Answer:
187.5
Explanation:
On top of a cliff of height h, a spring is compressed 5m and launches a projectile perfectly horizontally with a speed of 75 m s . It hits the ground with speed 90 m s . How high above the ground was the cliff? (Hint: use energy conservation to make the problem easier!)
Answer:
The height of the cliff is 121.276 m
Explanation:
Given;
initial velocity of the projectile, v₁ = 75 m/s
final velocity of the projectile, v₂ = 90 m/s
spring compression = 5 m
Apply the law of conservation of energy;
mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²
gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²
gh₁ - gh₂ = ¹/₂v₂² - ¹/₂v₁²
g(h₀ - h₂) = ¹/₂ (v₂² - v₁²)
h₀ - h₂ = ¹/₂g (v₂² - v₁²)
h₀ = h(cliff) + 5m
when the projectile hits the ground, Final height, h₂ = 0
[tex]h_o - 0 = \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} + 5= \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} = \frac{1}{2g}(v_2^2-v_1^2) - 5\\\\h_{cliff} = \frac{1}{2*9.8}(90^2-75^2) - 5\\\\h_{cliff} = 121.276 \ m[/tex]
Therefore, the height of the cliff is 121.276 m
What are the standard international (si) units of distance
Answer:
meter
Explanation:
Answer: The International System of Units is a system of measurement based on 7 base units
Explanation: the metre, kilogram, second, ampere, Kelvin, mole, and candela. These base units can be used in combination with each other.
what chemical diverged from trees
a resin and turpentine
b sodium
c lead
d marcotting
Answer:
a. resin and turpentine
Explanation:
The chemicals that diverged from trees are resin and turpentine.
Resin are produced by special cells in trees, most times we see them when a tree is damaged or cut. They are usually derived from pines and firs.
Turpentine is obtained by distilling resin.
Turpentine has an antiseptic property that has different uses. They are used as cleansing agents and for producing sanitary materials.
A ball of mass m is found to have a weight Wx on Planet X. Which of the following is a correct expression for the gravitational field strength of Planet X?
A. The gravitational field strength of Planet X is mg.
B. The gravitational field strength of Planet X is Wx/m.
C. The gravitational field strength of Planet X is 9.8 N/kg.
D. The gravitational field strength of Planet X is mWx.
Answer: B. The gravitational field strength of Planet X is Wx/m.
Explanation:
Weight is a force, and as we know by the second Newton's law:
F = m*a
Force equals mass times acceleration.
Then if the weight is:
Wx, and the mass is m, we have the equation:
Wx = m*a
Where in this case, a is the gravitational field strength.
Then, isolating a in that equation we get:
Wx/m = a
Then the correct option is:
B. The gravitational field strength of Planet X is Wx/m.
A 126 N force is applied at an angle of 25.00 to a 8.50 kg block pressed against a rough vertical wall and the block slides down the wall at constant velocity. Calculate the coefficient of kinetic friction between the block and the wall.
Answer:
μk = 0.58
Explanation:
If the block is sliding down at constant speed, this means that no net force is acting upon it in the vertical direction.As the block is pressed on the wall, this means that it doesnt accelerate in the horizontal direction either, so no net force acts upon it in this direction also.In this direction, we have only two forces acting, equal and opposite each other, one is the normal force (exerted by the wall) and the other is the horizontal component of the applied force.If the applied force forms an angle of 25º with the wall (which is vertical), this means that we can get its projection along the horizontal direction, using simple trigonometry , as follows:[tex]F_{apph} = F_{app} * sin\theta = 126 N * sin 25 = 53.3 N[/tex]
⇒ [tex]F_{n} = - F_{apph} = -53.3 N[/tex]
In the vertical direction, we have three forces acting on the block: the weight pointing downward, the kinetic friction force (as we know that the block is sliding), and the vertical component of the applied force, in the same direction as the friction one.As we have already said, the sum of these forces must be 0.[tex]F_{g} + F_{appv} + F_{ff} = 0 (1)[/tex] where Fg is the weight of the block, Fappv is the vertical component of the applied force, and Fff is the kinetic friction force.Replacing these forces by their mathematical expressions, we have:[tex]F_{g} = m_{b} * g = 8.5 Kg * (-9.8 m/s2) = -83.3 N[/tex]
[tex]F_{appv} = F_{app}* cos\theta = 126 N * cos 25 = 114.2 N[/tex]
[tex]F_{ff} = \mu_{k}* F_{n} =\mu_{k} * (-53.3 N)[/tex]
Replacing in (1), and solving for μk, we finally get:μk = 0.58
Which statement best describes an atom? (2 points)
оа
Protons and neutrons grouped in a specific pattern
Ob
Protons and electrons spread around randomly
ос
A group of protons and neutrons that are surrounded by electrons
Od
A ball of electrons and neutrons surrounded by protons
Answer:
A group of protons and neutrons that are surrounded by electrons I think that's the answer...
Explanation:
A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to:
29 m/s2
31 m/s2
33 m/s2
30 m/s2
32 m/s2
Explanation:
The question is incomplete. Here is the complete question.
A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of 30m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to...
Given
initial velocity of rocket u = 0m/s
final velocity of rocket = 30m/s
Height reached by the rocket = 98m
Required
upward acceleration of the rocket
Using the equation of motion below to get the acceleration a:
[tex]v^2 = u^2+2as\\30^2 = 0^2 + 2(a)(98)\\900 = 196a\\a = \frac{900}{196}\\a = 4.59m/s^2[/tex]
Hence upward acceleration of the rocket during the burn phase is closest to 5m/s²
Note that the velocity used in calculation was assumed.
Calculate the force a 75 kg high jumper must exert in order to produce an acceleration that is 3.2 times the acceleration due to gravity.
Answer:
Explanation
According to Newton's second law of motion,
F = ma
m is the mass
a is the acceleration
If the acceleration is 3.2 times the acceleration due to gravity, then a = 3.2g
The formula becomes;
F = m(3.2g)
F = 3.2mg
m= 75kg
g = 9.81m/s²
F = 3.2(75)(9.81)
F = 2,354.4N
Hence the force exerted by the jumper is 2,354.4N
The scientific method is the only way of learning about Nature used by scientist today *
A. true
B. false
Answer:
false
Explanation:
A man with a mass of 86.5 kg stands up in a 61-kg canoe of length 4.00 m floating on water. He walks from a point 0.75 m from the back of the canoe to a point 0.75 m from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move?
Answer:
The displacement of the canoe is 1.46 m
Explanation:
Given that,
Mass of canoe = 61 kg
Mass of man = 86.5 kg
Length = 4 m
Let the the displacement of the canoe is x'
We need to calculate the displacement of the man
Using formula of displacement
[tex]x=x_{2}-x_{1}[/tex]
Put the value into the formula
[tex]x=4-(0.75+0.75)[/tex]
[tex]x=2.5\ m[/tex]
We need to calculate the displacement of the canoe
Using conservation of momentum
[tex]M_{m}v_{m}=(M_{c}+M_{m})v_{c}[/tex]
[tex]M_{m}\dfrac{x}{t}=(M_{c}+M_{m})\dfrac{x'}{t}[/tex]
[tex]86.5\times2.5=(61+86.5)\times x'[/tex]
[tex]x'=\dfrac{86.5\times2.5}{61+86.5}[/tex]
[tex]x'=1.46\ m[/tex]
Hence, The displacement of the canoe is 1.46 m
A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes the ground 1330 m from the base of the cliff. What is the speed of the shell as it hits the ground
Answer:
V = 331.59m/s
Explanation:
First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.
S = ut + 1/2at²
Given height of the cliff S = 80m
initial velocity u = 0m/s²
a = g = 9.81m/s²
Substitute
80 = 0+1/2(9.81)t²
80 = 4.905t²
t² = 80/4.905
t² = 16.31
t = √16.31
t = 4.04s
Next is to get the vertical velocity
Vy = u + gt
Vy = 0+(9.81)(4.04)
Vy = 39.6324
Also calculate the horizontal velocity
Vx = 1330/4.04
Vx = 329.21m/s
Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.
V² = Vx²+Vy²
V² = 329.21²+39.63²
V² = 329.21²+39.63²
V² = 108,379.2241+1,570.5369
V² = 109,949.761
V = √ 109,949.761
V = 331.59m/s
Hence the speed of the shell as it hits the ground is 331.59m/s
Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 light-years away. Their message encourages us to send a radio answer, which we decide to do. Suppose our governing bodies take 2 years to decide whether and how to answer. When our answer arrives there, their governing bodies also take two of our years to frame an answer to us. How long after we get their first message can we hope to get their reply to ours? Enter your answer in years.
Answer:
The duration is [tex]T =72 \ years /tex]
Explanation:
From the question we are told that
The distance is [tex]D = 35 \ light-years = 35 * 9.46 *10^{15} = 3.311 *10^{17} \ m [/tex]
Generally the time it would take for the message to get the the other civilization is mathematically represented as
[tex]t = \frac{D}{c}[/tex]
Here c is the speed of light with the value [tex]c = 3.08 *10^{8} \ m/s[/tex]
=> [tex]t = \frac{3.311 *10^{17} }{3.08 *10^{8}}[/tex]
=> [tex]t = 1.075 *10^9 \ s[/tex]
converting to years
[tex]t = 1.075 *10^9 * 3.17 *10^{-8} [/tex]
[tex]t = 1.075 *10^9 * 3.17 *10^{-8} [/tex]
[tex]t = 34 \ years [/tex]
Now the total time taken is mathematically represented as
[tex]T = 2* t + 2 + 2[/tex]
=> [tex]T = 2* 34 + 2 + 2[/tex]
=> [tex]T =72 \ years /tex]
Astronomers have proposed the existence of a ninth planet in the distant solar system. Its semi-major axis is suggested to be approximately 600 AU. If this prediction is correct, what is its orbital period in years
Answer:
T = 1.4696 10⁴ years
Explanation:
For this exercise we must use Kepler's laws, specifically the third law which is the application of the universal law of gravitation to Newton's second law
F = ma
G m M / r² = m a_c = m v² / r
G M / r = v²
the speed of the circular orbit is
v = 2π r / T
we substitute
G M / r = 4π² r² / T²
T² = (4π² / G M) r³
Kepler proved that this expression is the same if the radius is changed by the semi-major axis of an ellipse
T² = (4π² /GM) a³
the constant is worth
(4π² / GM) = 2.97 10⁻¹⁹ s² / m³
let's reduce the distance to SI units
AU is the distance from the Earth to the Sun
a = 600 AU = 600 AU (1.496 10¹¹ m / 1 AU)
a = 8.976 10¹³ m
T² = 2.97 10⁻¹⁹ (8.976 10¹³)³
T² = 21.4786 10²²
T = 4.63 10¹¹ s
Let's reduce to years
T = 4.63 10¹¹s (1 h / 3600s) (1 day / 24 h) (1 year / 365 days)
T = 1.4696 10⁴ years
Part D
Next, we'll examine magnetic force. Bring the ends of your two magnets together. Explore the three
possible combinations. In two of the combinations, the two ends are the same. In one combination, the
two ends are different. Describe the force you feel in each combination
Answer:
i. The magnetic force of repulsion.
ii. The magnetic force of attraction.
Explanation:
A magnet is a material that has the attraction and repulsion capability. Magnets has two poles, north and south, thus would attract or repel another magnet in its neighborhood. It can either be a permanent or temporal magnet, and attracts ferrous metals.
i. In the case of two combinations where two ends are the same, it could be observed that the two ends (poles) repels each other. Thus since like poles repels, magnetic force of repulsion is felt.
ii. In the case of one combination in which the two ends are different, the two ends (poles) attract. Since unlike poles attracts, magnetic force of attraction is observed.
If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s than the recoil speed of the cannon will be *
A. 0.2 m/s
B. 2 m/s
C. 4 m/s
D. 10 m/s
Answer:
D. 10 m/s
Explanation:
In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 4.7 kg book is pushed from rest through a distance of 0.85 m by the horizontal 42 N force from the broom and then has a speed of 1.0 m/s, what is the coefficient of kinetic friction between the book and floor
Answer:
μ_k = 0.851
Explanation:
We are given;
Mass of book; m_book = 4.7 kg
Horizontal force; F_horiz = 42 N
Distance; d = 0.85 m
Speed; v = 1 m/s
First of all let's find the acceleration using Newton's equation of motion;
v² = u² + 2ad
u is initial velocity and it's 0 m/s in this case.
Thus;
1² = (2 × 0.85)a
1 = 1.7a
a = 1/1.7
a = 0.5882 m/s²
Now, resolving forces along the vertical direction, we have;
W - N = 0
Thus,W = N
Where W is weight = mg and N is normal force
Thus; N = mg = 4.7 × 9.81 = 46.107 N
Now, resolving forces along the horizontal direction, we have;
F_horiz - ((μ_k)N) = ma
Where μ_k is coefficient of kinetic friction.
Thus;
42 - 46.107(μ_k) = 4.7 × 0.5882
42 - 46.107(μ_k) = 2.76454
μ_k = (42 - 2.76454)/46.107
μ_k = 0.851
cameron drives his car 15 km north. He stops for lunch and then drives 12 km south. What is his displacement?
Answer:
Displacement is 3 km North
Explanation:
The current is suddenly turned off. How long does it take for the potential difference between points a and b to reach one-half of its initial value
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told that
The original voltage is [tex]V_o[/tex]
The new voltage is [tex]V =\frac{V_o}{2}[/tex]
The capacitance is [tex]C = 150\ nF = 150 *10^{-9} \ F[/tex]
The first resistance is [tex]R_i = 26 \Omega[/tex]
The second resistance is [tex]R_E = 200 \Omega[/tex]
Generally the equivalent resistance is
[tex]R_e = R_1 + R_E[/tex]
=> [tex]R_e = 26 +200 [/tex]
=> [tex]R_e = 226 \ \Omega [/tex]
Generally the time constant is mathematically represented as
[tex]\tau = RC[/tex]
=> [tex]\tau = 226 * 150 *10^{-9}[/tex]
=> [tex]\tau = 3.39 *10^{-5} \ s [/tex]
Generally the voltage is mathematically represented as
[tex]V = V_o e^{-\frac{t}{\tau} }[/tex]
=> [tex]\frac{V_o}{2} = V_o e^{-\frac{t}{\tau} }[/tex]
=> [tex]0.5 = e^{-\frac{t}{\tau} }[/tex]
=> [tex]ln(0.5) = {-\frac{t}{ 3.39 *10^{-5} } }[/tex]
=> [tex]ln(0.5) * 3.39 *10^{-5} = -t [/tex]
=> [tex]t = 2.35*10^{-5} \ s [/tex]
See Conceptual Example 6 to review the concepts involved in this problem. A 12.0-kg monkey is hanging by one arm from a branch and swinging on a vertical circle. As an approximation, assume a radial distance of 86.4 cm is between the branch and the point where the monkey's mass is located. As the monkey swings through the lowest point on the circle, it has a speed of 1.33 m/s. Find (a) the magnitude of the centripetal force acting on the monkey and (b) the magnitude of the tension in the monkey's arm.
Answer:
(a) 24.56 N
(b) 142.28 N
Explanation:
(a)
The designation assigned to something like the net force pointed toward the middle including its circular route seems to be the centripetal force. The net stress only at lowest point constitutes of the strain throughout the arm projecting upward towards the middle as well as the weight pointed downwards either backwards from the center.
The centripetal function is generated from either scenario by Equation:
⇒ [tex]Fc = \frac{mv^2}{r}[/tex]
On putting the values, we get
⇒ [tex]=\frac{12\times 1.33^2}{0.864}[/tex]
⇒ [tex]=24.56 \ N[/tex]
(b)
Use T to denote whatever arm stress we can get at the bottom including its circle:
⇒ [tex]Fc = T - mg =\frac{ mv^2}{r}[/tex]
⇒ [tex]T = mg + Fc[/tex]
⇒ [tex]=12\times 9.81+24.56[/tex]
⇒ [tex]=142.28 \ N[/tex]
in a controlled experiment do none of the variables change?
Answer:
Yes
Explanation:
The variables change in and experiment.
Answer:
If you are carefully enough to control everything, then everything that could change the result of your experiment won't happen.
Explanation:
A 1000-turn solenoid is 50 cm long and has a radius of 2.0 cm. It carries a current of 18.0 A. What is the magnetic field inside the solenoid near its center
Answer:
The value is [tex]B = 0.0452 \ T [/tex]
Explanation:
From the question we are told that
The number of turns is N = 1000
The length is L = 50 cm = 0.50 m
The radius is r = 2.0 cm = 0.02 m
The current is I = 18.0 A
Generally the magnetic field is mathematically represented as
[tex]B = \mu_o * \frac{N }{L} * I[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]B = 4\pi * 10^{-7} * \frac{1000}{0.50} * 18.0[/tex]
=> [tex]B = 0.0452 \ T [/tex]
A 50 kg bicyclist starts his ride down the road with an acceleration of 1m/s2 in air with a density of 1.2 kg/m3. If his velocity at a given moment is 2m/s, how much force is he exerting? Assume the area of his body is 0.5m^2.
a. The bicyclist is exerting 1.1 N of force.
b. The bicyclist is exerting 49 N of force.
c. The bicyclist is exerting 50 N of force.
d. The bicyclist is exerting 51 N of force.
Answer:
b. The bicyclist is exerting 49 N of force
Explanation:
Given;
mass of bicyclist start, m = 50 kg
acceleration, a = 1 m/s²
density of air, ρ = 1.2 kg/m³
velocity, v = 2 m/s
Area of the bicyclist body, A = 0.5 m²
The drag force on the bicyclist is given by ;
Fd = 0.5CρAv²
where;
C is drag coefficient = 0.9 for bicycle
Fd = 0.5 x 0.9 x 1.2 x 0.5 x 2²
Fd = 1.1 N
The force of the bicyclist is given by;
F = ma
F = 50 x 1
F = 50 N
The effective force exerted by the bicyclist is given by;
Fe = F - Fd = 50 N - 1.1 N
Fe = 49 N
Therefore, the force exerted by the bicyclist is 49 N
for an emitted wavelength of 500 nanometers and a redshift of 0.4 what will be the observed wavelength g
Answer:
The observed wavelength is [tex] \lambda = 700nm[/tex] (color - Red)
Explanation:
From the question we are told that
The wavelength of the emitter is [tex]\lambda_ e = 500 nm = 500 *10^{-9} \ m[/tex]
The redshift is R = 0.4
Generally red shift is mathematically represented as
[tex]R = \frac{ \lambda - \lambda_e }{\lambda_e}[/tex]
=> [tex]0.4 = \frac{ \lambda - 500 *10^{-9} }{500 *10^{-9} }[/tex]
=> [tex] \lambda - 500*10^{-9} = 200*10^{-9} [/tex]
=> [tex] \lambda = 700 *10^{-9}[/tex]
=> [tex] \lambda = 700nm[/tex]
Hollywood and video games often depict the bad guys being "blown away" when they’re shot by a bullet (i.e. once hit, their feet leave the ground and they fly backwards). Assuming that even if a handgun cartridge did generate enough momentum for the bullet to do this, why is it still nonsense on-screen?
Answer:
Taking a look at Newton's third law of motion which states "for every force exerted, their is an opposite force equal in magnitude and opposite in direction on the first force".
Similarly if a bullet had enough forces behind it to hurl someone through the air when they were hit, a similar force would act on the person holding the gun that fired the bullet.
What we load into the gun is called a 'cartridge' Each piece is composed of four basic substance the casing, the bullet, the primer, and the powder.
The primer explodes lighting the powder which causes a buildup of pressure behind the bullet. This powder can be used in rifle cartages because the bullet chamber is designed to withstand greater pressures.
It is difficult in practice to measure the forces within a gun bagel, but the one easily measured parameter is the velocity with which the bullet exits muzzle velocity, therefore assuming that even if a handgun cartridge which generate enough momentum for the bullet to do this, it is still nonsense on screen in Hollywood and video.
To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d = 4.00 m apart. A listener observes maximum constructive interference while standing in front of the loudspeakers, equidistant from both of them. The distance from the listener to the point halfway between the speakers is l = 5.00 m . One of the loudspeakers is then moved directly away from the other. Once the speaker is moved a distance r = 60.0 cm from its original position, the listener, who is not moving, observes destructive interference for the first time. Find the speed of sound v in the air if both speakers emit a tone of frequency 700 Hz .
Complete Question
The compete question is shown on the first uploaded question
Answer:
The speed is [tex] v = 350 \ m/s [/tex]
Explanation:
From the question we are told that
The distance of separation is d = 4.00 m
The distance of the listener to the center between the speakers is I = 5.00 m
The change in the distance of the speaker is by [tex]k = 60 cm = 0.6 \ m[/tex]
The frequency of both speakers is [tex]f = 700 \ Hz[/tex]
Generally the distance of the listener to the first speaker is mathematically represented as
[tex]L_1 = \sqrt{l^2 + [\frac{d}{2} ]^2}[/tex]
[tex]L_1 = \sqrt{5^2 + [\frac{4}{2} ]^2}[/tex]
[tex]L_1 = 5.39 \ m [/tex]
Generally the distance of the listener to second speaker at its new position is
[tex]L_2 = \sqrt{l^2 + [\frac{d}{2} ]^2 + k}[/tex]
[tex]L_2 = \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}[/tex]
[tex]L_2 = 5.64 \ m [/tex]
Generally the path difference between the speakers is mathematically represented as
[tex]pD = L_2 - L_1 = \frac{n * \lambda}{2}[/tex]
Here [tex]\lambda[/tex] is the wavelength which is mathematically represented as
[tex]\lambda = \frac{v}{f}[/tex]
=> [tex] L_2 - L_1 = \frac{n * \frac{v}{f}}{2}[/tex]
=> [tex] L_2 - L_1 = \frac{n * v}{2f}[/tex]
=> [tex] L_2 - L_1 = \frac{n * v}{2f}[/tex]
Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves
=> [tex] 5.64 - 5.39 = \frac{1 * v}{2*700}[/tex]
=> [tex] v = 350 \ m/s [/tex]
The speed of sound in air is 350 m/s
Since the distance between both speakers is 4 m, and the listener is standing 5 m away from halfway between them, the distance L from each loudspeaker to the listener, since the arrangement forms a right-angled triangle, using Pythagoras' theorem,
L = √[(5 m)² + (4/2 m)²]
= √[25 m² + (2 m)²]
= √[25 m² + 4 m²]
= √29 m² = 5.39 m.
Now, when one speaker is moved 60 cm = 0.6 m away from its original position, its distance from the listener is now
L' = √[(5 m)² + (4/2 + 0.6 m)²]
= √[25 m² + (2 m + 0.6 m)²]
= √[25 m² + (2.6 m)²]
= √[25 m² + 6.76 m²]
= √31.76 m²
= 5.64 m.
Now, the path difference when we first have destructive interference is
ΔL = L' - L
= 5.64 - 5.39
= 0.25
Since we have destructive interference for the first time when the speaker is moved, the path difference, ΔL = (n + 1/2)λ where λ = wavelength = v/f where v = speed of sound in air and f = frequency = 700 Hz.
Now, since we have destructive interference for the first time, n = 0.
So, ΔL = (n + 1/2)λ
ΔL = (0 + 1/2)v/f
ΔL = v/2f
Making v subject of the formula, we have
v = 2fΔL
Substituting the values of the variables into the equation, we have
v = 2fΔL
v = 2 × 700 Hz × 0.25 m
v = 350 m/s
So, the speed of sound in air is 350 m/s
Learn more about interference of sound here:
https://brainly.com/question/1346741
Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?
Answer:
Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?
my heart strike him to dead.what figure of speech is that?
Answer:
Hyperbole
Explanation:
this is an extreme exaggeration or overstatement/ magnification
Which is a belief held by sociologists who work from a social-conflict
perspective?
O A. The best approach for a study is from a micro-level orientation.
O B. Personal background has little impact on how individuals react
with one another.
C. Some social patterns are helpful, while others are harmful.
D. Data are irrelevant to the study of sociology.
SUBMIT
Answer:
C. Some social patterns are helpful, while others are harmful.
Explanation:
Hope this was helpful, Have an amazing,spooky Halloween!!
which statement is correct about the strength of forces?
-Electrostatic forces are exactly 10 times stronger than gravitational forces.
-Electrostatic forces are exactly 10 times weaker than gravitational forces.
-Electrostatic forces are trillions of times stronger than gravitational forces.
-Electrostatic forces are trillions of times weaker than gravitational forces.
Answer:
Thanks!!!!! adding this so it doesn’t get deleted.
Explanation:
1. Electrostatic forces are trillions of times stronger than gravitational forces. 2. normal force and friction 3. contact forces 4. The electrostatic forces from the contact of the hands with the paper causes the paper molecules to separate. 5. The electrostatic forces between the molecules of the board prevent the force of gravity from breaking the board apart.
The correct statement over here is that electrostatic forces are trillions of times stronger than gravitational forces. Hence, option C is correct.
What is an Electrostatic Force?One of the basic forces in the cosmos is electrostatic force. In the universe, there are four basic forces. These include gravitational force, electromagnetic force, weak nuclear force, and strong nuclear force. Under the umbrella of electromagnetic force is electrostatic force. Two charges placed apart are subject to the electrostatic force. The size of each charged and the separation between them determines how much electrostatic force there will be.
When two charges of the same type are brought together, whether positive or negative, they repel one another. It is known as the electrostatic force of repelling when it operates among two charges that are similar.
Therefore, the electrostatic forces are trillions of times stronger than the gravitational forces.
To know more about Electrostatic Force:
https://brainly.com/question/9774180
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Radio station KBOB broadcasts at a frequency of 85.7 MHz on your dial using radio waves that travel at 3.00 × 108 m/s. Since most of the station's audience is due south of the transmitter, the managers of KBOB don't want to waste any energy broadcasting to the east and west. They decide to build two towers, transmitting in phase at exactly the same frequency, aligned on an east-west axis. For engineering reasons, the two towers must be AT LEAST 10.0 m apart. What is the shortest distance between the towers that will eliminate all broadcast power to the east and west?
Answer:
12.5 m
Explanation:
The first thing we would do is to calculate the wavelength. To do this, we use the formula
v = fλ, where
v = wave speed
f = frequency
λ = wavelength
If we make wavelength the formula, we have
wavelength = speed / frequency
Now, we substitute the values we had been given and we have
wavelength = (3 * 10^8 m/s) / (85.7 * 10^6 Hz) wavelength = 3.50 m
half of this said wavelength will be
= 3.50 / 2
= 1.75 m
As a result of the engineering constraints with the towers being more than 10 m apart, the distance can't be 1.75 m and as such, it has to be a multiple of 1.75m. So we say,
(10 / 1.75) = 5.7
So the separation will have to be 7 half wavelengths
= (7 * 1.75) = 12.5 m