Answer:
31250 meters
Explanation:
Given data
Intitially at rest, the velocity will be
u= 0m/s
acceleration a= 25m/s^2
Time= 50s
We know that the expression for the displacement is given as
S=U+ 1/2at^2
S= 0+ 1/2*25*50^2
S= 12.5*2500
S=31250 meters
Hence the displacement is 31250 meters
You are outdoors when you hear the constant chirp of a still cricket. You start walking toward the cricket and at some point you are able to detect that the intensity of the chirp of the cricket has increased by a factor of 4. What of the following statements is true at your new position with respect to the cricket?
a. The power delivered by the sound wave you hear has doubled.
b. The speed of the sound wave emitted by the cricket has decreased by a factor of 4.
c. The distance between you and the cricket has decreased by a factor of 2
Answer:
C
Explanation:
intensity = Power delivered by the sound (Watt)/ Surounding Area (m²)
I = P/A
A = πr²
r = is the distance between you and the cricket.
so in other form we can get
I = P/πr²
let take I(1) as first intensitilynyou heard and I(2) as the increased intensity.
I(1) / I(2) = r(2)² / r(1)²
1/4 = r(2)²/r(1)²
1/2 = r(2) / r(1)
r(2) = ½ r(1)
or r(2) is decreaases by a factor of 2.
A wheel is rotating freely at angular speed 530 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with 9 times the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels
Answer: [tex]53\ rev/min[/tex]
Explanation:
Given
angular speed of wheel is [tex]\omega_1 =530\ rev/min[/tex]
Another wheel of 9 times the rotational inertia is coupled with initial wheel
Suppose the initial wheel has moment of inertia as I
Coupled disc has [tex]9I[/tex] as rotational inertia
Conserving angular momentum,
[tex]\Rightarrow I\omega_1=(I+9I)\omega_2\\\\\Rightarrow \omega_2=\dfrac{I}{10I}\times 530\\\\\Rightarrow \omega_2=53\ rev/min[/tex]
A car is traveling at 118 km/h when the driver sees an accident 85 m ahead and slams on the brakes. What minimum constant deceleration is required to stop the car in time to avoid a pileup
Answer:
The constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
Explanation:
From the question, the car is traveling at 118 km/h, that is the initial velocity, u = 118km/h
The distance between the car and the accident at the moment when the driver sees the accident is 85 m, that is s = 85 ,
Since the driver slams on the brakes and the car will come to a stop, then the final velocity, v = 0 km/h = 0 m/s
First, convert 118 km/h to m/s
118 km/h = (118 × 1000) /3600 = 32.7778 m/s
∴ u = 32.7778 m/s
Now, to determine the deceleration, a, required to stop,
From one of the equations of motion for linear motion,
v² = u² + 2as
Then
0² = (32.7778)² + 2×a×85
0 = 1074.3841 + 170a
∴ 170a = - 1074.3841
a = - 1074.3841 / 170
a = - 6.3199
a ≅ - 6.32 m/s²
Hence, the constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²
A parallel plate air capacitor has a circular disc of diameter 0.1 m, 2 mm apart and potential difference of 300 V is connected between the plates Calculate: (i) Energy of the capacitor and (ii) Electric intensity between the plates.
Your Answer Is in this attachment
A diffraction grating has 6000 lines per centimeter ruled on it. What is the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe when the grating is illuminated with a beam of light of wavelength 500 nm
Explanation:
Hope it Will help he hsuejwoamxgehanwpalasmbwfwfqoqlmdbehendalmZbgevzuxwllw. yeh we pabdvddxhspapalw. X
The angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
What is diffraction?Waves spreading outward around obstructions are known as diffraction. Sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as very small moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities all exhibit diffraction.
Given:
The number of lines = 6000 per cm,
The Wavelength, λ = 500 nm = 500 × 10 ⁻⁹ m
Calculate the diffraction grating,
[tex]d = 1 / no\ of\ lines[/tex]
d = 10⁻² / 6000 m,
Calculate the second-order maxima angle and third-order maxima angle by the formula given below,
[tex]dsin\theta_1 = n_1 \lambda[/tex]
[tex]sin\theta_1 = n_1\lambda / d[/tex]
[tex]\theta _1 = sin^{-1}[2\times 500\times 10 ^{-9}/10^{-2}\times 6000][/tex]
θ₁ = sin⁻¹(0.6)
θ₁ = 36.87°
Similarly, for θ₂,
θ₂ = sin⁻¹(3 × 500 × 10 ⁻⁹ / 10⁻² × 6000)
θ₂ = sin⁻¹(0.9)
θ₂ = 64.16°
Calculate the separation as follows,
θ₂ - θ₁ = 64.16° - 36.87°
θ₂ - θ₁ = 27.29°
Therefore, the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
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A mass is tired to spring and begins vibration periodically the distance between it's lowest position is 48cm what is the Amplitude of the vibration
Answer:
The amplitude of vibration of the spring is "24 cm"
The periodic vibrating body's motion follows a sinusoidal path. This sinusoidal path is illustrated in the attached picture.
From the picture, it can be clearly seen that the amplitude of the periodic vibration motion is the distance from its mean position to the highest point.
Since the distance of both the highest and the lowest points from the mean position is the same. Therefore, the distance between the lowest and the highest point must be equal to two times the amplitude of the wave.
Amplitude = 24 cm
why do atom absorb photon since it makes it unstable??
[tex]\textsf{When an electron is hit by a }[/tex] [tex]\textsf{photon of light, it absorbs the quanta}[/tex] [tex]\textsf{of energy the photon was carrying}[/tex] [tex]\textsf{and moves to a higher energya}[/tex] [tex]\textsf{ state. Electrons therefore have to }[/tex] [tex]\textsf{jump around within the atom as }[/tex] [tex]\textsf{they either gain or lose energy. }[/tex]
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. Electrons therefore have to jump around within the atom as they either gain or lose energy.
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A photon has an energy of 2.09×10^-18 kJ .What is its wavelength?
Answer:
wavelength= 1.05 × 10^ -46 m
Explanation:
the formula : λ= hc/E
where; "h" = Planck's constant [6.626 × 10^ -34]
c= speed of light [3.0 × 10^ 8]
you first have to convert the energy of the photon to Joules by dividing the constant by 1000
2.09 × 10^ -18 / 1000 = 2.09 × 10^ -21
then you replace you data into the equation
λ= 6.626 × 10^ -34 × 3.0 × 10^ 8 / 2.09 × 10 ^ -21
first multiply the Planck's constant and the speed of light then divide it by the energy which is in "Joules"
:. λ = 1.05 × 10^ -46
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There is more than 1 answer,
The picture is down
Answer:
test her prototype and collect data about its flight
A projectile is launched straight upwards at 75 m/s. Three seconds later, its velocity is...?
Answer:
V = V0 + a t
V = 75 - 9.8 * 3 = 45.6 m/s
The final velocity of the projectile after 3 seconds is equal to 45.6 m/s.
What is the equation of motion?The equations of motion can be defined as the relation of the motion of a physical system as the function of time and set up the relationship between the displacement (s), acceleration, velocity (v & u), and time of a moving system.
Given, the initial velocity of the projectile, u = 75 m/s
The time taken by the projectile, t = 3 sec
The acceleration due to gravity upward, g = - 9.8 m/s²
From the first equation of motion we can calculate the final velocity of the projectile:
v = u + at
v = u - gt
v = 75 - 9.8 ×(3)
v = 75 - 29.4
v = 45.6 m/s
Therefore, the final velocity of the projectile after three seconds is 45.6 m/s.
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If a conducting loop of radius 10 cm is onboard an instrument on Jupiter at 45 degree latitude, and is rotating with a frequency 2 rev/s; What is the maximum emf induced in this loop? If its resistance is 0.00336 ohms, how much current is induced in this loop? And what is the maximum power dissipated in the loop due to its rotation in Jupiter's magnetic field?
Answer:
a) fem = - 2.1514 10⁻⁴ V, b) I = - 64.0 10⁻³ A, c) P = 1.38 10⁻⁶ W
Explanation:
This exercise is about Faraday's law
fem = [tex]- \frac{ d \Phi_B}{dt}[/tex]
where the magnetic flux is
Ф = B x A
the bold are vectors
A = π r²
we assume that the angle between the magnetic field and the normal to the area is zero
fem = - B π 2r dr/dt = - 2π B r v
linear and angular velocity are related
v = w r
w = 2π f
v = 2π f r
we substitute
fem = - 2π B r (2π f r)
fem = -4π² B f r²
For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T
we reduce the magnitudes to the SI system
f = 2 rev / s (2π rad / 1 rev) = 4π Hz
we calculate
fem = - 4π² 428 10⁻⁶ 4π 0.10²
fem = - 16π³ 428 10⁻⁶ 0.010
fem = - 2.1514 10⁻⁴ V
for the current let's use Ohm's law
V = I R
I = V / R
I = -2.1514 10⁻⁴ / 0.00336
I = - 64.0 10⁻³ A
Electric power is
P = V I
P = 2.1514 10⁻⁴ 64.0 10⁻³
P = 1.38 10⁻⁶ W
Determine the magnitude as well as direction of the electric field at point A, shown in the above figure. Given the value of k = 8.99 × 1012N/C.
Answer:
Electric field at A = 9.28 x 10¹² N/C
Explanation:
Given:
K = 8.99 x 10¹² N/C
Missing information:
Length = 11 cm = 11 x 10⁻² m
q = 12.5 C
Find:
Electric field at A
Computation:
Electric field = Kq / r²
Electric field at A = [(8.99 x 10¹²)(12.5)] / [11 x 10⁻²]²
Electric field at A = 9.28 x 10¹² N/C
The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88
is the correct answer
what effect does the force of gravity have on a stone thrown vertically upwards
Answer:
rock go down
Explanation:
what comes up must come down.
Meaning of power in physics
Answer:
The rate of doing work is called power.
Answer:
The amount of energy transported or transformed per unit time is referred to as power in physics. The watt, which is equal to one joule per second in the International System of Units, is the unit of power.
OAmalOHopeO
an alternating voltage of 100V, 50HZ Is Applied across an impedance of (20-j30) calculate the resistance, the capacitance, current, the phase angle between current and voltage
The resistance R = 20 Ω
The capacitance C = 106.1 μF
The current, I is 2.773 A at 56.31°.
The phase angle of the between the current and the voltage is 56.31° leading.
Since the impedance Z = 20 - j30 Ω, the resistance, R is the real part of the impedance. So R = ReZ = 20 Ω
So, the resistance R = 20 Ω
To find the capacitance, we need first to find the reactance of the capacitor X. Since the impedance Z = 20 - j30, the reactance of the capacitor X. is the imaginary part of the impedance. So X = ImZ = 30 Ω.
Now the reactance of the capacitor X = 1/ωC where ω = angular frequency of the circuit = 2πf where f = frequency of the circuit = 50 Hz and C = capacitance
So, C = 1/ωX = 1/2πfX
Substituting the values of the variables into the equation, we have
C = 1/2πfX
C = 1/(2π × 50 Hz × 30 Ω)
C = 1/3000π
C = 1/9424.778
C = 1.061 × 10⁻⁴ F
C = 106.1 × 10⁻⁶ F
C = 106.1 μF
So, the capacitance is 106.1 μF
The current I = V/Z where V = voltage = 100 V at 0° and Z = impedance.
The magnitude of Z = √(20² + (-30)²)
= √(400 + 900)
= √1300
= 36.06 Ω
and its angle Φ = tan⁻¹(ImZ/ReZ)
= tan⁻¹(-30/20)
= tan⁻¹(-1.5) = -56.31°
So, V = 100 ∠ 0° and Z = 36.06 ∠ -56.31°
So, the current, I = V/Z = (100 ∠ 0°)/36.06 ∠ -56.31°
= 100/36.06 ∠(0° - (-56.31° ))
= 2.773 ∠ 56.31° A
So, the current is 2.773 A at 56.31°.
Since the current is 2.773 A at 56.31°, the phase angle of the between the current and the voltage is 56.31° leading.
So, the phase angle of the between the current and the voltage is 56.31° leading.
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Vặt nhỏ được ném lên từ điểm A trên mặt đất với vận tốc đầu 20m/s theo phương thẳng đứng. Xác định độ cao của điểm O mà vật đạt được. Bỏ qua ma sát
Explanation:
mặt đất với vận tốc ban đầu 20m/s. Bỏ qua mọi ma sát, lấy g = 10 m/s2. Độ cao cực đại mà vật đạt được là.
An organ pipe of length 3.0 m has one end closed. The longest and next-longest possible wavelengths for standing waves inside the pipe are
Answer:
The longest wavelength for closed at one end and open at the other is
y / 4 where y is the wavelength - that is node - antinode
The next possible wavelength is 3 y / 4 - node - antinode - node -antinode
y / 4 = 3 m y = 12 meters the longest wavelength
3 y / 4 = 3 m y = 4 meters 1 / 3 times as long
A 1-cm long wire carrying 15 A is inside a solenoid 4 cm in radius with 800 turns/m carrying a current of 40 mA. The wire segment is oriented perpendicularly to the axis of the solenoid. What is the magnitude of the magnetic force on this wire segment in ???? N?
Answer:
the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N
Explanation:
Given;
length of the conductor, L = 1 cm = 0.01 m
current carried by the solenoid, I₁ = 15 A
radius of the solenoid, r = 4 cm
number of turns per length of the solenoid, n = 800 turns/m
current carried by the solenoid, I₂ = 40 mA = 0.04 A
The magnetic field of the solenoid is calculated as;
B = μnI₂
where;
μ is the permeability of free space = 4π x 10⁻⁷ Tm/A
B = ( 4π x 10⁻⁷) x (800) x (0.04)
B = 4.022 x 10⁻⁵ T
The magnitude of the magnetic force on the wire segment is calculated as;
F = BI₁L sinθ
where
θ is the angle made by the wire segment against the solenoid = 90⁰
F = (4.022 x 10⁻⁵) x (15) x (0.01) x sin(90)
F = 6.03 x 10⁻⁶ N
Therefore, the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N
A block of mass 2 kg starts from rest at the top of a friction quarter of a circle of radius R. The block then slides over frictionless curved surface in the shape of a eventually comes to rest 8 m from the beginning s a horizontal rough surface where e of the horizontal surface. The coefficient kinetic friction between the rough surface and the block is 0.4 . determine the acceleration of the block over the rough surface length 8m
The acceleration of the block over the rough surface is 1.22625 m/s²
The process through which the acceleration is obtained is presented as follows of approach to
The given parameters are;
Mass of block, m = 2 kg
Nature of the surface of the quarter circle = Frictionless
The length of the horizontal, d = 8 m
The coefficient of friction of the horizontal surface, μ = 0.4
The unknown parameter;
The acceleration of the block over the rough surface
Method;
Find the work done by friction to stop the block and divide the result by the mass of the block
The work done by friction, [tex]W_f[/tex] = (Force of friction) × (Distance the block moves on the rough surface before coming to rest)
[tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d
[tex]F_f[/tex] = Normal reaction of surface on block, [tex]N_r[/tex] × μ
Normal reaction on block, [tex]\mathbf{N_r}[/tex] = Weight of block
[tex]\mathbf{N_r}[/tex] ≈ 2 kg × 9.81 m/s² = 19.62 N
Therefore;
The work done by friction [tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d = [tex]\mathbf{N_r}[/tex] × μ × d
[tex]\mathbf{W_f}[/tex] = 19.62 N × 0.4 × 8 m = 62.784 J
The work done by the block, W = Force, F × d
Force, F = m × a
Where;
a = The acceleration of the block
According to the principle of conservation of energy, we have;
[tex]\mathbf{W_f}[/tex] = W
∴ 19.62 J = 2 kg × a × 8 m
a = 19.62/(2 kg × 8 m) = 1.22625 m/s²
The acceleration of the block over the rough surface, a = 1.22625 m/s²
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help me I m stuck on this question
Answer:
Answer is in the picture.
Explanation:
Answer is in the picture.
2. What is the average speed of an athlete who runs 1500 m in 4 minutes?
Answer:
375 is the answer.
Explanation:
Speed : Distance / Time taken
S: m/ s
s: 1500/4
375 m / s answer
Answer:
375m per minute
Explanation:
if you are looking for a diffrent unit just multiply your answer by however many minutes are in that time frame
Light takes 1.2 sec to get from the moon to the Earth. Assume you are looking at the moon with noticeable earth shine. If the Sun burned out, you would eventually see the crescent of the moon disappear. The earth shine part of the moon would disappear Answer 2.4 s after the crescent disappeared.
Answer:
1.2 seconds
Explanation:
Answer to the following question is 1.2 seconds
Because light from the moon takes 1.2 seconds to reach Earth, the light released from the crescent immediately before it vanishes will also take 1.2 seconds to reach Earth. As a result, the earth-shine portion of the moon will vanish 1.2 seconds after the crescent has vanished.
A sound wave made up of large number of unrelated frequencies superposted on each other is
Since the frequencies are unrelated, and there are a large number of them, I'll say this represents an example of noise.
When a charged particle moves at an angle of 26.1 with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90o) with respect to this field will this particle, moving at the same speed?
Answer:
The angle is 153.9 degree.
Explanation:
Let the magnetic field is B and the charge is q. Angle = 26.1 degree
The force is F.
Let the angle is A'.
Now equate the magnetic forces
[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]
Your little sister (mass 25 kg) is sitting in her little red wagon (mass
8.5 kg) at rest. You begin pulling her forward, accelerating her with a
constant force for 2.35 s to a speed of 1.8 m/s. Calculate the impulse
you imparted to the wagon and its passenger.
Answer:
p = 60.6N*s
Explanation:
v_f = v_0+a*t
a = (v_f-v_0)/t
a = (1.8m/s)/2.35s
a = 0.77m/s²
F = m*a
F = (25kg+8.5kg)*0.77m/s²
F = 25.8N
^p = F*t
p = 25.8N*2.35s
p = 60.6N*s
Fast please!!. Serious answers only.
Considering that Susan is 38 years old, what is the probability that she lives to the age of 85?
Answer:
The probability that Susan turns 85 years old is 45,436/97,825= 0.4644.
Explanation:
Edmentum
Answer:
45,436/97,825= 0.4644.
what is the major difference between the natural frequency and the damped frequency of oscillation.
Answer:
This causes the amplitude of the oscillation to decay over time. The damped oscillation frequency does not equal the natural frequency. Damping causes the frequency of the damped oscillation to be slightly less than the natural frequency
: A fan is placed on a horizontal track and given a slight push toward an end stop 1.80 meters away. Immediately after the push, the fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2. What is the maximum possible velocity (magnitude) the cart can have after the push so that the cart turns around just before it hits the end-stop
Answer:
The initial velocity is 1.27 m/s.
Explanation:
distance, s = 1.8 m
acceleration, a = - 0.45 m/s^2
final velocity, v = 0
let the initial velocity is u.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times 0.45\times 1.8\\\\u = 1.27 m/s[/tex]
We have that the Initial velocity is mathematically given as
u=1.27m/s
Maximum possible velocity
Question Parameters:
a slight push toward an end stop 1.80 meters away
he fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2
Generally the equation for the third equation of motion is mathematically given as
Vf^2 = Vi^2 + 2ad
Therefore
0=u^2+0.45*1.8
u=1.27m/s
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A/An is a type of blood cell that's also called a red blood cell. a) Jeukocyte O b) thrombocyte c) plasma d) erythrocyte
Answer:
red blood cell, also called erythrocyte
Explanation:
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