Answer:
The answer is below
Explanation:
Given that:
The area of the plates is 6 m by 0.030 m, Therefore the area = 6 m × 0.03 m = 0.18 m²
the relative permittivity of dielectric (εr) is 7.0
Permittivity of free space (εo) = 8.854 × 10^(-12)
capacitance of 100uF
potential difference (V) of 12V
d = separation between plate
The capacitance (C) of a capacitor is given by:
[tex]C=\frac{\epsilon_o \epsilon_r A }{d}\\ 100*10^{-6}=\frac{8.854*10^{-12}*7*0.18}{d}\\ d=\frac{8.854*10^{-12}*7*0.18}{100*10^{-6}}=1.11*10^{-7}\ m[/tex]
The electric field between plates is given as:
E = V /d
[tex]E = 12 / 1.11*10^{-7}=10.75*10^7\ V/m[/tex]
g A tube open at both ends, resonated at it's fundamental frequency, to a sound wave traveling at 330m/s. If the length of the tube is 4cm, find the frequency of the sound wave.
Answer:
frequency =4125Hz
Explanation:
L = 4cm = 0.04m
f =v/2L
f = 330/2 x 0.04
f = 4125Hz
A hoop, a solid disk, and a solid sphere, all with the same mass and the same radius, are set rolling without slipping up an incline, all with the same initial kinetic energy.
Which goes furthest the incline?
a. The hoop
b. The disk
c. The sphere
d. They all roll to the same height
Answer:
The sphere
Explanation:
Because it has a smaller inertia (I) value in the explanation in the attached file
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the wavelength of the illuminating light is doubled
Answer:
Doubling the wavelength of the diffracting doubles the angle of diffraction. So, the width of the central bright spot pattern formed on the screen will also be doubled.
Explanation:
For a single slit diffraction, the path length difference is related to the wavelength of the light leaving the slit onto the screen by
D sin θ = mλ
where D sin θ is the path length of the waves, each.
mλ is the wavelength of the wavelet
where m is the the order of each minimum
m = m = 1,−1,2,−2,3, . . .
The wavelength of each wavelet is always a multiple of the wavelength of the light source, and from the equation, we can see that the angle of diffraction depend on the wavelength of the light. From this we can see that increasing the wavelength of the light increases the angle of diffraction, and hence we can say that doubling the wavelength will double the diffraction angle. Also, the width of the central bright spot of the screen will spread or increase with the angle of diffraction, so doubling the wavelength doubles the central bright spot on the screen.
A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is the velocity of the exiting water? Ignore all orificelosses.
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
[tex]\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2[/tex]
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
[tex]\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}[/tex]
= 4.7476 m/sec
= 4.75 m/s
On a certain planet a body is thrown vertically upwards with an initial speed of 40 m / s. If the maximum height was 100 m, the acceleration due to gravity is
a) 15 m / s 2
b) 12.5 m / s 2
c) 8 m / s 2
d) 10 m / s 2
Answer:
C) 8 m/s²
Explanation:
Given:
v₀ = 40 m/s
v = 0 m/s
Δy = 100 m
Find: a
v² = v₀² + 2aΔy
(0 m/s)² = (40 m/s)² + 2a (100 m)
a = -8 m/s²
Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?
Answer:
I₂ = 0.25 I₀
Explanation:
To know the light transmitted by a filter we must use the law of Malus
I = I₀ cos² θ
In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.
I₁ = I₀ cos² 45
I₁ = I₀ 0.5
this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂
I₂ = I₁ cos² 45
I₂ = (I₀ 0.5) 0.5
I₂ = 0.25 I₀
this is the intensity of the light transmitted by the set of polarizers
Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determine:
(a) the magnitude and direction of the vector D=A+B+C and
(b) the magnitude and direction of E=-A - B + C.
Answer:
(a) [tex]\vec D = 2\,i - 2\,j[/tex], (b) [tex]\vec E = -6\,i + 12\,j[/tex]
Explanation:
Let be [tex]\vec A = 3\,i - 3\,j\,[m][/tex], [tex]\vec B = i - 4\,j\,[m][/tex] and [tex]\vec C = -2\,i + 5\,j \,[m][/tex], each resultant is found by using the component method:
(a) [tex]\vec D = \vec A + \vec B + \vec C[/tex]
[tex]\vec D = (3\,i - 3\,j) + (i-4\,j) + (-2\,i+5\,j)\,[m][/tex]
[tex]\vec D = (3\,i + i -2\,i)+(-3\,j-4\,j+5\,j)\,[m][/tex]
[tex]\vec D = (3 + 1 -2)\,i + (-3-4+5)\,j\,[m][/tex]
[tex]\vec D = 2\,i - 2\,j[/tex]
(b) [tex]\vec E = -\vec A - \vec B + \vec C[/tex]
[tex]\vec E = -(3\,i-3\,j)-(i - 4\,j)+(-2\,i+5\,j)[/tex]
[tex]\vec E = (-3\,i + 3\,j) +(-i+4\,j) + (-2\,i + 5\,j)[/tex]
[tex]\vec E = (-3\,i-i-2\,i) + (3\,j+4\,j+5\,j)[/tex]
[tex]\vec E = (-3-1-2)\,i + (3+4+5)\,j[/tex]
[tex]\vec E = -6\,i + 12\,j[/tex]
Since you analyzed the charging of a capacitor for a DC charging voltage, how is it possible that you
Answer:
I = E/R e^{-t/RC}
Explanation:
In a capacitor charging circuit you must have a DC power source, the capacitor, a resistor, and a switch. When closing the circuit,
E -q / c-IR = 0
we replace the current by its expression and divide by the resistance
I = dq / dt
dq / dt = E / R -q / RC
dq / dt = (CE -q) / RC
we solve the equation
dq / (Ce-q) = -dt / RC
we integrate and evaluate for the charge between 0 and q and for the time 0 and t
ln (q-CE / -CE) = -1 /RC (t -0)
eliminate the logarithm
q - CE = CE [tex]e^{-t/RC}[/tex]
q = CE (1 + 1/RC e^{-t/RC} )
In general the teams measure the current therefore we take the derivative to find the current
i = CE (e^{-t/RC} / RC)
I = E/R e^{-t/RC}
This expression is the one that describes the charge of a condensate in a DC circuit
A stone is thrown vertically upward with a speed of 29.0 m/s and when it reaches a height of 13 m, the velocity is 24.2 m/s.
Using the formula x = v0 • t + ½ • a • t2, find the time it takes to reach this height? Why do you get two values for time? Explain.
Answer:
the value of t = 0.49 seconds shows that its upward journey
and
at t = 5.43 seconds shows in downward journey
Explanation:
Given:
initial speed, u = 29 m/s
acceleration due to gravity, g = - 9.8 m/s^2
h = 13 m
Let it is moving with velocity v at a height of 13 m.
Use third equation of motion
v² = u² + 2gh
By substituting the values
v² = 29² - (2 * 9.8 * 13)
v = sqrt 585.94
v = 24.2 m/s
Let it takes time t to reach at height 13 m
Use second equation of motion
s = u * t + 1/2 * g * t²
13 = 29t - 4.9t²
4.9t² - 29t + 13 = 0
using quadratic equation to solve time
29 ± [tex]\sqrt{29^2 - 4 * 4.9 * 13}\\[/tex]
t = ------------------------------------
2 * 4.9
t = 5.43 second or t = 0.49 second
Therefore...
the value of t = 0.49 seconds shows that its upward journey
and
at t = 5.43 seconds shows in downward journey
The cost of buying shirts is partly constant and partly varies with the number of shirts bought. When the number of shirts is 5 the cost is #240, also, 10 shirts costs #400. find the cost when 300 shirts were bought
Answer:
The cost of the buying the shirts is #9680
Explanation:
let the cost of buying shirt = C
let the number of shirt bought = N
The following equation can be generated based on the statement above;
C = k + Nb
When the cost, C = #240, the number of shirt = 5
240 = k + 5b ------ equation (1)
where;
k and b are constants
When the cost, C = #400, the number of shirt = 10
400 = k + 10b ------ equation (2)
From equation (1), make k the subject of the formula;
k = 240 - 5b ---- equation (3)
Substitute in the value of k into equation (2)
400 = k + 10b
400 = (240 - 5b) + 10b
400 = 240 - 5b + 10b
400 - 240 = -5b + 10b
160 = 5b
b = 160 / 5
b = 32
From equation (3), calculate k
k = 240 - 5b
k = 240 -5(32)
k = 240 - 160
k = 80
When the number of shirts bought = 300, the cost of the buying the shirts =
C = k + Nb
C = 80 +32N
Where;
N is the number of shirts
C = 80 + 32(300)
C = 80 + 9600
C = #9680
Therefore, the cost of the buying the shirts is #9680
If a system has 4.50×102 kcal of work done to it, and releases 5.00×102 kJ of heat into its surroundings, what is the change in internal energy (ΔE or Δ????) of the system?
The change in internal energy (ΔE) of the system is equal to -18823 Kilojoules.
Given the following data:
Quantity of heat = [tex]5.00 \times 10^2 \;kJ[/tex]Work done = [tex]4.50 \times 10^2 \;kcal[/tex]Conversion:
1 kcal = 4.184 kJ
[tex]4.50 \times 10^2 \;kcal[/tex] = [tex]4.50 \times 10^2 \times 4.184 = 18828 \; kJ[/tex]
To determine the change in internal energy (ΔE) of the system, we would apply the first law of thermodynamics.
Mathematically, the first law of thermodynamics is given by the formula:
[tex]\Delta E = Q - W[/tex]
Where;
[tex]\Delta E[/tex] is the change in internal energy.Q is the quantity of heat released.W is the work done.Substituting the given parameters into the formula, we have;
[tex]\Delta E = 5 - 18828\\\\\Delta E = -18823[/tex]
Change in internal energy, E = -18823 Kilojoules
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Two point charges attract each other with an electric force of magnitude F. If one charge is reduced to one-third its original value and the distance between the charges is doubled, what is the resulting magnitude of the electric force between them
Answer:
F' = F/12
Therefore, the electrostatic force is reduced to one-twelve of its original value.
Explanation:
The electrostatic force of attraction or repulsion between to charges is given by Coulomb's Law:
F = kq₁q₂/r² ---------- equation 1
where,
F = Electrostatic Force
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between charges
Now, if we double the distance between charges and reduce one charge to one-third value, then the force will become:
F' = kq₁'q₂'/r'²
where,
q₁' = (1/3)q₁
q₂' = q₂
r' = 2r
Therefore,
F' = k(1/3 q₁)(q₂)/(2r)²
F' = (1/12)kq₁q₂/r²
using equation 1:
F' = F/12
Therefore, the electrostatic force is reduced to one-twelve of its original value.
A sharp edged orifice with a 50 mm diameter opening in the vertical side of a large tank discharges under a head of 5m. If the coefficient of contraction is 0.62 and the coefficientof velocity is 0.98, what is thedischarge?
Answer:
0.24
Explanation:
See attached file
Consider the following three objects, each of the same mass and radius:
(1) a solid sphere
(2) a solid disk
(3) a hoop
All three are released from rest at the top of an inclined plane. The three objects proceed down the incline undergoing rolling motion without slipping. Use work-kinetic energy theorem to determine which object will reach the bottom of the incline first.
a) 1, 2, 3
b) 2, 3, 1
c) 3, 1, 2
d) 3, 2, 1
e) All three reach the bottom at the same time.
Answer:
Explanation:a 1
Two capacitors, CA and CB, are such that CA > CB. These are connected with a battery in various ways: each individually, series, and parallel. Rank these four cases according to the total amount of charge, greatest first.
a. (CA) > (C) > (CA and CB in series) > (CA and Co in parallel)
b. (CA)>(Cb)> (CA and CB in parallel) > (CA and CB in series)
c. (CA and CB in series) > (CA) > (CB) > (CA and CB in parallel)
d. (CA and Cg in parallel) > (CA) > (CB) > (CA and Cg in series)
Answer:
b. (CA)>(Cb)> (CA and CB in parallel) > (CA and CB in series)
Explanation:
This is because capacitors in series is the sum of the reciprocal of the capacitances while that of parallel is the sum of the individual capacitances
A car travels at 100 km / h, collides head-on against a pole. Assuming the vehicle stopped at 2.2 seconds after impact, calculate the magnitude of the deceleration suffered by the driver.
Answer:
12.6 m/s²
Explanation:
First, convert to m/s.
100 km/h × (1000 m/km) × (1 hr / 3600 s) = 27.8 m/s
a = Δv / Δt
a = (0 m/s − 27.8 m/s) / 2.2 s
a = -12.6 m/s²
Which of the following represents a concave mirror? +f,-f,-di,+di
Answer:
fully describes the concave mirror is + f
Explanation:
A concave mirror is a mirror where light rays are reflected reaching a point where the image is formed, therefore this mirror has a positive focal length, the amount that fully describes the concave mirror is + f
This allows defining a sign convention, for concave mirror + f, the distance to the object is + d0 and the distance to the image is + di
Answer:
+f
Explanation:
because you have to be really dumb to get an -f
Diamagnetic materialsA) have small negative values of magnetic susceptibility.B) are those in which the magnetic moments of all electrons in each atom cancel.C) experience a small induced magnetic moment when placed in an external magnetic field.D) exhibit the property of diamagnetism independently of temperature.E)are described by all
Answer:
C) experience a small induced magnetic moment when placed in an external magnetic field.
Explanation:
Diamagnetics materials are those that experience a small induced magnetic moment when placed in an external magnetic field. These materials, such as bismuth, copper, silver and lead, have elementary magnets in their compositions. When they are exposed to an external magnetic cap, these elemental magnets tend to follow an orientation contrary to the external magnetic field. As a result, a magnetic field is created in the opposite direction to the external magnetic field.
In a shipping yard, a crane operator attaches a cable to a 1,390 kg shipping container and then uses the crane to lift the container vertically at a constant velocity for a distance of 33 m. Determine the amount of work done (in J) by each of the following.
a) the tension in the cable.
b) the force of gravity.
Answer:
a) A = 449526 J, b) 449526 J
Explanation:
In this exercise they do not ask for the work of different elements.
Note that as the box rises at constant speed, the sum of forces is chorus, therefore
T-W = 0
T = W
T = m g
T = 1,390 9.8
T = 13622 N
Now that we have the strength we can use the definition of work
W = F .d
W = f d cos tea
a) In this case the tension is vertical and the movement is vertical, so the tension and displacement are parallel
A = A x
A = 13622 33
A = 449526 J
b) The work of the force of gravity, as the force acts in the opposite direction, the angle tea = 180
W = T x cos 180
W = - 13622 33
W = - 449526 J
with a speed of 75 m sl. Determine
1) A vehicle of mass 1600 kg moves
the magnitude of its momentum.
Answer:
120000 kgxm/s
Explanation:
momentum is mass times velocity so just multiply 1600 kg times 75 m/s and you get 120000 kgxm/s
If an X-ray beam of wavelength 1.4 × 10-10 m makes an angle of 30° with a set of planes in a crystal causing first order constructive interference, what is the plane spacing?
Answer:
Plane spacing, [tex]d=1.4\times 10^{-10}\ m[/tex]
Explanation:
It is given that,
Wavelength of x-ray, [tex]\lambda=1.4\times 10^{-10}\ m[/tex]
Angle the x-ray made with a set of planes in a crystal causing first order constructive interference is 30 degrees
We need to find the plane spacing. It is based on Bragg's law such that,
[tex]2d\sin\theta=n\lambda[/tex]
d is plane spacing
n = 1 here
[tex]d=\dfrac{\lambda}{2\sin\theta}\\\\d=\dfrac{1.4\times 10^{-10}}{2\times \sin (30)}\\\\d=1.4\times 10^{-10}\ m[/tex]
So, the plane spacing is [tex]1.4\times 10^{-10}\ m[/tex].
The inner and outer surface temperature of a glass window 10 mm thick are 25 and 5 degree-C, respectively. What is the heat loss through a 1 m x 3 m window
Answer:
The heat loss is [tex]H = 8400\ W[/tex]
Explanation:
From the question we are told that
The thickness is [tex]t = 10 \ mm = 0.01 \ m[/tex]
The inner temperature is [tex]T_i = 25 ^oC[/tex]
The outer temperature is [tex]T_o = 5 ^oC[/tex]
The length of the window is L = 1 m
The width of the window is w = 3 m
Generally the heat loss is mathematically represented as
[tex]H = \frac{k * A * \Delta T}{t}[/tex]
Where k is the thermal conductivity of glass with value [tex]k = 1.4\ W/m \cdot K[/tex]
and A is the area of the window with value
[tex]A = 1 * 3[/tex]
[tex]A = 3 \ m^2[/tex]
substituting values
[tex]H = \frac{1.4 * 3 * (23-5)}{0.01}[/tex]
[tex]H = 8400\ W[/tex]
The sound intensity at a distance 2.00 m from a sound source is 5.00 Find the total sound energy emitted by the source in each second.
Answer:
P = 251, 3 W
Explanation:
The intensity is defined as the power emitted per unit area
I = P / A
Since sound is distributed in all directions spherical shape, the area of a sphere is
A = 4π r²
let's clear the power and replace
P = I A
P = I (4π r²)
let's calculate
P = 5.00 (4π 2²)
P = 251, 3 W
Light with a frequency of 5.70×10^14 Hz travels in a block of glass that has an index of refraction of 1.56. What is the wavelength of the light in the glass?
✔First you have to calculate the light's speed in the glass,
You know that in the air and in the void (where the refraction index n is zero) the light's speed C corresponds to 3,0 x 10^8 m/s
So We have :
V = C/n
V = 3,0 x 10^8/1,56 V ≈ 1,92 x 10^8 m/s✔ Now, you know the light's speed in glass, and you know that : the wavelength λ is the quotient of light's speed V on its frequency ν, so :
λ = V/ ν
λ = 1,82 x 10^8/5,70 x 10^14 λ ≈ 3.40 x 10^-7 m λ ≈ 340 nmA proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 45000 m/s.
What will be the final speed of an electron released from rest at the negative plate?
Express your answer to two significant figures and include the appropriate units
V=_________
Answer:
2.1x10^6m/s
Explanation:
One electron has a charge of –1.602e-19 C
mass of electron is 9.1e-31 kg
mass of proton is 1.6726e−27 kg
mass ratio is 1.6726e−27 / 9.1e-31 = 1838
The force is constant, F
distance is constant, d
a = F/m
a increases by a factor 1838, as m decreases by that factor
a = a₀1838
v₀² = 2a₀d
v² = 2a₀d1838
v²/v₀² = 2a₀d1838 / 2a₀d = 1838
v² = 1838v₀² = 1838(45000)²
v = 45000√1838 = 2.1e6 m/s
Unlike a roller coaster, the seats in a Ferris wheel swivel so that the rider is always seated upright. An 80-ft-diameter Ferris wheel rotates once every 24 s.What is the apparent weight of a 40 kg passenger at the lowest point of the circle?What is the apparent weight of a 40 kg passenger at the highest point of the circle?
Answer:
a
[tex]F_A =425.42 \ N[/tex]
b
[tex]F_A_H = 358.58 \ N[/tex]
Explanation:
From the question we are told that
The diameter of the Ferris wheel is [tex]d = 80 \ ft = \frac{80}{3.281} = 24.383[/tex]
The period of the Ferris wheel is [tex]T = 24 \ s[/tex]
The mass of the passenger is [tex]m_g = 40 \ kg[/tex]
The apparent weight of the passenger at the lowest point is mathematically represented as
[tex]F_A_L = F_c + W[/tex]
Where [tex]F_c[/tex] is the centripetal force on the passenger, which is mathematically represented as
[tex]F_c =m * r * w^2[/tex]
Where [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{2* \pi }{T}[/tex]
substituting values
[tex]w = \frac{2* 3.142 }{24}[/tex]
[tex]w = 0.2618 \ rad/s[/tex]
and r is the radius which is evaluated as [tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{24.383}{2}[/tex]
[tex]r = 12.19 \ ft[/tex]
So
[tex]F_c = 40 * 12.19* (0.2618)^2[/tex]
[tex]F_c = 33.42 \ N[/tex]
W is the weight which is mathematically represented as
[tex]W = 40 * 9.8[/tex]
[tex]W = 392 \ N[/tex]
So
[tex]F_A = 33.42 + 392[/tex]
[tex]F_A =425.42 \ N[/tex]
The apparent weight of the passenger at the highest point is mathematically represented as
[tex]F_A_H = W- F_c[/tex]
substituting values
[tex]F_A_H = 392 - 33.42[/tex]
[tex]F_A_H = 358.58 \ N[/tex]
Blood is pumped from the heart at a rate of 5.0 L/min into the aorta (of radius 1.0 cm). Determine the speed of blood through the aorta in units of m/s
Answer:
The speed of blood through the aorta is 0.265 m/s
Explanation:
Given;
volumetric flow rate, Q = 5.0L/min = 0.005 m³/min x 1min/60s = 8.333 x 10⁻⁵ m³/s
radius of the aorta, r = 1.0 cm = 0.01 m
Area of the aorta = πr²
Area of the aorta = π(0.01)² = 3.142 x 10⁻⁴ m²
Volumetric flow rate is given by;
Q = Av
where;
v is the speed of blood through the aorta
v = Q /A
v = (8.333 x 10⁻⁵ ) / (3.142 x 10⁻⁴)
v = 0.265 m/s
Therefore, the speed of blood through the aorta is 0.265 m/s
The blood's speed through the aorta will be "0.265 m/s". To understand the calculation, check below.
Blood and AortaAccording to the question,
Volumetric flow rate, Q = 5.0 L/min or,
= 0.005 × [tex]\frac{1 \ min}{60 \ s}[/tex]
= 8.333 × 10⁻⁵ m³/s
Aorta's radius, r = 1.0 cm
We know the formula,
→ Q = A × v
or,
Speed, v = [tex]\frac{Q}{A}[/tex]
By substituting the values,
= [tex]\frac{8.333\times 10^{-5}}{3.142\times 10^{-4}}[/tex]
= 0.265 m/s
Thus the above answer is correct.
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3) If a ball launched at an angle of 10.0 degrees above horizontal from an initial height of 1.50 meters has a final horizontal displacement of 3.00 meters, what is its launch velocity
Answer:
35.6 m
Explanation:
The given ball possesses a projectile motion from its initial height. So, the required launch velocity of the ball is 6.55 m/s.
What is launch velocity?The horizontal component of velocity during the projection of an object is known as launch velocity. It is obtained when the horizontal range is known.
Given data -
The angle of projection is, [tex]\theta = 10.0 {^\circ}[/tex].
The initial height of the projection is, h = 1.50 m.
The horizontal displacement is, R = 3.00 m.
The mathematical expression for the horizontal displacement (Range) of the projectile is given as,
[tex]R = \dfrac{u^{2} \times sin2 \theta}{g}[/tex]
here,
u is the launch velocity.
g is the gravitational acceleration.
Solving as,
[tex]u =\sqrt{\dfrac{R \times g}{sin2 \theta}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin(2 \times 10)}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin20^\circ}}\\\\\\u=6.55 \;\rm m/s[/tex]
Thus, we can conclude that the required launch velocity of the ball is 6.55 m/s.
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A man is at a car dealership, looking for a car to buy. He looks at the sticker on the driver’s window of a car and sees that the price of the car is $32,540. Why is the long shadow of scarcity visible at the car dealership? Check all that apply.
Answer:
he will see the sticker because its behind a window bruh and thats a big daddy stack of greens
Explanation:
A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down
Answer:
The angular acceleration is [tex]\alpha = 0.4418 \ rad /s^2[/tex]
Explanation:
From the question we are told that
The angular speed is [tex]w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s[/tex]
The angular displacement is [tex]\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad[/tex]
From the first equation of motion we can define the movement of the record as
[tex]w_f ^2 = w_o ^2 + 2 * \alpha * \theta[/tex]
Given that the record started from rest [tex]w_o = 0[/tex]
So
[tex]4.713^2 = 2 * \alpha * 25.14[/tex]
[tex]\alpha = 0.4418 \ rad /s^2[/tex]