Answer:
a) θ = 14.23º, b) θ₂ = 75.77, c) t = 0.6019 s
Explanation:
This is a missile throwing exercise.
a) the reach of the ball is the distance traveled for the same departure height
R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]
sin 2θ = [tex]\frac{Rg}{v_o^2}[/tex]
sin 2θ = 7.00 9.8 / 12.0²
2θ = sin⁻¹ (0.476389) = 28.45º
θ = 14.23º
the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º
θ ’= 90 -14.23
θ’= 75.77º
b) the two angles that give the same range are
θ₁ = 14.23
θ₂ = 75.77
the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.
C) the time of the pass can be calculated with the expression
x = v₀ₓ t
t = x / v₀ₓ
t = 7 / 11.63
t = 0.6019 s
The standard unit of brightness is called the candela.
True
False
Answer:
TRUE
Explanation:
Electromagnetic radiation with a wavelength of 525 nm appears as green light to the human eye. Calculate the frequency of this light. Be sure to include units in your answer.
Answer:
5.71×10¹⁴ Hz
Explanation:
Applying,
v = λf................. Equation 1
Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency
make f the subject of the equation
f = v/λ............. Equation 2
From the question,
Given: λ = 525 nm = 5.25×10⁻⁷ m,
Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s
Substitute these values into equation 2
f = (3.0×10⁸)/(5.25×10⁻⁷)
f = 5.71×10¹⁴ Hz
Hence the frequency of light is 5.71×10¹⁴ Hz
De que esta hecho el sol? plisss ayuda.no necesito un texto de 100 reglones, puede ser resumido en solo 2 renglones
What is the de Broglie wavelength of a red blood cell with a mass of 1.00 * 10-11 g that is moving with a speed of 0.400 cm/s? Do we need to be concerned with the wave nature of the blood cells when we describe the flow of blood in the body?
Answer:
The wavelength is "[tex]=16.5675\times 10^{-18} \ m[/tex]".
Explanation:
Given:
Mass,
m = [tex]1\times 10^{-11} \ g[/tex]
Speed,
V = [tex]0.400 \ cm/s[/tex]
or,
= [tex]0.4\times 10^{-2}[/tex]
According to De Broglie,
The wavelength will be:
⇒ [tex]\lambda = \frac{h}{mV}[/tex]
[tex]=\frac{6.627\times 10^{-34}}{1\times 10^{-11}\times 10^{-3}\times 0.4\times 10^{-2}}[/tex]
[tex]=16.5675\times 10^{-18} \ m[/tex]
So, blood cells move these wavelength.
Se lanza un cohete en un ángulo de 53° sobre la horizontal con una rapidez inicial de 100 m/s. El cohete se mueve por
3.00 s a lo largo de su línea inicial de movimiento con una aceleración de 30.0 m/s2
. En este momento, sus motores fallan,
y el cohete procede a moverse como un proyectil. Determine: (a) la altitud máxima que alcanza el cohete, (b) su tiempo
total de vuelo y (c) su alcance horizontal
Answer:
Explanation:
v = u + at
v₃ = 100 +30.0(3.00) = 190 m/s
s = vt + ½at²
y₃ = (100sin53)(3.00) + ½(30sin53)(3.00²) = 347.4 m
x₃ = (100cos53)(3.00) + ½(30cos53)(3.00²) = 261.8 m
a) v² = u² + uas
s = (v² - u²) / 2a
ymax = 347.4 + (0² - (190sin53)²) / (2(-9.80)) = 1,522 m
b) t₁ = 3.00 s
t₂ = (190sin53) / 9.80 = 15.5 s
t₃ = √(2(1522) / 9.80) = 17.6 s
t = 3.00 + 15.5 + 17.6 = 36.1 s
c) xmax = 261.8 + (190cos53)( 15.5 + 17.6) = 4,047 m
why did Rita's hands get hot when she rubbed them ?
Answer:
due to production of heat through friction
Explanation:
because of the friction produce between her hands
Use the DC Construction kit to build a simple circuit to perform the following task:
You are asked to use a single resistor and a 110 V DC battery for the purpose of boiling a litter of water (4,184 Joule/Kg*degree Celsius), with a starting temperature of 20 C, in exactly 4 minutes.
Answer:
The resistance is 8.7 ohm.
Explanation:
Voltage, V = 110 V
mass, m = 1 kg
change in temperature, T = 100 - 20 = 80 C
time, t = 4 min = 4 x 60 = 240 s
specific heat, c = 4184 J/kg C
let the resistance is R.
The heat generated by the heater is used to the heat the water.
[tex]\frac{V^2}{R} t = m c T \\\\\frac{110^2}{R}\times 240 = 1\times 4184\times 80\\\\R = 8.7 ohm[/tex]
Accommodation of the eye refers to its ability to __________. see on both the brightest days and in the dimmest light see both in air and while under water move in the eye socket to look in different directions focus on both nearby and distant objects
Answer:
to adjust from distant to the near objects
Explanation:
The process of accommodation is achieved by changing in the shape and position of the eye ball. Just like adjusting the lens of the camera.Answer:
The ability of eye lens to change the focal length of eye lens is called accommodation power of eye.
Explanation:
The human eye is the optical instrument which works on the refraction of light.
The ability of eye lens to change its focal length is called accommodation power of eye.
The focal length of eye lens is changed by the action of ciliary muscles.
When the ciliary muscles are relaxed then the thickness of lens is more and thus the focal length is small. When the ciliary muscles is stretched, the lens is thin and then the focal length is large.
state the laws of reflection
Answer:
Explanation:
The law of reflection says that the reflected angle (measured from a vertical line to the surface called the normal) is equal to the reflected angle measured from the same normal line.
All other properties of reflection flow from this one statement.
Question 5 of 10
What must be the same for two resistors that are connected in parallel?
Answer:
in parallel combination : potential difference between two terminal of resistors are always constant. ... hence, potential difference ( voltage ) must be same across each resistor .
Explanation:
3. Some guitarists like the feel of a set of strings that all have the same tension. For such a guitar, the G string (196 Hz) has a mass density of 0.31 g/m. What is the mass density of the A string (110 Hz)
Answer:
0.98 g/m
Explanation:
Note: Since Tension and frequency are constant,
Applying,
F₁²M₁ = F₂²M₂............... Equation 1
Where F₁ = Frequency of the G string, F₂ = Frequency of the A string, M₁ = mass density of the G string, M₂ = mass density of the A string.
make M₂ the subject of the equation
M₂ = F₁²M₁/F₂²............... Equation 2
From the question,
Given: F₁ = 196 Hz, M₁ = 0.31 g/m, F₂ = 110 Hz
Substitute these values into equation 2
M₂ = 196²(0.31)/110²
M₂ = 0.98 g/m
CHEGG Over the course of a multi-stage 4820-km bicycle race, the front wheel of an athlete's bicycle makes 2.40x106 revolutions. How many revolutions would the wheel have made during the race if its radius had been 1.4 cm larger?
Answer:
θ' = 14.44 × [tex]10^{6}[/tex]
Explanation:
given data
total distance is d = 4820
radius = 1.4 cm
solution
we get here total angle by which the wheel rotates traveling is express as
⇒ [tex]\theta=2.40\times10^6\ \rm{rev}=2.40\times 2\pi\times10^6\ \rm{rad}[/tex] ................1
and
total angle (θ) and the total distance (d) express as
⇒ d = r × θ ...............2
here r is radius
and here rotated through some other angle θ' so put value in given equation and find revolutions
⇒ d = (r+r)θ' ........3
here r = d/θ
so
⇒ [tex]d = ( \frac{d}{\theta}+r) \theta'[/tex]
so put value and get θ'
⇒ θ' = 2.40 × 2π × [tex]10^{6}[/tex] × [tex]\frac{4820 \times 10^3}{4820 \times 10^3 +0.014 \times 2.40 \times 2 \times \pi \times 10^6}[/tex]
⇒ θ' = 14.44 × [tex]10^{6}[/tex] revAt room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder arrive 7.5 seconds later. How many miles away was the lighting strike? (Assume the light takes essentially no time to reach you).
Answer:
1.6031 miles
Explanation:
Given the following data;
Speed = 344 m/s
Time = 7.5 seconds
To find how many miles away was the lighting strike;
Mathematically, the distance travelled by an object is calculated by using the formula;
Distance = speed * time
Distance = 344 * 7.5
Distance = 2580 meters
Next, we would have to convert the value of the distance travelled in meters to miles;
Conversion:
1609.344 metres = 1 mile
2580 meters = X mile
Cross-multiplying, we have;
X * 1609.344 = 2580
X = 2580/1609.344
X = 1.6031 miles
In a single-slit diffraction pattern, the central fringe is 360 times as wide as the slit. The screen is 14,000 times farther from the slit than the slit is wide. What is the ratio /W, where is the wavelength of the light shining through the slit and W is the width of the slit
Answer:
0.01286
Explanation:
In a given single-slit, the central fringe (Y) is 360 times as wide as the slit (a). Then
2Y₁ = 360a
Y₁ = 360a/2
= 180a
The distance D = 14000a
In a given single-slit diffraction, the ratio = [tex]\dfrac{\lambda }{W}[/tex]
and since the angle is infinitesimally small;
sin θ ≅ tan θ = [tex]\dfrac{Y}{D}[/tex]
∴
For the first dark fringe;
Suppose: [tex]\dfrac{a}{2}sin \theta = \dfrac{\lambda }{2}[/tex]
then,
[tex]\dfrac{a}{2} \ \dfrac{Y_1}{D} = \dfrac{\lambda }{2}[/tex]
[tex]aY_1 = \lambda D[/tex]
[tex]\dfrac{\lambda }{a} = \dfrac{Y_1}{D}\\ \\ \\ \implies \dfrac{180\ a}{14000 \ a} \\ \\ \mathbf{\dfrac{\lambda }{a} = 0.01286}[/tex]
) The velocity function is v(t)=−t2+3t−2v(t)=−t2+3t−2 for a particle moving along a line. Find the displacement (net distance covered) of the particle during the time interval [−2,5][−2,5].
Answer:
89.87m/s
Explanation:
Given the velocity function
v(t)=−t²+3t−2
In order to get the displacement function, we will integrate the velocity function as shown:
[tex]\int\limits^5_{-2} {v(t)} \, dt \\d(t)= \int\limits^5_{-2}{(-t^2+3t+2)} \, dt \\\\d(t)=[\frac{-t^3}{3}+\frac{3t^2}{2}+2t ]^5_{-2}\\[/tex]
at t = 5
[tex]d(5)=[\frac{-5^3}{3}+\frac{3(5)^2}{2}+2(5) ]\\d(5)=[\frac{-125}{3}+\frac{75}{2}+10 ]\\d(5)=-41.7+37.5+10\\d(5)=89.2m/s[/tex]
at t = -2
[tex]d(-2)=[\frac{-(-2)^3}{3}+\frac{3(-2)^2}{2}+2(-2) ]\\d(-2)=[\frac{-8}{3}+\frac{12}{2}+(-4) ]\\d(-2)=-2.67+6-4\\d(-2)=-0.67m/s[/tex]
Required displacement = d(5) - d(-2)
Required displacement = 89.2 - (-0.67)
Required displacement = 89.2 + 0.67
Required displacement = 89.87m/s
An electron is released from rest at a distance of 9.00 cm from a fixed proton. How fast will the electron be moving when it is 3.00 cm from the proton
Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]
Apply the principle of conservation of energy;
ΔK.E = ΔU
[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]
Therefore, the speed of the electron at the given position is 106.2 m/s
An airplane propeller is 2.16 m in length (from tip to tip) with mass 100 kg and is rotating at 2900 rpm (rev/min) about an axis through its center. You can model the propeller as a slender rod. What is its rotational kinetic energy? Suppose that, due to weight constraints, you had to reduce the propeller's mass to 75.0% of its original mass, but you still needed to keep the same size and kinetic energy. What would its angular speed have to be, in rpm?
a) The rotational kinetic energy of the airplane propeller is 1792152.287 joules.
b) The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute.
How to determine the angular speed of a airplane propeller
Let consider the airplane propeller a rigid body, the rotational kinetic energy of the propeller (K), in joules, is described by the following formula:
K = 0.5 · I · ω² (1)
Where:
I - Moment of inertia of the airplane propeller, in kilogram-square meters.ω - Angular speed, in radians per secondIn addition, the moment of inertia of a slender rod rotating around its center is:
I = 0.0833 · M · L² (2)
Where:
M - Mass of the propeller, in kilogramsL - Length of the propeller, in metersa) If we know that M = 100 kg, L = 2.16 m and ω = 303.687 rad/s, then the rotational kinetic energy of the propeller is:
K = 0.5 · [0.0833 · (100 kg) · (2.16 m)²] · (303.687 rad/s)²
K = 1792152.287 J
The rotational kinetic energy of the airplane propeller is 1792152.287 joules. [tex]\blacksquare[/tex]
b) By (1) and (2) we know that the mass of the propeller is inversely proportional to the square of the angular speed. Therefore, we have the following relationship:
[tex]M_{o}\cdot \omega_{o}^{2} = M_{f}\cdot \omega_{f}^{2}[/tex]
[tex]\omega_{f} = \sqrt{\frac{M_{o}}{M_{f}} }\cdot \omega_{o}[/tex] (3)
If we know that [tex]\omega_{o} = 2900\,\frac{rev}{min}[/tex], [tex]M_{o} = 100\,kg[/tex] and [tex]M_{f} = 75\,kg[/tex], then the angular speed of the airplane propeller is:
[tex]\omega_{f} = \left(2900\,\frac{rev}{min} \right)\cdot \sqrt{\frac{100\,kg}{75\,kg} }[/tex]
[tex]\omega_{f} \approx 3348.631\, \frac{rev}{min}[/tex]
The angular speed of the airplane propeller is approximately 3348.631 revolutions per minute. [tex]\blacksquare[/tex]
To learn more on rotational kinetic energy, we kindly invite to check this verified question: https://brainly.com/question/20261989
The primary purpose of a switch in a circuit is to ___________.
A)either open or close a conductive path
B)change a circuit from parallel to series
C)change a circuit from series to parallel
D)store a charge for later use
Answer:
store a charge for later use
A 40 kg boy is standing on the edge of a stationary 30-kg platform that is free to rotate without friction. The boy tries to walk around the platform in a counterclockwise direction. As he does:
a. the platform doesn't rotate.
b. the platform rotates in a clockwise direction just fast enough so that the boy remains stationary relative to the ground.
c. the platform rotates in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground.
d. both go around with equal angular velocities but in opposite directions
Answer:
the correct one is C
Explanation:
To find the answer, let's propose the solution of the problem
We create a system formed by the child and the platform so that all the forces have been internal and the angular momentum is conserved.
Initial instant. Before starting to walk
L₀ = 0
Final moment. After the child is walking
L_f = I₁ w₁ + m r v₂
where index 1 is used for the platform and index 2 for the child
linear and angular velocity are related
v₂ = w₂ r
angular momentum is conserved
0 = I₁ w₁ + m r (w₂ r)
w₁ = [tex]- \frac{m r^2}{I1} \ w_2[/tex]
the moment of inertia of the platform bringing it closer to a disk or cylinder
I₁ = [tex]\frac{1}{2}[/tex] M r²
sustitute
w₁ = [tex]- \frac{2 m }{M} \ w_2[/tex]
W₁ = - [tex]- \frac{2 40}{30} \ w_2 = - \frac{8}{3} \ w_2[/tex]
from here we can see that the platform and the child rotate in the opposite direction and with different angular speeds
when examining the answers the correct one is C
Answer:
Option C (the platform rotates in a clockwise direction while the boy goes around in a counterclockwise direction relative to the ground)Explanation:
relative to the ground the boy moves in a counter clockwise motion , now the boy and the wheel are one system
so by conservation of angular momentum their net sum of angular momentum relative to a point outside the system(say ground) should be zero
so the wheel moves in a clockwise direction , their angular velocity may or may not be same depending on I. so option D is wrong
option B is wrong because relative to ground their angular momentum should be equal and opposite
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Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.00 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound.
Required:
What is the lowest possible frequency of sound for which this is possible?
Answer:
The lowest possible frequency of sound for which this is possible is 1307.69 Hz
Explanation:
From the question, Abby is standing 5.00m in front of one of the speakers, perpendicular to the line joining the speakers.
First, we will determine his distance from the second speaker using the Pythagorean theorem
l₂ = √(2.00²+5.00²)
l₂ = √4+25
l₂ = √29
l₂ = 5.39 m
Hence, the path difference is
ΔL = l₂ - l₁
ΔL = 5.39 m - 5.00 m
ΔL = 0.39 m
From the formula for destructive interference
ΔL = (n+1/2)λ
where n is any integer and λ is the wavelength
n = 1 in this case, the lowest possible frequency corresponds to the largest wavelength, which corresponds to the smallest value of n.
Then,
0.39 = (1+ 1/2)λ
0.39 = (3/2)λ
0.39 = 1.5λ
∴ λ = 0.39/1.5
λ = 0.26 m
From
v = fλ
f = v/λ
f = 340 / 0.26
f = 1307.69 Hz
Hence, the lowest possible frequency of sound for which this is possible is 1307.69 Hz.
Two masses of 3 kg and 5 kg are connected by a light string that passes over a smooth polley as shown in the Figure.
QL
Determine:
i. the tension in the string,
ii. the acceleration of each mass, and
iii. the distance each mass moves in the first second of motion if they start from rest
i. [tex]T = 36.8\:\text{N}[/tex]
ii. [tex]a = 2.45\:\text{m/s}^2[/tex]
iii. [tex]x = 1.23\:\text{m}[/tex]
Explanation:
Let's write Newton's 2nd law for each object. We will use the sign convention assigned for each as indicated in the figure. Let T be the tension on the string and assume that the string is inextensible so that the two tensions on the strings are equal. Also, let a be the acceleration of the two masses. And [tex]m_1 = 3\:\text{kg}[/tex] and [tex]m_2 = 5\:\text{kg}[/tex]
Forces acting on m1:
[tex]T - m_1g = m_1a\:\:\:\:\:\:\:(1)[/tex]
Forces acting on m2:
[tex]m_2g - T = m_2a\:\:\:\:\:\:\:(2)[/tex]
Combining Eqn(1) and Eqn(2) together, the tensions will cancel out, giving us
[tex]m_2g - m_1g = m_2a + m_1a[/tex]
or
[tex](m_2 - m_1)g = (m2 + m_1)a[/tex]
Solving for a,
[tex]a = \left(\dfrac{m_2 - m_1}{m_2 + m_1}\right)g[/tex]
[tex]\:\:\:\:= \left(\dfrac{5\:\text{kg} - 3\:\text{kg}}{5\:\text{kg} + 3\:\text{kg}}\right)(9.8\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:= 2.45\:\text{m/s}^2[/tex]
We can solve for the tension by using this value of acceleration on either Eqn(1) or Eqn(2). Let's use Eqn(1).
[tex]T - (3\:\text{kg})(9.8\:\text{m/s}^2) = (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]
[tex]T = (3\:\text{kg})(9.8\:\text{m/s}^2) + (3\:\text{kg})(2.45\:\text{m/s}^2)[/tex]
[tex]\:\:\:\:= 29.4\:\text{m/s}^2 + 7.35\:\text{m/s}^2 = 36.8\:\text{N}[/tex]
Assuming that the two objects start from rest, the distance that they travel after one second is given by
[tex]x = \frac{1}{2}at^2 = \frac{1}{2}(2.45\:\text{m/s}^2)(1\:\text{s})^2 = 1.23\:\text{m}[/tex]
find the upward force in Newton when each of these is under water(density of 1g/cm3) a lump of iron of volume 2000cm3
Answer:
Upthrust = 19.6 N
Explanation:
When an object is immersed under water, the upward force it experience is called an upthrust. An upthrust is a force which is applied on any object in a fluid which acts in an opposite direction to the direction of the weight of the object.
Upthrust = density of liquid x gravitational force x volume of object
i.e U = ρ x g x vol
Given: ρ = 1g/[tex]cm^{3}[/tex] (1000 kg/[tex]m^{3}[/tex]), volume = 2000 c[tex]m^{3}[/tex] (0.002 [tex]m^{3}[/tex]) and g = 9.8 m/[tex]s^{2}[/tex]
So that;
U = 1000 x 9.8 x 0.002 (kg/[tex]m^{3}[/tex] x [tex]m^{3}[/tex] x m/[tex]s^{2}[/tex])
= 19.6 Kg m/[tex]s^{2}[/tex]
U = 19.6 Newtons
The upthrust on the iron is 19.6 N.
In a rolling race, two objects are released from the top of two identical ramps. They then roll without slipping to the bottom of the ramp. If the two objects are 2 hoops of the same radius but different masses, which reaches the bottom first?
a. The lighter one reaches the bottom first
b. The heavier one reaches the bottom first
c. We don’t have enough information
d. They reach the bottom at the same time
Answer:
b. The heavier one reaches the bottom first.
Answer:
B
Explanation:
The answer is B the heavier item has more g force pushing it making it roll faster reaching the bottom of the ramp first.
Which technological device makes an energy conversion in the same way that a human ear makes an energy conversion?
a.) a loudspeaker
b.) a headphone
c.) a light bulb
d.) a microphone
I think it's c because of the concept of mechanical energy to electrical energy but I'm not sure
Answer:
I THINK C
Explanation:
BECAUSE A Light Emitting Diode (LED) glows even when a weak electric current passes through it.
A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and we can model the situation to see why. Assume that the man has a mass of 90 kg, with a center of gravity 1.0 m above the ground. The action of the wind on his torso, which we approximate as a cylinder 50 cm wide and 90 cm long centered 1.2 m above the ground, produces a force that tries to tip him over backward. To keep from falling over, he must lean forward.
A. What is the magnitude of the torque provided by the wind force? Take the pivot point at his feet. Assume that he is standing vertically. Assume that the air is at standard temperature and pressure.
B. At what angle to the vertical must the man lean to provide a gravitational torque that is equal to this torque due to the wind force?
Answer:
a) [tex]t=195.948N.m[/tex]
b) [tex]\phi=13.6 \textdegree[/tex]
Explanation:
From the question we are told that:
Density [tex]\rho=1.225kg/m^2[/tex]
Velocity of wind [tex]v=14m/s[/tex]
Dimension of rectangle:50 cm wide and 90 cm
Drag coefficient [tex]\mu=2.05[/tex]
a)
Generally the equation for Force is mathematically given by
[tex]F=\frac{1}{2}\muA\rhov^2[/tex]
[tex]F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2[/tex]
[tex]F=163.29[/tex]
Therefore Torque
[tex]t=F*r*sin\theta[/tex]
[tex]t=163.29*1.2*sin90[/tex]
[tex]t=195.948N.m[/tex]
b)
Generally the equation for torque due to weight is mathematically given by
[tex]t=d*Mg*sin90[/tex]
Where
[tex]d=sin \phi[/tex]
Therefore
[tex]t=sin \phi*Mg*sin90[/tex]
[tex]195.948=833sin \phi[/tex]
[tex]\phi=sin^{-1}\frac{195.948}{833}[/tex]
[tex]\phi=13.6 \textdegree[/tex]
what aspect of the US justice system has its roots in Jewish scripture?
The aspect of the US justice system that has its roots in Jewish scripture is:
the idea that all people are subject to the same rules and laws.
It is the doctrine of "equality before the law." Equality before the law means that every individual is equal in the eyes of the law, whether the individual is a lawmaker, a judge, a law enforcement officer, etc. Equality before the law is also known as equality under the law, equality in the eyes of the law, legal equality, or legal egalitarianism. It is a legal principle that treats each independent being equally and subjects each to the same laws of justice and due process.
Answer:
answer is C
the idea that all people are subject to the same rules and laws
Explanation:
hope this helps!
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.
Learn more about the equation of motion, here:
brainly.com/question/16982759
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Action and Reaction are equal in magnitude and opposite in direction.Then,Why do they not balance each other?
Answer:
Newton's third law of motion states that every action has an equal and opposite reaction. This indicates that forces always act in pairs. Reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don't cancel each other out.
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]
what do we mean by thrust?
Answer:
the answer is push example: she thrust her hand into her pocket
