Answer:
The values is [tex]f_b =14.9 \ beats/s[/tex]
Explanation:
From the question we are told that
The speed of the fire engine is [tex]v = 5\ m/s[/tex]
The frequency of the tone is [tex]f = 500 \ Hz[/tex]
The speed of sound in air is [tex]v_s = 340 \ m/s[/tex]
The beat frequency is mathematically represented as
[tex]f_b = f_a - f[/tex]
Where [tex]f_a[/tex] is the frequency of sound heard by the people in the fire engine and is is mathematically evaluated as
[tex]f_a = [\frac{v_s + v }{v_s -v} ]* f[/tex]
substituting values
[tex]f_a = [\frac{340 + 5 }{340 -5} ]* 500[/tex]
[tex]f_a = 514.9 \ Hz[/tex]
Thus
[tex]f_b =514.9 - 500[/tex]
[tex]f_b =14.9 \ beats/s[/tex]
A charged particle has mass 1.20kg and charge negative 0.002C. It is moving. There is a uniform magnetic field of strength 0.25T pointing in the positive z direction present everywhere in the room. The particle experiences a magnetic force of 0.015N in the positive x direction. What is the velocity of the particle
Answer:
The velocity of the particle is 30 m/s.
Explanation:
The velocity can be found using Lorentz force:
[tex] F = q*v \times B [/tex]
Where:
q: is the charge of the particle = 0.002 C
v: is the velocity of the particle
B: is the magnetic field = 0.25 T
F: is the magnetic force = 0.015 N
Since the magnetic field and the particle's velocity are orthogonal, the cross product is equal to the product of the absolutes values of v and B.
We have:
[tex] F = q|v||B| [/tex]
[tex] v = \frac{F}{qB} = \frac{0.015 N}{0.002 C*0.25 T} = 30 m/s [/tex]
Therefore, the velocity of the particle is 30 m/s.
I hope it helps you!
If you wish to detect details of the size of atoms (about 1 ✕ 10−10 m) with electromagnetic radiation, it must have a wavelength of about this size.
(a) What is its frequency?
(b) What type of electromagnetic radiation might this be?
Answer:
a) 3×10^18 Hz
b) The electromagnetic wave is an x-ray
Explanation:
We know that the speed of an electromagnetic wave is given by;
c= λf
Where;
c= speed of electromagnetic waves = 3×10^8 ms-1
f= frequency of electromagnetic waves= the unknown
λ = wavelength of the electromagnetic wave= 1 ×10^-10 m
Hence;
f = c/λ= 3×10^8/ 1×10^-10
f= 3×10^18 Hz
b) The electromagnetic wave is an x-ray.
(a) The frequency of the electromagnetic radiation is [tex]3\times 10^{18} \ Hz[/tex]
(b) The electromagnetic radiation is ultra violet ray.
The given parameters;
distance of the atom, λ = 1 x 10⁻¹⁰ mThe frequency of the electromagnetic radiation is calculated as follows;
[tex]c =f \lambda \\\\f = \frac{c}{\lambda} \\\\f = \frac{3\times 10^8}{1\times 10^{-10}} \\\\f = 3\times 10^{18} \ Hz[/tex]
The frequency range of ultra violet ray is between 10¹⁴ to 10¹⁸ m.
Thus, the electromagnetic radiation is ultra violet ray.
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A 40-turn coil has a diameter of 17 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.40 T so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf induced in the coil (in V) if the magnetic field is reduced to zero uniformly in the following times.
Complete question:
A 40-turn coil has a diameter of 17 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.40 T so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf induced in the coil (in V) if the magnetic field is reduced to zero uniformly in the following times. (a) 0.10 s, (b) 1.0 s, and (c) 60 s.
Answer:
(a) E = 3.632 V
(b) E = 0.3632 V
(c) E = 6.053 mV
Explanation:
Given;
number of turn of the coil, N = 40 turn
diameter of the coil, d = 17 cm = 0.17 m
radius of the coil, r = d /2 = 0.17 /2 = 0.085 m
initial magnitude of magnetic field, B₁ = 0.4 T
final magnitude of magnetic field, B₂ = 0
The induced emf in the coil is given by;
[tex]E = -N\frac{dB}{dt} A = \frac{-NA(B_2 - B_1)}{dt} =\frac{NA(B_1-B_2)}{dt} =\frac{NAB_1}{t}[/tex]
A is area of the coil = πr² = π(0.085)² = 0.0227 m²
(a) when the time is 0.1 s
[tex]E= \frac{40*0.0227*0.4}{0.1} = 3.632 \ V[/tex]
(b) when the time is 1 s
[tex]E = \frac{40*0.0227*0.4}{1} \\\\E = 0.3632 \ V[/tex]
(c) when the time is 60 s
[tex]E = \frac{40*0.0227*0.4}{60} \\\\E = 6.053 \ mV[/tex]
Water flowing through a 1.6-cm-diameter pipe can fill a 500 L bathtub in 6.5 min.What is the speed of the water in the pipe?
Answer:
v = 6.38 m/s
Explanation:
discharge Q = 500 L / 6.5 min
flow through pipe is 1.6 cm x (1 m/100 cm ) = 0.016 m
using the flow rate formula Q = A * v
where A = area, v = velocity
the speed of water in the pipe = v = Q / A
500 L 1 m³ 1 min π (0.016 m)²
v = ---------- x -------- x ------------ ÷ --------------------
6.5 min 10³ L 60 sec 4
v = 6.38 m/s
The speed of the water is "12.95 m/s".
According to the question,
Volume,
V = 500 L= [tex]500\times 10^{-3} \ m^3[/tex]
Time,
t = 3.2 min= [tex]3.2\times 60[/tex]
= [tex]192 \ s[/tex]
Diameter,
d = 1.6 cm= 0.016 m
As we know,
Area of cross-section of pipe will be:
→ [tex]A = \pi (\frac{d}{2} )^2[/tex]
[tex]= 3.14\times (\frac{0.016}{2} )^2[/tex]
[tex]= 2.01\times 10^{-4} \ m^2[/tex]
Flow rate is:
→ [tex]\frac{V}{t} = Av[/tex]
[tex]v = \frac{V}{At}[/tex]
[tex]= \frac{500\times 10^{-3}}{2.01\times 10^{-4}\times 192}[/tex]
[tex]= 12.95 \ m/s[/tex]
Thus the above approach is right.
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How much work did the movers do (horizontally) pushing a 46.0-kgkg crate 10.5 mm across a rough floor without acceleration, if the effective coefficient of friction was 0.50
Answer:
7.1 J
Explanation:
From the question,
Work done by the mover = work done in pushing the crate + work done against friction
W = W'+Wf................. Equation 1
W = mgd+mgμd............ Equation 2
W = mgd(1+μ)................ Equation 3
Where m = mass of the crate, g = acceleration due to gravity, d = distance, μ = coefficient of friction.
Given: m = 46 kg, d = 10.5 mm = 0.0105 m, μ = 0.5
constant: g = 9.8 m/s²
Substitute these values into equation 3
W = 46×9.8×0.0105(1+0.5)
W = 7.1 J
The 60.0 g mass, attached to a light spring with a 40.0 N/m force constant, vibrates with an amplitude of 5.00 cm on a horizontal, frictionless plane. find (a) the total energy of the pulsating System, (b) the speed of mass when the displacement is 2.00 cm. When the displacement is 2.50 cm, (c) the kinetic energy and (d) the potential energy
Explanation:
(a) The total energy is the elastic potential energy at maximum displacement.
E = ½ kx²
E = ½ (40.0 N/m) (0.0500 m)²
E = 0.05 J
(b) At this displacement, there is both elastic potential energy and kinetic energy.
E = EE + KE
0.05 J = ½ kx² + ½ mv²
0.05 J = ½ (40.0 N/m) (0.0200 m)² + ½ (0.060 kg) v²
v = 1.18 m/s
(c) E = EE + KE
0.05 J = ½ kx² + ½ mv²
0.05 J = ½ (40.0 N/m) (0.0250 m)² + KE
KE = 0.0375 J
(d) EE = ½ kx²
EE = ½ (40.0 N/m) (0.0250 m)²
EE = 0.0125 J
The first step when doing
an
investigation is the observe a situation. True or false?
Answer:
True! First step is to make objective observations.
Two monatomic ideal gases are in thermal equilibrium with each other. Gas A is composed of molecules with mass m while gas B is composed of molecules with mass 4m. The ratio of the average molecular kinetic energy KA / KB is
Answer: [tex]\frac{K_{A}}{K_{B}}[/tex] = 1
Explanation: Average molecular kinetic energy for monatomic idela gases is given by:
[tex]E=\frac{3}{2}RT[/tex]
where
R is gas constant of ideal gas
T is temperature
Which means kinetic energy depnends only on temperature.
Since gas A and gas B are at the same temperature, kinetic energy will be the same. Therefore:
[tex]\frac{K_{A}}{K_{B}}[/tex] = 1
Ratio of this average molecular kinetic energy is 1.
HELP ASAP PLZ
What is the most likely reason for using the following model? A diagram that shows the different layers inside the sun Image © 2010 NASA The object is too small. The object represents a set of data. The object cannot be directly observed. The object represents a prediction about the future.
Answer:
Option C, The object cannot be directly observed.
Explanation:The temperature of sun is extremely high due to which it is almost impossible to land on its surface and explore the depth of it. Thus a prototype of it is required to predict its probable internal structure and associated feature which effect our planet "earth".
This prototype/model is based on the deduction arrived after analyzing the satellite information available in the form of high resolution images.
Hence, option C is correct
Answer:
The answer is c the object cannot be directly observed.
Explanation:
I took the test and got it right.
If a ball is accelerating down through the air with no horizontal motion, what must be true about the net forces acting on the ball?(1 point) The net force on the ball is directed upward. The net force on the ball is zero. The gravitational force is greater than the drag. The drag is greater than the gravitational force.
Answer:
The net force on the ball is zero.
Explanation:
Newton's first law of motion states that an object will remain in the same state of motion or at rest unless a resultant force acts on it, and if the net force on an object is zero, then that means that the object will keep moving at the same acceleration or the object stays stationary.
So, the truth about the net forces acting on the ball is that the net force on the ball is zero.
Answer:
The gravitational force is greater than the drag.
Explanation:
I took the quick check and got it wrong when I chose The net force on the ball is zero but I was right the first time with the third answer choice listed in the question.
Drag and drop each description into the correct category
They copy the original sound
Analog signals
Digital signals
They change sound into numbers.
They reduce unwanted noise.
They may pick up unwanted noise.
They flow continuously.
They do not flow continuously
Intro
Done
Explanation:
Analog signals :
Analog signals are continuous signal. These are like wave signals that change with time period.The main disadvantage of analog signal is that it contains noise. .They flow continuously.Digital signals :
Digital signals do not flow continuously. It has finite number of values. They form in the form of numbers. They reduce unwanted noise. It is the advantage of digital signals.Answer:
Analog -
They flow continuously
They may pick up unwanted noise
They copy the original sound
Digital -
They do not flow continuously
They reduce unwanted noise.
They change sound into numbers.
Mark Brainliest Please a) ¿En qué posición es mínima la magnitud de la fuerza sobre la masa de un sistema masa-resorte? 1) x 0, 2) x A o 3) x A. ¿Por qué? b) Con m 0.500 kg, k 150 N/m y A 0.150 m, calcule la magnitud de la fuerza sobre la masa y la aceleración de la masa en x 0, 0.050 m y 0.150 m.
Answer:
a) the correct answer is 1 , b) x=0 F=0, a=0
x= 0.050 F= -7.5 N, a= -15 m/s²
x= 0.150 F= 22.5 N, a=- 45 m/s²
Explanation:
a) In a mass - spring system the force is given by the Hooke force,
F = - k x
Analyzing this equation we see that the outside is proportional to the elongation from the equilibrium position, therefore the force is zero when the spring is in its equilibrium position
the correct answer is 1
b) we assume that the given values are from the equilibrium position of the spring.
Let's calculate the force
x = 0
F = 0
x = 0.050
F = - 150 0.050
F = - 7.50 N
x = 0.150
F = - 150 0.150
F = - 22.5 N
let's use Newton's second law to find the acceleration
F = m a
a = F / m
x = 0 m
a = 0
x = 0.050 m
a = -7.50 / 0.50
a = - 15 m / s²
x = 0.150 m
a = - 22.5 / 0.50
a = - 45 m/s²
TRASLATE
a) En un sistema masa – resorte la fuerza es dada por la fuerza de Hoke,
F= - k x
analizando esta ecuación vemos que la fuera es proporcional a la elongación desde la posición de equilibrio, por lo tanto la fuerza es cero cuando el resorte esta en su posición de equilibrio
la respuesta correcta es 1
b)suponemos que los valores dados son desde la posición de equilibrio del resorte.
Calculemos la fuerza
x=0
F= 0
x=0.050
F = - 150 0.050
F= - 7.50 N
x= 0.150
F= - 150 0.150
F= - 22.5 N
usemos la segunda ley de Newton para encontrar la aceleración
F = m a
a = F/m
x =0 m
a = 0
x= 0.050 m
a = -7.50/ 0.50
a =- 15 m/s²
x= 0.150 m
a= - 22.5 / 0.50
a= - 45 m/s²
From what maximum height could a 75 kg person jump and land rigidly upright on both feet without breaking their legs
Answer:
h is 0.27metres without breaking their leg
Explanation:
A body jumping from any heights has potential energy mgh and maximum energy which can be converted by human body from the jump is 200J derived from experiment so using
200J = mgh
h = 200/ 9.8* 75
h= 0.27 m
Explanation:
The magnetic flux through a coil of wire containing two loops changes at a constant rate from -83 Wb to 82 Wb in 0.39 s .
What is the magnitude of the emf induced in the coil?
Answer:
423v
Explanation:
Using
E= -N ∆န/ ∆ t
= ( -1) X (-83 - 82)/0.39
= 423volts
A vertical tube one meter long is open at the top. It is filled with 75 cm of water. If the velocity of sound is 344 m/s, what will the fundamental resonant frequency be (in Hz)?
Answer:
344Hz
Explanation:
First we need to more that the frequency will inky resonate in the portion not containing water. If the total length is 1m and the filled with water up to 75cm, the length of the air column will be 100cm - 75cm = 25cm
Fundamental frequency of a closed pipe fo = V/4L
V is the velocity of sound in air
L is the length of the air column
Given V = 344m/s
L = 25cm = 0.25m
fundamental resonant frequency = 344/4(0.25)
= 344/1
= 344Hz
Which of the following is not an example of matter?
A. sand
B. heat from a fire,
C. helium in a balloon
D. fog paint on a canvas
Answer:
B. heat from fire
Explanation:
It is not an example of matter because heat is energy and not matter
Two helium-filled balloons are released simultaneously at points A and B on the x axis in an earth-based reference frame. Balloon A is to the left of balloon B. Which one of the following statements is true for an observer moving in the +x direction?
A. The observer sees balloon B released before balloon A.
B. The observer always sees the balloons released simultaneously.
C. The observer could see either balloon released first depending on her speed and the distance between A and B.
D. The observer sees balloon A released before balloon B.
Answer:
The correct answer is C
Explanation:
In this exercise you are asked to analyze the following situation: for a fixed observer, the two balloons are launched at the same time.
If the observer is mobile moving towards the positive side of axis ax we have several
possibilities when starting the movement
* the observer is to the left of the two balloons
* the observer is between the two balloons
* the observer is to the right of the two baloomls
The time it takes for the signal to arrive to know which ball goes first is
v = d / t
t = d / v
If the signal goes at the speed of light, the speed is a constant and the time will depend only on the distance, so we see that the trigger changes depending on the relative position between a given ball and the observer.
Consequently, it will be seen which comes out first, depending on the relative position with the observer.
The correct answer is C
A block sliding along a horizontal frictionless surface with speed v collides with a spring and compresses it by 1.0 cm . What will be the compression if the same block collides with the spring at a speed of 4v
Answer:
The compression is [tex]x = 0.05 \ m[/tex]
Explanation:
From the question we are told that
The first velocity is v
The compression of the spring is [tex]d = 1.0 \ cm = 0.01 \ m[/tex]
The second velocity is 4v
Generally according to the law of energy conservation
The kinetic energy of the block is equal to the energy stored in the spring that is
[tex]\frac{1}{2} * m* v^ 2 = \frac{1}{2} * k * d^2[/tex]
For first speed
[tex]m* v^ 2 = k * 0.01^2[/tex]
=> [tex]m* v^ 2 = k * 0.0001[/tex]
=> [tex]k = \frac{ m v^2 }{ 0.0001}[/tex]
For second speed
[tex]\frac{1}{2} * m* (5v)^ 2 = \frac{1}{2} * k * x^2[/tex]
=> [tex]12.5mv^2 = 0.5 k x^2[/tex]
substituting for k
=> [tex]12.5mv^2 = 0.5 (\frac{mv^2}{0.0001} ) x^2[/tex]
=> [tex]12.5 = 5000x^2[/tex]
=> [tex]x = 0.05 \ m[/tex]
g Let the orbital radius of a planet be R and let the orbital period of the planet be T. What quantity is constant for all planets orbiting the sun, assuming circular orbits?
Explanation:
Kepler's third law gives the relationship between the orbital radius and the orbital period of the planet. Its mathematical form is given by :
[tex]T^2=\dfrac{4\pi ^2}{GM}a^3[/tex]
Here,
G is gravitational constant
M is mass of sun
It means that the mass of Sun is constant for all planets orbiting the sun, assuming circular orbits.
(b) A piece of wood of volume 0.6 m² floats in water. Find the volume
exposed. What force is required to immerse it completely under water?
(Density of wood = 600 kg/m3, water = 1000 kg/m3)
[8]
Answer:
Explanation:
Let the volume below water be v . Then
buoyant force = v d g where d is density of water , g is acceleration due to gravity
= v x 1000 x g
weight of wood piece = volume x density of wood x g
= .6 x 600 x g
for equilibrium while floating
buoyant force = weight
= v x 1000 x g = .6 x 600 x g
v = .36 m²
volume above water or volume exposed = .6 - .36
= .24 m²
When immersed completely ,
buoyant force = .6 x 1000 x 9.8
= 5880 N
weight of wood
= .6 x 600 x g
= 3528 N
buoyant force is more than the weight . In order to equalise them for floating with full volume in water
weight required = 5880 - 3528
= 2352 N.
When you squeeze together the coils of a spring and then release them, you are creating a ____ wave? transverse compressional water seismic
Answer:
Hey there!
Because you are compressing the spring, these are compressional waves.
Let me know if this helps :)
A flatbed truck is carrying a 20-kg crate up a sloping road. The coefficient of static friction between the crate and the bed is 0.40, and the coefficient of kinetic friction is 0.30. What is the maximum angle of slope that the truck can climb at constant speed if the crate is to stay in place
Answer:
The angle is [tex]\theta =21.8 ^o[/tex]
Explanation:
From the question we are told that
The mass of the crate is [tex]m_c = 20 \ kg[/tex]
The coefficient of static friction is [tex]\mu_s = 0.40[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.30[/tex]
Generally for the the crate not to slip , the static frictional must be equal to the force driving the truck
i.e
[tex]F_f = F[/tex]
Now since we are considering a slope that static frictional force is mathematically represented as
[tex]F_f = mg * cos(\theta) * \mu_s[/tex]
While the force driving the truck is mathematically represented as
[tex]F = mg * sin (\theta )[/tex]
Here mg is the weight of the crate so
So
[tex]mg * cos (\theta ) \mu_s = mg * sin (\theta )[/tex]
=> [tex]\frac{sin (\theta )}{cos (\theta)} = \mu_s[/tex]
=> [tex]\theta = tan ^{-1} [\mu_s ][/tex]
=> [tex]\theta = tan ^{-1} [0.40 ][/tex]
=> [tex]\theta =21.8 ^o[/tex]
A diagram show an illustration is on the first uploaded image
Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron would the proton have to be? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2, e = 1.6 × 10-19 C, mproton = 1.67 × 10-27 kg, melectron = 9.11 × 10-31 kg)
Answer:
r = 5.08 m
Explanation:
The electric force of attraction or repulsion is given by :
[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]
We need to find how far above the electron would the proton have to be if you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it.
So, the force from the proton is balanced by the mass of the electron.
[tex]\dfrac{kq_pq_e}{r^2}=mg[/tex]
r is distance
[tex]r=\sqrt{\dfrac{kq_pq_e}{mg}} \\\\r=\sqrt{\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{9.11\times 10^{-31}\times 9.8}} \\\\r=5.08\ m[/tex]
So, proton have to be at a distance of 5.08 meters above the electron.
If an ancient campfire were analyzed, was found to have only about one-eighth the carbon-14 that is normally found in living things, how long ago was that campfire extinguished? Answer in units of y.
Answer:
17047.54 years or 17048 y
Explanation:
From;
0.693/t1/2 = 2.303/t log (Ao/A)
Where;
t1/2=half-life of the carbon-14 =5670 years
t= time elapsed
Ao= activity of living C-14
A= activity if the sample under study
But A= 1/8 Ao
Hence;
0.693/5670= 2.303/t log(Ao/1/8Ao)
1.22×10^-4 = 2.303/t log 8
1.22×10^-4 = 2.0798/t
t= 2.0798/1.22×10^-4
t= 17047.54 years
A boat sailing against the current experience a negative acceleration of 11 m/s^2. If the boat's initial velocity is 44m/s upstream, how long until it comes to a stop?
Answer:
4 s
Explanation:
Given:
v₀ = 44 m/s
v = 0 m/s
a = -11 m/s²
Find: t
v = at + v₀
0 m/s = (-11 m/s²) t + 44 m/s
t = 4 s
If the boat's initial velocity is 44m/s upstream, the time it takes to stop is found to be 4 seconds.
What is Acceleration?Acceleration may be defined as the rate of change of velocity with respect to time. It is a vector quantity as it has both magnitude and direction. It is also the second derivative of position and the first derivative of velocity with respect to time.
According to the question,
v₀ = 44 m/s
v = 0 m/s
a = -11 m/s²
Now, you have to calculate the time, t.
So, you calculate the time, t with the help of the following formula:
v = at + v₀0 m/s = (-11 m/s²) t + 44 m/s
t = 4 s.
Therefore, if the boat's initial velocity is 44m/s upstream, the time it takes to stop is found to be 4 seconds.
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Find "x" at: x = Cos4377π + 8Sen750 ° Select one:
a. 1/3
b. 2
C. 1/2
d. 3
Answer:
d. 3
Explanation:
x = cos(4377π) + 8 sin(750°)
A full circle is 2π radians or 360°.
x = cos(4377π − 4376π) + 8 sin(750° − 720°)
x = cos(π) + 8 sin(30°)
x = -1 + 8 (1/2)
x = 3
A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T. If the bob's mass is doubled, approximately what will the pendulum's new period be?
A: T/2
B: T
C: sqrt(2)*T
D: 2T
Answer:
B. T
Explanation:
This is because the mass of the Bob does no relate to the period as given by the relationship
T = 2π(√L/g) so double mass is still T
Answer:
The correct option is (B).
Explanation:
The time period of a pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is the length of the pendulum
g is acceleration due to gravity
It is very clear that the time period is independent of the mass of the bob. If the mass of the bob is doubled, the time period of the pendulum remains the same i.e. T. Hence, the correct option is (B).
Can someone please solve these 6 questions and explain briefly? I really do not understand them. Thank you
Explanation:
Use SOH-CAH-TOA.
Sine = Opposite / Hypotenuse
Cosine = Adjacent / Hypotenuse
Tangent = Opposite / Adjacent
For example, in 2(a), we are given an angle and the hypotenuse. We want to find the side adjacent to the angle. Therefore, we should use cosine.
cos 58° = y / 32.3
y = 32.3 cos 58°
y ≈ 17.1
In 3(a), we are given the adjacent side and opposite side. We want to find the angle. So use tangent.
tan θ = 3.6 / 6.2
θ = tan⁻¹(3.6/6.2)
θ = 30.1°
In 4(a), we are given an angle and the hypotenuse. We want to find the side opposite of the angle. Therefore, we should use sine.
sin 47° = x / 29
x = 29 sin 47°
x ≈ 21.2
An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next second of time? Explain.
a. 2h
b. 4h
c. h
d. h^2
e. 3h
Answer:
E. 3h
Explanation:
We know that
u = 0 m/s.
velocity after t = 1s
v = u+gt = 0+9.81 x 1s= 9.81 m/s
distance covered in 1st sec
= =>> ut+0.5 x g x t²
=>>0 + 0.5x 9.81 x 1 = 4.90m
Let 4.90 be h
distance travelled in 2nd second will now be used
So velocity after t = 1s
=>>1 x t+ 0.5 x g x t²
=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90
So since h= 4.90
Then the ans is 3x h = 3h
When reading a digital volt-ohmmeter (DVOM), you have a reading of 2168 mV, which is the same as:__________
A. 2168 millivolts.
B. 2.168 volts.
C. 1000 mV.
D. Both A and B
Answer:
D, both A and B
Explanation:
2168 mV is the SI unit for potential difference and the Voltmeter.
The primary unit is Volt, represented as V. Due to the fact that there can be a much higher reading, or an even much more smaller one, comes the need for variants of the same unit.
10^-3 is called milli and represented as m
10^3 is called killo and represented as k
10^-6 is called micro and represented as µ
10^6 is called mega and represented as M
and even much higher variants of up to 10^12 and 10^-12
As we can see from the aforementioned example, 10^-3 is milli and represented as m
And our question gave us the unit in mV, which stands for millivolts.
Also, if we look at option B, it states, 2.168 volts. This 2.168 volts is also the same thing as A. Take a look at it this way, I said mV is 10^-3, right?
So, 2168*10^-3 is also 2168/100 which is 2.168. The only difference here is, once we make this conversion from mV, we have to drop the milli tag, because we have already made a conversion, and thus, leave it as V.
2168 mV = 2.168V
Hence why we picked option D, Both A & B as the right one