]v_{1} = 16.2\,\frac{m}{s}[/tex], [tex]v_{2} = 0\,\frac{m}{s}[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then:
[tex]z_{2} = \frac{v_{1}^{2}-v_{2}^{2}}{2\cdot g} +z_{1}[/tex]
[tex]z_{2} = \frac{\left(16.2\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)} +0\,m[/tex]
[tex]z_{2} = 13.380\,m[/tex]
The maximum height to which water could be squirted with the hose is 13.380 meters.
What do we measure sound intensity in?
Answer:
we measure sound intensity in Decibels.
Answer:
DecibelsExplanation:
We measure sound power or sound pressure in decibels.
They were named in honour of Alexander Graham Bell,( the inventor of both the telephone and the audiometer).
Why is the forward biased voltage for a typical red LED so much greater than the forward biased voltage for a typical silicon diode?
Answer:
This is because as we know the magnitude of a forward voltage drop depends on the material used for constructing the diode and also the magnitude of the band gap, so since LED material are generally gallium arsenide and gallium phosphide which has the higher band gap than silicon and since the higher the band gap the higher the voltage drop so LED has the greater forward voltage drop then the silicon diode
The voltage for the following cell is +0.731 V. Find Kb for the organic base RNH2. Use E_SCE = 0.241V.
Pt(s)│H2 (1.00 atm)│RNH2 (0.100M), RNH3 + (0.0500M)║SCE
Answer: Kb = 1.89 × 10⁻⁶
Explanation:
R-NH₃+ (aq) <----------> R-NH₂ (aq) + H^+ (aq)
Ecell = - ( 0.0592 / n ) log Kₐ
Where, Ecell = 0.731 - 0.241 = 0.490 V
Therefore, 0.490 = - (0.0592 / 1 ) log Kₐ
Therefore, Kₐ = 5.30 × 10⁻⁹
Thus, Kb = Kw / Ka = ( 1.0 × 10⁻¹⁴ / 5.30 × 10⁹ ) = 1.89 × 10⁻⁶
Understanding how behavior has changed over time to help humans adapt to their environment is known as the __________ approach to psychology.
biological
evolutionary
sociocultural
behavioral
The behavioral approach's main objective is to describe how leaders mix those same two types of actions to influence followers throughout their attempts to achieve those goals.Behaviorism seems to be a teaching strategy, whereas theoretical approaches are concerned with how humans are conditioned just to respond to events as well as stimuli. Several ideas describe how conduct is influenced by experience.
Thus the response above is correct.
Learn more about behavioral approach here:
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You don't learn any movement concepts until high school. A. True B. False
Answer:
B
Explanation:
Answer:
false
Explanation:
A railroad train is traveling at a speed of 26.0 m/s in still air. The frequency of the note emitted by the locomotive whistle is 420 Hz.
1. What is the wavelength of the sound waves in front of the locomotive? Use 344 m/s for the speed of sound in air.
2. What is the wavelength of the sound waves behind the locomotive? Use 344 m/s for the speed of sound in air.
3. What is the frequency of the sound heard by a stationary listener in front of the locomotive? Use 344 m/s for the speed of sound in air.
4. What is the frequency of the sound heard by a stationary listener behind the locomotive? Use 344 m/s for the speed of sound in air.
Answer: 0.757m; 0.881m; 432.70Hz; 371.89Hz
Explanation:
Give the following :
Velocity of train (Vt) = 26m/s
Frequency of sound (Fs) = 420Hz
Speed of sound (Vs) = 344m/s
1) wavelength = (Vs - Vt) / Fs
Wavelength = (344 - 26) / 420 = 318/420 = 0.757m
11) Wavelength = (Vs + Vt) / Fs
Wavelength = (344 + 26) / 420 = 370/420 = 0.881m
111) According to the doppler effect :
Fl = [(V + Vl) / (V + Vs)] * fs
Fl = frequency of listener ; fs = frequency of sound source ; V = speed of sound ; Vl = Velocity of listener ; Vs = speed of sound source
Vs = - ve (train moving towards listener)
Fl = [(V + Vl) / (V + Vs)] * fs
Fl = [(344 + 0) / (344 - 26)] * 400
Fl = (344 / 318) * 400 = 432.70Hz
1V) Vs = + ve (train moving away listener)
Fl = [(V + Vl) / (V + Vs)] * fs
Fl = [(344 + 0) / (344 + 26)] * 400
Fl = (344 / 370) * 400 = 371.89Hz
Here are the positions at three different times for a bee in flight (a bee's top speed is about 7 m/s). Time 6.6 s 6.9 s 7.2 s Position 1.8, 5.0, 0 m 0.5, 6.9, 0 m −0.4, 9.5, 0 m (a) Between 6.6 s and 6.9 s, what was the bee's average velocity? Be careful with signs. vavg, a = 7.6739 (b) Between 6.6 s and 7.2 s, what was the bee's average velocity? Be careful with signs. vavg, b = 4.58557 (c) Of the two average velocities you calculated, which is the best estimate of the bee's instantaneous velocity at time 6.6 s? (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s? Δr = m
Answer:
(A.) (- 4.33, 6.33 , 0); (B.) (- 3.66, 7.5, 0); (C.) average at (A) (- 4.33, 6.33 , 0) ; (D.) (- 0.2165, 0.3165, 0)
Explanation:
Given the following :
Time - - - - - - - 6.6s - - - - - - - - - 6.9s - - - - - 7.2s
Position - (1.8,5.0,0) - (0.5,6.9,0) - - (−0.4,9.5,0)
(a) Between 6.6 s and 6.9 s, what was the bee's average velocity?
Vavg = Distance / time
[(0.5,6.9,0) - (1.8,5.0,0)] / 6.9 - 6.6
Vavg = [(0.5 - 1.8), (6.9 - 5.0), (0 - 0)] / 0.3
Vavg = - 1.3 / 0.3, 1.9/0.3, 0/3
Vavg = (- 4.33, 6.33 , 0)
b) Between 6.6 s and 7.2 s, what was the bee's average velocity?
Vavg = [(−0.4,9.5,0) - (1.8,5.0,0)] / 7.2 - 6.6
Vavg = - 2. 2/0.6, 4.5/0.6, 0/0.6
Vavg = (- 3.66, 7.5, 0)
c.) Of the two averages (- 4.3, 6.3 , 0) is closer to the instantaneous Velocity at 6.6s
D.) (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s?
Displacement = Velocity * time
Vavg between 6.6 to 6.9 ; time = (6.65 - 6.6) = 0.05 s
= (- 4.33, 6.33 , 0) * 0.05
= (- 0.2165, 0.3165, 0)
Aisha used a compass to walk 59° south of west for 3.50 miles, then stopped for lunch. Assume that the point from which she started her hike corresponds to the origin of the coordinate system.
(a) Find the polar coordinates corresponding to the place where Aisha stopped for lunch. (Enter your answer for r in miles and counterclockwise from the +x-axis. Round your answers to at least th ree significant figures. as the smallest positive angle in degrees r =26
The radial polar coordinate does not coincide with x. mi =
(b) Find the Cartesian coordinates (in miles) for the same point.
x = 2.6
Use the relations between polar and Cartesian coordinates to find the coordinate x. mi
y = 4.24
Use the relations between polar and Cartesian coordinates to find the coordinate y. mi
Answer:
Explanation:
a )
59° south of west means (180 + 59 )° counterclockwise from the +x-axis.
so angle θ = 239 ° .
r = 3.5 miles .
polar coordinates ( r , θ )
= ( 3.50 , 239° )
As the smallest positive angle in degree is 26 then 26 x 9 = 234
so rounding off 239 to 234 , the angle will be 234
polar coordinates ( 3.50 , 234 )
b )
Cartesian coordinates (in miles) for the same point
x coordinate = r cosθ = 3.5 cos 234 = - 2.06 mi
y coordinate = r sinθ = 3.5 sin 234 = - 2.83 mi