Answer:
A. Using
Fl= ( v+vl/v+vz)fz
= (340+19/340+31) x 2020
= 1954.7Hz
Then to find the frequency of sound when reflected from the truck such that the driver becomes the listener
we use
F"= ( v+vz/v+vl) fz
= 340+31/340+19 x 2020
2087.5Hz
B to find the wavelength of sound we use
Wavelength= V+vl/ F"
= 340+31/2087.5= 0.18m
Find the tension between the 4 kg and the 6 kg blocks once the system is released from rest and the 9 kg block accelerates downward.
Answer:
Tension on the string is 48 N
Explanation:
Mass of A = 6 kg
mass of B = 4 kg
If the 9 kg accelerates downwards then the equation for the 6 kg mass side of the pulley will be
6g - T = 6a ....1
where
g is acceleration due to gravity = 10 m/s^2
a is the acceleration of the body
T is the tension in the string.
For the 4 kg mass side of the string, we have
T - 4g = 4a ....2
transposing, we have
T = 4a + 4g ....3
substitute equation 3 in equation 1, we have
6g - (4a + 4g) = 6a
6g - 4a - 4g = 6a
6g - 4g = 6a + 4a
2g = 10a
but g = 10 m/s^
2 x 10 = 10a
20 = 10a
acceleration a = 20/10 = 2 m/s^2
substitute value of a in equation 1
6g - T = 6a
6(10) - T = 6(2)
60 -T = 12
60 - 12 = T
T = 48 N
What challenges are faced by scientist investigating effects of radiation in Chernobyl and Eniwetok Atoll in the Chernobyl accident? 3 sentences long.
Answer:
The health hazard of directly going to these places to take direct readings and observations.
Explanation:
The Chernobyl accident was a catastrophic radioactive accidents, with an immediate fatal effect on some victims (mostly firemen and law enforcement officers). Scientists investigating the effects of the radiation left on the venues of the accident face a health risk if they intend to carry out a full scale on-field testing and experiment. Some of the relocated occupants of these places, that have long been relocated, have shown some health effect of the radiation. And scientist studying and working on these places for too long stand a risk of developing health complications if they are exposed to the radiations for too long.
define polarization in reference in the simple cell
Answer:polarization pertains to the act or process of producing a positive electrical charge and a negative electrical charge such that between a nerve cell internal electrical charge..................
Explanation:
Robert pushes the box to the left at constant speed. In doing so, Robert does ______ work on the box. A. positive B. negative C. zero
Answer:
He does postive work
Explanation:
Because W= F xS sinစ
So since work is done such that Force and distance have the Sam direction
Thus စ=0
Then work is positive in value
Is there any induced current in the pickup coil when the polarity of the bar magnet is flipped?
Answer:
Faraday's law , he direction of the magnetic field changes by 180º, in the polarity inversion processes, induces a voltage.
Explanation:
For this exercise let's use Faraday's law
E = - dФ / dt
Ф = B.A = B A cos θ
where B is the magnetic field, A is the area and θ is the angle between the field line and the normal to the area.
We can see that an electromotive force (voltage) is indexed when there is a variation of the field B, a variation of the area and change of the angle or when there is a combinational of them.
In this case, the magnitude of the field is constant, as the wire is rigid metal, the area is constant, but the direction of the magnetic field changes by 180º, in the polarity inversion processes, for which reason each change induces a voltage.
If a voltage is created in the ring, which has a resistance, a current is also generated in it.
Therefore the answer is If a current is created in the hoop
If the steam in the engine is able to raise a 30 kg piston to a height of 0.68 m, how much work has been done by the steam?
Answer:
199.92joules
Explanation:
workdone = Mgh
mass x height x acceleration due to gravity
How much power is used by a hair dryer if it does 40,000 J of work in 40 seconds?
Answer:
40000÷40=1000 joules is required to work in 40 seconds
Find the magnetic field strength at 1.50 m from the center of the circular region (e.g., outside the electric-field region).
Answer:
[tex]$ 1.81 \times 10^{-7} \ T$[/tex]
Explanation:
Given :
r = 1.50 m
R = 1 m
[tex]$\frac{dE}{dt}$[/tex] = [tex]$ 4.88 \times 10^{10} $[/tex] V / m s
Therefore the displacement current is
[tex]$ I_d = \epsilon_0. \frac{dE}{dt} . A $[/tex]
= [tex]$ \epsilon_0. \frac{dE}{dt} . \pi R^2 $[/tex]
Now according to law
[tex]$ B= \frac{\mu_0I_d.\frac{dE}{dt}.\pi R^2}{2 \pi r}$[/tex]
= [tex]$ \frac{\mu_0I_d.\frac{dE}{dt}. R^2}{2 r}$[/tex]
= [tex]$ \frac{4 \pi \times 10^{-7} \times 8.85 \times 10^{-12} \times 4.88 \times 10^{10} \times 1^2}{2 \times 1.5} $[/tex]
= [tex]$ 1.81 \times 10^{-7} \ T$[/tex]
Therefore, the magnetic field strength at 1.50 m from the center of the ring is [tex]$ 1.81 \times 10^{-7} \ T$[/tex].
the effect on any force depends on what two things
Answer: mass and distance
Explanation: Two major factors, mass and distance, affect the strength of gravitational force on an object. You witness the first factor in everyday life - more massive objects are heavier. The second factor, distance, is less familiar, as it takes a distance of many thousands of kilometers to significantly weaken the pull of Earth's gravity.
Answer:
The factor of a force depends on two things:
1. the amount of a force applied. The more the force applied the greater the effect.
2.the area on which the force is applied. The smaller the area the force is applied on, the greater the effect.
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Effect of thunder storm
Answer:
Some of the most severe weather occurs when a single thunderstorm affects one location for an extended time. Thunderstorms can bring heavy rains (which can cause flash flooding), strong winds, hail, lightning, and tornadoes. Severe thunderstorms can cause extensive damage to homes and property.
Explanation:
A toy police car moves across the room in 10 seconds. The distance the car traveled is
450 centimeters. At what speed did the car travel?
Answer:
45cm/s
Explanation:
If the distance traveled is 450 centimeters we can assume the speed is 45cm/s by taking the distance and dividing by the time, in this case that is 450 and 10
so our answer would be 45cm/s
what magnetic field strength will allow the electron to pass between the plates without being deflected
Complete Question
An electron travels with speed 2.0 X 10^7 m/s between two parallel charged plates. The plates are separated by 2.0 cm and are charged by a 400V battery. What magnetic field strength and direction will allow the electron to pass between the plates without being deflected.?
Answer:
The value [tex]B = 0.001 \ T[/tex]
Explanation:
From the question we are told that
The speed of the electron is [tex]v = 2.0 *10^{7} \ m/s[/tex]
The voltage is [tex]V = 400 \ V[/tex]
The distance of separation is [tex]d= 2.0 \ cm = 0.02 \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{V}{d }[/tex]
=> [tex]E = \frac{400}{0.02 }[/tex]
=> [tex]E = 20000 V/m[/tex]
Generally for the electron to pass without deflection then
[tex]F_B = F_E[/tex]
Where [tex]F_B[/tex] is the maximum magnetic force acting on the electron which is mathematically represented as
[tex]F_B = evBsin (\theta )[/tex]
Since we are considering maximum magnetic field then [tex]\theta = 90^o[/tex]
=> [tex]F_B = evB[/tex]
And
[tex]F_E[/tex] is the electric force on the electron which is mathematically represented as
[tex]F_E = e * E[/tex]
Here e is the charge on the electron with the value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]F_E = 1.60*10^{-19} * 20000[/tex]
=> [tex]F_E = 3.2*10^{-15} \ N[/tex]
So
[tex]3.2*10^{-15} = q* v B[/tex]
=> [tex]B = \frac{3.2*10^{-15}}{ 2.0*10^7 * 1.6*10^{-19}}[/tex]
=> [tex]B = 0.001 \ T[/tex]
A cannonball falls at a rate of about 10 m/s2 due to gravity. It has a mass of 200 kg. What is the total force of the
cannonball?
[tex]\star[/tex][tex]\blue\star[/tex]
[tex]\blue\star[/tex]Givenmass = 200kgacceleration = 10m\s^2Star coding
[tex]\blue\star[/tex]As we know that force = mass × accelerationF = m×aF = 200 ×10F = 2000N[tex]\blue\star[/tex]so mate here is ur ans force is equal to 2000N.Hope it helps
What is the toy's total energy at any point of its motion? Express your answer with the appropriate units.]
The complete question is as follows:
A 0.150-kg toy is undergoing SHM on the end of a horizontal spring with force constant k = 300 N/m. When the toy is 0.0120 m from its equilibrium position, it is observed to have a speed of 0.400 m/s.
Answer:
The correct answer is 0.034 J.
Explanation:
Given :
mass of the toy is m = 0.15 kg.
The force constant of restoring force k = 300 Nm⁻¹
When the position of the toy from the equilibrium is x = 0.012m, then the
speed of the toy vx = 0.4m s
The total mechanical energy in SHM is given by
E= 1/2 (mv²+ kx²) = 1/2 kA²
(here, m = mass of the object, vx = velocity, k = force constant
of restoring force, and A = amplitude of SHM.)
Hence by substituting the numerical values in equation 1, we get
[tex]E= \frac{1}{2} (0.15* 0.4) + \frac{1}{2} (300* 0.012)[/tex]
= 0.034 J
Thus, the correct answer is 0.034 J.
Acceleration of 1.5 ms expressed in km /hr2?
What operation do you apply to the position function of a particle to compute the particles velocity
Answer:
the derivative with respect to time
Explanation:
This is an exercise in kinematics, where the velocity is defined as a function of the position of a body of the form
v = dx/dt
where v is the velocity of the body, x is the position that we assume is a continuous and differentiable function.
The function written in the equation is the derivative with respect to time
A particle moves along a line so that its velocity at time t is ( | |. Find the displacement of the particle during the time period .
Complete question is;
A particle moves along a line so that its velocity at time t is
v(t) = t² − t − 6
(measured in meters per second).
(a) Find the displacement of the particle during the time period 1 ⩽ t ⩽ 4.
(b) Find the distance traveled during this time period.
Answer:
A) -4.5 m
B) 10.17 m
Explanation:
We are given;
v(t) = t² − t − 6
We know that v = ds/dt
Thus,
S = v dt
s(4) - s(1) = (4,1)∫t² − t − 6 dt
= (4,1)[(t³/3) - (t²/2) - 6t]
= ((4³/3) - (4²/2) - 6(4)) - ((1³/3) - (1²/2) - 6(1))
= 64/3 - 8 - 24 - 1/3 + 1/2 + 6
= -4.5 m
B) Since v(t) = t² − t − 6, then factorizing we can write it as;
v(t) = (t + 2)(t - 3)
Thus, v(t) ⩽ 0 at the interval (1, 3) and v(t) ≥ 0 at the interval (3, 4)
Thus;
-x1 = (1, 3)∫t² − t − 6 dt
-x1 = (1,3)(t³/3 - t²/2 - 6t)
-x1 = (3³/3 - 3²/2 - 6(3)) - (1³/3 - 1²/2 - 6(1))
-x1 = 9 - 9/2 - 18 - 1/3 + 1/2 + 6
-x1 = -3 - 4 - 1/3
-x1 = -22/3 m
x1 = 22/3 m
x2 = (3, 4)∫t² − t − 6 dt
x2 = (3, 4)(t³/3 - t²/2 - 6t)
x2 = ((4³/3) - (4²/2) - 6(4)) - (3³/3 - 3²/2 - 6(3))
x2 = 64/3 - 8 - 24 - 9 + 9/2 + 18
x2 = -23 + 64/3 + 9/2
x2 = (-138 + 128 + 27)/6
x2 = 17/6 m
Thus, total distance = x1 + x2 = 22/3 + 17/6 = 10.17m
A grating has 460 rulings/mm. What is the longest wavelength for which there is a 6.0th-order diffraction line
Answer:
λ = 3.62 x 10⁻⁷ m = 362 nm
Explanation:
The grating equation gives the relationship between the wavelength, the diffraction line order and the diffraction angle. The grating equation is written as follows:
mλ = d Sinθ
where,
m = order of diffraction = 6
λ = longest wavelength = ?
d = 1/(460 rulings/mm)(1000 mm /1 m) = 2.17 x 10⁻⁶ m/ruling
θ = Diffraction angle = 90° (for longest wavelength)
(6)λ = (2.17 x 10⁻⁶ m/ruling) Sin 90°
λ = (2.17 x 10⁻⁶ m/rulings)/6
λ = 3.62 x 10⁻⁷ m = 362 nm
It takes Serina 4.44 hours to drive to school. Her route is 15 km long. What is Serina’s average speed on her drive to school?
Answer: 3.38 km per hour
Explanation:
Which of the following does NOT cause acceleration?
A change in the magnitude of the velocity
A change in the direction of the velocity
Constant speed
Increasing speed
Decreasing spped
Answer:
Constant speed
Explanation:
Acceleration is a change of magnitude and (or) direction of velocity, it will cause increase or decrease, or changing of the direction of the velocity.
think of a teacher you have enjoyed/liked, and tell me what made you think they were a good teacher.
Answer:
a teacher that ive enjoyed is my 5th grade reading teacher. I enjoyed her because she always made class fun
Explanation:
A worker pushes horizontally on a large crate with a force of 265 n and the crate is moved 4.5 M how much work was done
Answer:
the work done is 1192.5Nm
Explanation:
work done = force × distance
w.d = 265N × 4.5m
w.d = 1192.5Nm
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The north pole of a bar magnet points towards a thin circular coil of wire containing 40 turns. The magnet is moved away from the coil, so that the ux through one turn inside the coil decreases by ∆Φ = 0.3 T.m2 in a time ∆t = 0.2 s. What is the average EMF induced in the (whole) coil during this time interval? Viewed from the side opposite of the bar magnet (from the right), does the induced current run clockwise or counterclockwise? Explain briey. [2 p.]
Answer:
60 V
The direction will be anticlockwise from the right side of the magnet.
Explanation:
The change in magnetic flux = ∆Φ = 0.3 T.m^2
The change in time = ∆t = 0.2 s
number of turns = 40 turns
The induced emf E = N∆Φ/∆t =
E = (40 x 0.3)/0.2 = 60 V
If the magnet is moved away from the coil, the induced current on the coil will try to oppose the motion of the magnet by attracting the magnet towards the coil. For the magnet to be attracted towards the coil, it must possess the equivalent of a magnetic south pole. For the equivalent of a magnetic south pole, the current on the coil will flow in the clockwise direction when viewed from the left side of the magnet. When viewed from the right side of the magnet, the direction will appear as anticlockwise.
The x-component of vector R is Rx = −28.2 units and its y-component is Ry = 19.6 units. What are its magnitude and direction? Give the direction as an angle measured counterclockwise from the +x-direction.
Answer:
Explanation:
Rx = -28.2 units
Ry = 19.6 units
magnitude of R = √ [( - 28.2 )² + ( 19.6 ) ]
= √ ( 795.24 + 384.16 )
= 34.34 units
If θ be the angle measured counterclockwise from the +x-direction
Tanθ = 19.6 / - 28.2 = -0.695
θ = 180 - 34.8
= 145.2° .
g summarize what you know about how changes in frequency and amplitude determine the characteristics
Answer:
Loudness and pitch are the characteristics of a sound related to amplitude and frequency.
Explanation:
Loudness and pitch are the characteristics of a sound that are related with the frequency and amplitude of a sound. If any change occurs in the frequency and amplitude of a sound, change also occurs in the loudness and pitch of a sound. If a source has larger amplitude so it produces large loudness or volume of a sound. On the other hand when the frequency is higher than the pitch will also be higher and the sound produce will be high.
You wish to watch TV at exactly 85 dB and no louder to avoid long term damage to your hearing. You record the sound intensity level to be 125 dB from a 1,200 W amplifier. The source is way too loud so you hit the remote and reduce the power to make the new sound intensity level be 107 dB.
1. What is the new power coming from the amplifier?
2. How far away are you from the amplifier now?
3. The sound is still too loud so you decide to cut amplifier power in half. How much farther/closer do you need to move to get down to 85 dB?
Answer:
1) the new power coming from the amplifier is 19.02 W
2) The distance away from the amplifier now is 5.50 m
3) u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
Explanation:
Lets say that I am at a distance "u" from the TV,
Let I₁ be the corresponding intensity of the sound at my location when sound level is 125dB
SO
S(indB) = 10log (I₁/1₀)
we substitute
125 = 10(I₁/10⁻¹²)
12.5 = log (I₁/10⁻¹²)
10^12.5 = I₁/10^-12
I₁ = 10^12.5 × 10^-12
I₁ = 10^0.5 W/m²
Now I₂ will be intensity of sound when corresponding sound level is 107 dB
107 = 10log(I₂/10⁻²)
10.7 = log(I₂/10⁻¹²)
10^10.7 = I₂ / 10^-12
I₂ = 10^10.7 × 10^-12
I₂ = 10^-1.3 W/m²
Now since we know that
I = P/4πu² ⇒ p = 4πu²I
THEN P₁ = 4πu²I₁ and P₂ =4πu²I₂
Therefore
P₁/P₂ = I₁/I₂
WE substitute
P₂ = P₁(I₂/I₁) = 1200 × ( 10^-1.3 / 10^0.5)
P₂ = 19.02 W
the new power coming from the amplifier is 19.02 W
2)
P₁ = 4πu²I₁
u =√(p₁/4πI₁)
u = √(1200/4π × 10^0.5)
u = 5.50 m
The distance away from the amplifier now is 5.50 m
3)
Let I₃ be the intensity corresponding to required sound level 85 dB
85 = 10log(I₃/10⁻¹²)
8.5 = log (I₃/10⁻¹²)
10^8.5 = I₃ / 10^-12
I₃ = 10^8.5 × 10^-12
I₃ = 10^-3.5 w/m²
Now, I ∝ 1/u²
so I₂/I₃ = u₁²/u²
u₁ = √(I₂/I₃) × u
u₁ = √(10^-1.3 / 10^-3.5) × 5.50
u₁ = 69.24 m
Therefore have to move u₁ - u ( 69.24 - 5.50) = 63.74 farther
A-0.06 C charge that moves downward is in a uniform electric field with a strength of 200 N/C. What is the
magnitude and direction of the force on the charge?
O 12 Nup
O 12 N down
O 3333 Nup
O 3333 N down
Answer:
its B
Explanation:
i just took the test
The correct option is B. The magnitude of the force on the charge is 12N and the direction is downward.
The electric field is a region of space where electric force can be felt. The formula for calculating the magnitude of a force in an electric field is expressed as:
[tex]\overline F=qE[/tex] where:
q is the charge measured in CoulombsE is the electric field intensity measured in N/CGiven the following parameters:
q = -0.06C
E = 200N/C
Substitute the given parameters into the formula as shown:
[tex]\overline F = -0.06 * 200\\\overline F = -12N\\[/tex]
Since the modulus of the force is in the negative sense, hence the direction will be downward.
Learn more here: https://brainly.com/question/22735564
Alice and Bob are bicycle-racing, on a two-mile track. Bob rides the first mile at 50 mph,
and the second mile at 71 mph. Calculate the time in minutes) that he travels the two
miles.
Answer:
2.04 minutes
Explanation:
He travels the first mile at 50 mph so:
[tex]60/50=1.2[/tex]
so it takes 1.2 minutes to travel each mile when going 50 miles per hour
and for the second mile he travels at 71 mph so:
[tex]60/71=0.84[/tex] (this answer is rounded)
so it takes him 0.84 minutes to travels each mile while going 71 mph.
If you ad them together his total times while biking two miles was 2.04 minutes.
A variable is defined as anything that
An electron moving parallel to the x axis has an initial speed of 4.10 106 m/s at the origin. Its speed is reduced to 1.76 105 m/s at the point x = 2.00 cm.
Required:
a. Calculate the electric potential difference between the origin and that point.
b. Which point is at the higher potential?
Answer:
a
[tex]V = -47.65 N/C[/tex]
b
At the origin
Explanation:
From the question we are told that
The initial speed is [tex]v_1 = 4.10 *10^{6} \ m/s[/tex]
The speed at (x = 2.00 cm) is [tex]v_f = 1.76 *10^{5} \ m/s[/tex]
Generally the electric potential difference is mathematically represented as
[tex]V = \frac{W}{q}[/tex]
Here W is the work-done which is mathematically represented as
[tex]W = K_f - K_i[/tex]
Here [tex]K_f[/tex] is the kinetic energy at x = 2.00 cm mathematically expressed as
[tex]K_f = \frac{1}{2} * m* v^2_f[/tex]
and
[tex]K_i[/tex] is the kinetic energy at origin mathematically expressed as
[tex]K_f = \frac{1}{2} * m* v^2_i[/tex]
So
[tex]V = \frac{1}{q} [ \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2][/tex]
[tex]V = \frac{m}{2q} [v_f^2 - v_i^2][/tex]
Here m is the mass of electron with value [tex]m = 9.1*10^{-31} \ kg[/tex]
q is the charge on the electron with value [tex]q = 1.60*10^{-19} \ C[/tex]
So
[tex]V = \frac{(9.1 *10^{-31})}{2(1.60*10^{-19})} [(1.76*10^{5})^2 - (4.10*10^{6})^2][/tex]
[tex]V = -47.65 N/C[/tex]
So given that the difference is negative then potential is higher at the origin