Answer:
Models change to accommodate new discoveries
Explanation:
Owing to the dynamism of science, scientific models also change to accommodate new ideas and discoveries. Science is not a static body of knowledge. Research consistently brings new information to the fore. This propels a change in certain scientific models to accommodate the new discoveries and present a more accurate representation of phenomena in nature.
Hence, scientific models change to accommodate new discoveries.
Answer: Models change to accommodate new discoveries.
Explanation: I took the test
Juliette sets the initial velocity to +10 m/s and the acceleration to −2 m/s2, then clicks "start". How can Shakina describe the subsequent motion of the car?
Answer:
Shakina can sayThe car moves initially to the right with decreasing speed. Eventually, the car stopsand moves to the left with increasing speed.
Explanation:
Because initially the acceleration is negative during which the car is moving to the right and then it slows down and stops, and once it begins moving to the left, it begins to accelerate faster
A beam of gamma rays is incident on a block of lead. The intensity of gamma rays passing through a uniform medium varies with depth x, in a manner given by the expression I(x) = 1, e-ux Here, I, is the intensity of the radiation at the surface of the material (at x = 0) and u is called the linear absorption coefficient. A) Find an expression for the thickness of material that would absorb half of the gamma rays. B) Find an expression for the thickness of material that would reduce the radiation intensity to a fraction of the initial intensity.
Answer:
A. x = -㏑(1/2)/μ B. x = -㏑(I/I₀)/μ
Explanation:
A. Since the intensity I = I₀exp(-μx) where I₀ = intensity at x = 0.
When I = I₀/2,
I = I₀exp(-μx)
I₀/2 = I₀exp(-μx)
dividing through by I₀, we have
1/2 = exp(-μx)
taking natural logarithm of both sides, we have
㏑(1/2) = ㏑[exp(-μx)]
㏑(1/2) = -μx
dividing both sides by -μ
x = -㏑(1/2)/μ
where x is the thickness of the material that would absorb half of the gamma rays
B. Since the intensity I = I₀exp(-μx) at thickness x, where I₀ = intensity at x = 0
I = I₀exp(-μx)
dividing through by I₀
I/I₀ = exp(-μx)
taking natural logarithm of both sides, we have
㏑(I/I₀) = ㏑exp(-μx)
㏑(I/I₀) = -μx
dividing both sides by -μ
x = -㏑(I/I₀)/μ
where x is the thickness of the material that would reduce the radiation intensity to a fraction of the initial intensity.
A drone is launched with a velocity of 63 m/s, 29 . Three minutes after the drone is launched it suddenly changed its course to 22 m/s, 233 to avoid high trees on its way. Then, it finally goes back at a rate of 98 m/s, 321. Calculate for the total resultant vector of the drone.
Answer:
v = 127.66 m / s θ’= 337.59
Explanation:
For this exercise we must use the speed composition of the drone.
The first speed is 63 m / s in direction 29, three minutes later the speed reaches 22 m / s and direction 233, finally it returns to the launch point with 98 m / s in direction 321, in the attachment you can see a diagram of these speeds.
To find the resulting average velocity, the easiest thing is to decompose each velocity into the x and y coordinate system, then add each velocity
let's break down the speeds
cos 29 = v₁ₓ / v₁
sin 29 = [tex]v_{1y}[/tex] / v₁
v₁ₓ = v₁ cos 29
v_{1y} = v₁ sin 29
v₁ₓ = 63 cos 29 = 55.10 m / s
v_{1y} = 63 sin 29 = 30.54 m / s
speed 2
cos 22 = v₂ₓ / v₂
sin 22 = v_{2y} / v₂
v₂ₓ = v₂ cos 233
v_{2y} = v₂ sin 233
v₂ₓ = 22 cos 233 = -13.24 m / s
v_{2y} = 22 sin 233 = -17.57 m / s
speed 3
cos 321 = v₃ₓ / v₃
sin 321 = v_{3y} / v₃
v₃ₓ = v₃ cos 321
v_{1y} = vₐ sin 321
v₃ₓ = 98 cos 321 = 76.16 m / s
v_{3y} = 98 sin 321 = -61.67 m / s
We already have all the component of the speeds, the resulting speed is
vₓ = v₁ₓ + v₂ₓ + v₃ₓ
vₓ = 55.10 -13.24 +76.16
vₓ = 118.02 m / s
v_{y} = v_{1y} + v_{2y} + v_{3y}
v_{y} = 30.54 -17.54 - 61.67
v_{y} = -48.67 m / s
there are two ways to give the result
v = (118.02 i -48.67 j) m / s
or in the form of magnitud and angle.
We use the Pythagorean theorem for the module
v = √ (vₓ² + v_{y}²)
v = RA (118.02² + 48.67²)
v = 127.66 m / s
let's use trigonometry for the angle
tan θ = v_{y} / vₓ
θ = tan⁻¹ (v_{1} / vₓ)
θ = tan⁻¹ (-48.67 / 118.02)
θ = -22.41
if we want to measure the angles with respect to the positive side of the x axis
θ’= 360 - 22.41
θ’= 337.59
Tell uses of cancave mirror and convex mirror.
Answer:
Uses of concave mirror:
Shaving mirrors.
Head mirrors.
Ophthalmoscope.
Astronomical telescopes.
Headlights.
Solar furnaces.
Uses of convex mirror:
Convex mirrors always form images that are upright, virtual, and smaller than the actual object. They are commonly used as rear and side view mirrors in cars and as security mirrors in public buildings because they allow you to see a wider view than flat or concave mirrors.
please give me full points.
Work: If the force on an object is in the negative direction, the work it does on the object must be
Answer:
The work must be negative or positive depending on direction of motion of the object
Explanation:
This is because
W= force x distance
And = F x S cosစ
And this is positive when
Theta is less than π/2 and negative when theta is less than or equal to π/2
If a force acts in a specific direction on a thing, the work it puts on it can be positive or negative, regardless of the direction the object moves.
Define force.A force is indeed an influence which can alter an angular velocity. An object can alter its velocity, or speed, as a result of a force. Naturally, force can be characterized as a push or just a pull.
If a force acts in a specific direction on a thing, the work it puts on it can be positive or negative, regardless of the direction the object moves.
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A 0.40 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.5 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 63 N, what is the maximum speed the ball can have?
Answer:
The maximum speed the ball can have is 15.37 m/s
Explanation:
Given;
mass of the ball, m = 0.4 kg
radius of the chord, r = 1.5 m
maximum tension on the chord, T = 63 N
The maximum tension on the chord is given by;
[tex]T_{max} = \frac{mv_{max}^2}{r} \\\\v_{max}^2 = \frac{T_{max} *r}{m}\\\\ v_{max} = \sqrt{\frac{T_{max} *r}{m}} \\\\ v_{max} =\sqrt{\frac{63 *1.5}{0.4}}\\\\v_{max} = 15.37 \ m/s[/tex]
Therefore, the maximum speed the ball can have is 15.37 m/s
The force on an object is F⃗ =−17j⃗ . For the vector v⃗ =2i⃗ +3j⃗ , find: (a) The component of F⃗ parallel to v⃗
Answer:
(a) [tex]\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{103}{13}\,j[/tex] , (b) [tex]\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j[/tex], (c) [tex]W = -51[/tex]
Explanation:
The statement is incomplete:
The force on an object is [tex]\vec F = -17\,j[/tex]. For the vector [tex]\vec v = 2\,i +3\,j[/tex]. Find: (a) The component of [tex]\vec F[/tex] parallel to [tex]\vec v[/tex], (b) The component of [tex]\vec F[/tex] perpendicular to [tex]\vec v[/tex], and (c) The work [tex]W[/tex], done by force [tex]\vec F[/tex] through displacement [tex]\vec v[/tex].
(a) The component of [tex]\vec F[/tex] parallel to [tex]\vec v[/tex] is determined by the following expression:
[tex]\vec F_{\parallel} = (\vec F \bullet \hat {v} )\cdot \hat{v}[/tex]
Where [tex]\hat{v}[/tex] is the unit vector of [tex]\vec v[/tex], which is determined by the following expression:
[tex]\hat{v} = \frac{\vec v}{\|\vec v \|}[/tex]
Where [tex]\|\vec v\|[/tex] is the norm of [tex]\vec v[/tex], whose value can be found by Pythagorean Theorem.
Then, if [tex]\vec F = -17\,j[/tex] and [tex]\vec v = 2\,i +3\,j[/tex], then:
[tex]\|\vec v\| =\sqrt{2^{2}+3^{3}}[/tex]
[tex]\|\vec v\|=\sqrt{13}[/tex]
[tex]\hat{v} = \frac{1}{\sqrt{13}} \cdot(2\,i + 3\,j)[/tex]
[tex]\hat{v} = \frac{2}{\sqrt{13}}\,i+ \frac{3}{\sqrt{13}}\,j[/tex]
[tex]\vec F \bullet \hat{v} = (0)\cdot \left(\frac{2}{\sqrt{13}} \right)+(-17)\cdot \left(\frac{3}{\sqrt{13}} \right)[/tex]
[tex]\vec F \bullet \hat{v} = -\frac{51}{\sqrt{13}}[/tex]
[tex]\vec F_{\parallel} = \left(-\frac{51}{\sqrt{13}} \right)\cdot \left(\frac{2}{\sqrt{13}}\,i+\frac{3}{\sqrt{13}}\,j \right)[/tex]
[tex]\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j[/tex]
(b) Parallel and perpendicular components are orthogonal to each other and the perpendicular component can be found by using the following vectorial subtraction:
[tex]\vec F_{\perp} = \vec F - \vec F_{\parallel}[/tex]
Given that [tex]\vec F = -17\,j[/tex] and [tex]\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j[/tex], the component of [tex]\vec F[/tex] perpendicular to [tex]\vec v[/tex] is:
[tex]\vec F_{\perp} = -17\,j -\left(-\frac{102}{13}\,i-\frac{153}{13}\,j \right)[/tex]
[tex]\vec F_{\perp} = \frac{102}{13}\,i + \left(\frac{153}{13}-17 \right)\,j[/tex]
[tex]\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j[/tex]
(c) The work done by [tex]\vec F[/tex] through displacement [tex]\vec v[/tex] is:
[tex]W = \vec F \bullet \vec v[/tex]
[tex]W = (0)\cdot (2)+(-17)\cdot (3)[/tex]
[tex]W = -51[/tex]
An electron in a region where there is an electric field experiences a force of magnitude 3.1 10-16 N. What is the magnitude of the electric field at the location of the electron?
Answer:
1.938*10^3N/C
Explanation:
Using F= qE
E= F/q
Where q= 1.6x10-19
So 3.1*10^-16/1.6*10^-19
= 1.938*10^3N/C
Explanation:
You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.0dB . So you decide to move closer to give the conversation a sound level of 80.0dB instead. How close should you come?
Answer:
The distance is [tex]r_2 = 0.24 \ m[/tex]
Explanation:
From the question we are told that
The distance from the conversation is [tex] r_1 = 24.0 \ m[/tex]
The intensity of the sound at your position is [tex]\beta _1 = 40 dB[/tex]
The intensity at the sound at the new position is [tex]\beta_2 = 80.0dB[/tex]
Generally the intensity in decibel is is mathematically represented as
[tex]\beta = 10dB log_{10}[\frac{d}{d_o} ][/tex]
The intensity is also mathematically represented as
[tex]d = \frac{P}{A}[/tex]
So
[tex]\beta = 10dB * log_{10}[\frac{P}{A* d_o} ][/tex]
=> [tex]\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ][/tex]
From the logarithm definition
=> [tex]\frac{P}{A * d_o} = 10^{\frac{\beta}{10} }[/tex]
=> [tex]P = A (d_o ) [10^{\frac{\beta }{ 10} } ][/tex]
Here P is the power of the sound wave
and A is the cross-sectional area of the sound wave which is generally in spherical form
Now the power of the sound wave at the first position is mathematically represented as
[tex]P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ][/tex]
Now the power of the sound wave at the second position is mathematically represented as
[tex]P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ][/tex]
Generally power of the wave is constant at both positions so
[tex]A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ][/tex]
[tex]4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ][/tex]
[tex]r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}[/tex]
substituting value
[tex]r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}[/tex]
[tex]r_2 = 0.24 \ m[/tex]
When it comes to how ray lines are drawn, what makes the convex lens and concave mirror similar to each other?
Answer:
Convex lens and convex mirrors are similar that
1. They have the same image characteristics at various object positions
2. They possess a positive focal length
3. Both their ray lines converge to a particular focal point
Convex lens and concave mirror have certain similarities that are explained below.
The shape of concave mirror is spherical.
Convex lens is the combination of two concave mirrors.
When the ray lines are drawn at the convex lens, then the coverage of the rays leads parallel to its principal axis.
When the ray lines are drawn at the concave mirror, then the coverage of the rays leads parallel to its focal point.
The focal length of concave mirror and convex lens is positive.
Concave mirror and convex lens both project the real and inverted image of the object.
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determine the weight in newtons of a woman whose weight in pounds is 125. also, find her mass in slugs and in kilograms. DEtermine your own weight in newtons
Answer:
124 just subtract only one
Explanation:
Skater A skates toward Skater B and gently pushes Skater B away. If the skaters are in a closed system, which statement is correct about the total momentum of the system? A. Total momentum may increase or decrease depending on the duration of the collision. B. Total momentum remains the same. C. Total momentum increases. D. Total momentum decreases.
Answer:
option B is correct
Explanation:
according to law of conservation of momentum the total momentum in a closed system remains constant before and after collision
If the skaters are in a closed system the total momentum remains the same.
In a closed system the total momentum is always conserved.
By applying the principle of conservation of linear momentum, the total momentum before the push of skater A, will be equal to the total momentum after the push.
The equation is given as;
Initial momentum = final momentum
[tex]m_au_a + m_bu_b = m_av_a + m_bv_b[/tex]
where;
u and v represents the initial and final velocity of both skaters.Thus, we can conclude that if the skaters are in a closed system the total momentum remains the same.
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The World Snail Racing Championships have been held in the English village of Congham, Norfolk since the 1960s. The contestants are placed at the center of a circular table covered with a damp tablecloth and race outwards 13 inches (33 cm) towards the finish line. A contestant known only as Archie set the current world record of 2 minutes 20 seconds in 1995. Determine Archie's speed in m/s and km/h.
Speed = distance / time
Speed = (33 cm) / (2min 20sec)
Speed = (0.33 m) / (140 sec)
Speed = 0.0024 m/s
Speed = (0.0024 m/s) x (3600 sec / 1 hr) x (1 km / 1000 m)
Speed = (0.0024 x 3600 / 1000) (m-sec-km / sec-hr-m)
Speed = 0.0085 km/hr
For the time of 2 min and 20 seconds, Archie's speed in m/s will be 0.0024 m/s and Archie's speed in km/hr will be 0.0085 km/hr.
What is Speed?The "speed at which an object is traveling," or speed, is a scalar quantity. The speed of an object is the rate at which it moves through space. A fast-moving object covers a considerable distance in a short amount of time while traveling at a high speed. An object moving slowly, on the other hand, moves a comparatively modest distance in the same amount of time. An object with zero speed is one that is not moving at all.
According to the question, the given values are :
Distance, d = 33 inches or,
d = 33 cm
Time, t = 2 minutes 20 seconds, or,
t = 140 seconds.
Speed is the ratio of distance to time.
s = 33/140
s = 0.0024 m/s
Now the distance in km/hr will be :
s = 0.0024 m/s × 3600 sec / 1hr × 1 km/ 1000 m
s= 0.0085 km/hr
Hence, the speed in km/hr is 0.0085 km/hr and in m/s speed is 0.0024 m/s.
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Two cars, P and Q, travel in the same direction on a long, straight section of a highway. car P passes car Q, and is adjacent to car Q at time t (subscript 0).
a) Suppose that car P and car Q each move with constant speed. At time t (subscript 0), is the magnitude of the instantaneous velocity of car P greater than, less than, or equal to the magnitude of the instantaneous velocity of car Q? Explain.
b) Suppose instead that car P is moving with constant speed but car Q is speeding up. At time (subscript 0), is the magnitude of the instantaneous velocity of car P greater than, less than, or equal to the magnitude of the instantaneous velocity of car Q? Explain.
Answer:
the instantaneous velocity of P is greater that Q because P obviously passes Q at that instant in time t
2. P has greater velocity at that instant (to) because P will need to be at a greater velocity to pass Q
Kendra drove 300 km in 2 hours. Henry drove the same 300 km in 3 hours. If
Kendra and Henry had the same average speed for their trips, what must be
true?
a. Kendra must have stopped for one hour during her trip
b. henry must have stopped for one hour during his trip
c. kendra must have had a lower velocity than henry
d. henry must have had a lower velocity then kendra
Answer:
b. henry must have stopped for one hour during his trip
Question No 1 Find the voltage drop across 24 ohm resistor and current flowing through 22 ohm resistor in the given circuit as shown in figure.
Answer:
8.25 V
Explanation:
We can ignore the 22Ω and 122Ω resistors at the bottom. Since there's a short across those bottom nodes, any current will go through the short, and none through those two resistors.
The 2Ω resistor and the 44Ω resistor are in parallel. The equivalent resistance is:
1 / (1 / (2Ω) + 1 / (44Ω)) = 1.913Ω
This resistance is in series with the 12Ω resistor. The equivalent resistance is:
1.913Ω + 12Ω = 13.913Ω
This resistance is in parallel with the 24Ω resistor. The equivalent resistance is:
1 / (1 / (13.913Ω) + 1 / (24Ω)) = 8.807Ω
Finally, this resistance is in series with the 4Ω resistor. The equivalent resistance of the circuit is:
8.807Ω + 4Ω = 12.807Ω
The current through the battery is:
12 V / 12.807Ω = 0.937 A
The voltage drop across the 4Ω resistor is:
(0.937 A) (4Ω) = 3.75 V
So the voltage between the bottom nodes and the top nodes is:
12 V − 3.75 V = 8.25 V
Multiple choice! To be dimensionally consistent, velocity [L/T], force [ML/T^2], mass [M], and length [L] must be related as follows. Multiple Choice A. velocity^2 = force × length/mass B. velocity^2 = force × length/mass^2 C.velocity = force × length^2/mass D. velocity = force × length/mass
Answer:
velocity^2=force*length/mass
Explanation:
velocity^2=force*length/mass is dimensionally consistent
as we know that
unit of force=Newton
unit of length=meter
unit of mass=kilogram so
velocity^2=newton*meter/kilogram
newton=kg*m/sec^2 therefore
v^2=kg*m*m/sec^2*kilogram
v^2=kg*m^2/sce^2*kg
v^2=m^2/sec^2
taking square root on both side
√ v²=√m²/sec²
v=m/s now the dimensions of V is
v=(L/T)
so option A is dimensionally consistent
It should be related like velocity^2=force*length/mass
Calculation of the relationship:
Since we know that
unit of force=Newton
unit of length=meter
And,
unit of mass=kilogram
Therefore,
velocity^2=newton*meter/kilogram
Now
newton=kg*m/sec^2
So,
v^2=kg*m*m/sec^2*kilogram
v^2=kg*m^2/sce^2*kg
v^2=m^2/sec^2
Now we take square root on both side
So,
√ v²=√m²/sec²
v=m/s
v=(L/T)
Therefore, option A is correct
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How does the decrease in gravitational potential energy of a falling ball compare to its increase in kinetic energy? (Ignore air friction.)
Answer:
for every decrease in potential energy, a kinetic energy proportional to 2gΔh is gained.
Explanation:
Let us consider a ball falling from its maximum height
For a body falling from its maximum height to a point p
change in height = Δh
The potential energy decrease is then proportional to
ΔPE = mgΔh
where
ΔPE is the decrease in kinetic energy
m is the mass of the ball
g is acceleration due to gravity
Δh is the change in height
For a body falling from its maximum height, the increase change in velocity
Δv = u + 2gΔh (at maximum height u = 0)
where
u is the initial kinetic energy of the ball
Δv = 0 + 2gΔh
Δv = 2gΔh
The kinetic energy increases by
ΔKE = [tex]\frac{1}{2}[/tex]m(Δv)^2
but Δv = 2gΔh
therefore
ΔKE = [tex]\frac{1}{2}[/tex]m(2gΔh)^2 = 2m(gΔh)^2
comparing the increase in kinetic energy to the decrease in potential energy, we have
(2m(gΔh)^2)/(mgΔh) = 2gΔh
This means that for every decrease in potential energy, a kinetic energy proportional to 2gΔh is gained.
A
is a precise way to represent data, but it does not identify trends in the data.
O graph
O table
Ovariable
O diagram
Answer: table
Explanation:
Answer:Table
Explanation:
A table is a way to represent data , but does not identify trends in the data
If a planet has an angular size of 0.29 degrees, what will be its angular size in arcminutes?
Answer:
Angular size in arc-minutes = 17.4 arc-minutes
Explanation:
Given:
Angular size = 0.29
Find:
Angular size in arc-minutes
Computation:
1 degree = 60 arc-minutes
So,
0.29 degree = 60 × 0.29 arc-minutes
0.29 degree = 17.4 arc-minutes
Angular size in arc-minutes = 17.4 arc-minutes
Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds.
This question is incomplete, the complete question is;
A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t² + 8√t, where t is the time in seconds and s(t) is the distance in feet from the top of the building.
(a) Write a formula for the average velocity of the ball for t near 4.
(b) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds
(c) What is your estimate for the instantaneous velocity of the ball at t = 4
Answer:
a)
Average velocity, (Vavg) of the ball for t near 4.
Vavg = [s(4) - s(0)] / (4 - 0)
Where s(4) = -16 × 4² + 8 × √4= - 240 m
s(0) = -16 × 0 + 8 * 0 = 0
b)
duration = 1 sec
Vavg = [s(5) - s(4)] / (5 - 4)
s(5) = -16 × 52 + 8 × √5 = - 382 m
s(4) = -16 × 42 + 8 √4 = - 240 m
Vavg = (-382 - (-240)) / (5 - 4)
Vavg = - 142.1 m/s
duration = 0.5 sec
Vavg = [s(4.5) - s(4)] / (4.5 - 4)
s(4.5) = -16 × 4.52 + 8 × √4.5 = - 307 m
s(4) = -16 × 42 + 8 × √4 = - 240 m
Vavg = (-307 - (-240)) / (4.5 - 4)
Vavg= - 134.1 m/s
duration = 0.05 sec
Vavg = [s(4.05) - s(4)] / (4.05 - 4)
s(4.05) = -16 × 4.052 + 8 × √4.05 = - 246 m
s(4) = -16 × 42 + 8 × √4 = - 240 m
Vavg = (-246 - (-240)) / (4.05 - 4)
Vavg= - 126.8 m/s
c)
Instantaneous velocity, v = ds/dt
= - 16 × 2 × t + 8 ×× (0.5 / √t )
= - 32 × t + 4/√t
ds/dt at t = 4 is,
v = - 32 × 4 + 4 / √4
= - 126 m/s
A 150 lb diver stands at the end of a rigid 8 ft long diving board. What is the magnitude of the torque the diver exerts on the diving board
Answer:
1200 lb-ft
Explanation:
Weight of diver W = 150 lb
length of board L = 8 ft
This board is pivoted at one end therefore, the perpendicular distance from the pivot = 8 ft
Torque = weight x perpendicular distance from pivot
T = WL
Torque = 150 x 8 = 1200 lb-ft
Velocity head is the mechanical energy of a fluid per unit weight of the fluid measured relative to the bottom of the channel.
a. True
b. False
Answer:
b. False
Explanation:
The Bernoulli's equation is given by;
[tex]P + \rho gh + \frac{1}{2} \rho v^2 = constat[/tex]
Divide through by the density of the fluid and acceleration due to gravity (ρg), to obtain energy per unit weight of the fluid known as head;
[tex]\frac{P}{\rho g} + h + \frac{v^2}{2g} = constant[/tex]
where;
[tex]\frac{P}{\rho g}[/tex] is pressure head
h is datum head (mechanical energy due to height of the fluid)
[tex]\frac{v^2}{2g}[/tex] is velocity head (mechanical energy due to velocity of the fluid)
Thus, the mechanical energy of a fluid per unit weight of the fluid, measured relative to the bottom of the channel is datum head.
The correct option is "b. False"
Compare and contrast the inner and outer planets. (Must be in complete sentences) (At least 5 sentences).
To contrast inner and outer planets we will start with the climate of the planets and then move on to there lighting. To start the planets closet to the sun, mercury, venus, earth and mars, are all hot compared to the further one, jupiter, saturn, uranus, neptune. This distance also makes the farthe away planets darker than the ones closer. Now to compare all the planets vary from either gass or solid, rocky or icy. All of them spin around the sun and all have objects spinning around them, moons.
1. An insect inside a bus flies from the back toward the front at 2 m/s. The bus is moving in a
straight line at 20 m/s. What is the total speed of the insect as seen by a person standing on the
side of the road?
Answer:
Total speed of the insect as seen by a person standing on the side of the road is 22 m/s
Explanation:
There i a composition of velocities, since the air inside the bus is moving at the same speed as the bus, and the insect flying relative to it.
So for an observer on the side of the road, the insect is flying at a speed of :
20 m/s + 2 m/s = 22 m/s
What is the magnitude of the electric force on the charge at the bottom left-hand vertex of the triangle due to the other two charges
Answer:
[tex]\frac{kq^{2} }{2a^2}[/tex]
Explanation:
The magnitude on the charge at the bottom-left corner due to the charge on the top vertex of the triangle will act along the +ve x-axis and the +ve y-axis.
From Coulomb's law the magnitude of the forces on the charge at the bottom-left corner, due to the charge on the top vertex of the triangle are [tex]\frac{kq^{2} }{a^2}[/tex]cos 60, and
The magnitude on the charge due to the charge at the bottom-right corner will only act in the -ve x-axis, since they repel each other (like charges repel). The magnitude is [tex]\frac{kq^{2} }{a^2}[/tex]
The angle made by the upper charge to the charge we're considering is 60° with the horizontal.
The total force on the charge along the x-axis is
[tex]F_{x}[/tex] = [tex]\frac{kq^{2} }{a^2}[/tex]cos 60 -
[tex]F_{x}[/tex] = [tex]\frac{kq^{2} }{2a^2}[/tex] - [tex]\frac{kq^{2} }{a^2}[/tex]
==> -[tex]\frac{kq^{2} }{2a^2}[/tex]
For the y-axis, we have
[tex]F_{y}[/tex] = [tex]\frac{kq^{2} }{a}[/tex]sin 60
[tex]F_{y}[/tex] = [tex]\frac{\sqrt{3}* kq^{2} }{2a^2}[/tex]
The resultant force is
[tex]|F| = \sqrt{F_{x}^{2}+ F_{y}^2 }[/tex]
The common factors between the two x-axis force, and the y-axis force is
[tex]\frac{kq^{2} }{2a^2}[/tex], we put this outside the square root (squaring this and square rooting will give us the initial value)
[tex]|F|[/tex] = [tex]\frac{kq^{2} }{2a^2}[/tex][tex]\sqrt{(\frac{1}{2})^2 + (\frac{\sqrt{3} }{2})^2 }[/tex]
[tex]|F|[/tex] = [tex]\frac{kq^{2} }{2a^2}[/tex][tex]\sqrt{\frac{1}{4} +\frac{3}{4} }[/tex]
==> [tex]\frac{kq^{2} }{2a^2}[/tex][tex]\sqrt{1}[/tex]
the magnitude of the electric force on the charge at the bottom left-hand vertex of the triangle due to the other two charges is
[tex]|F|[/tex] = [tex]\frac{kq^{2} }{2a^2}[/tex]
E
Which is the downward pull on an object due to gravity?
volume
mass
weight
matter
Answer:
Weight
Explanation:
Weight is the downward pull on an object due to gravity.
For example, the moon has less gravity than Earth so we would weigh less on the moon. Our Mass and volume always stay the same but our weight could change.
Answer:
weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight weight
Explanation:
Re-arrange the equation
x = vit + 1/2at2 v2f = vi2 + 2ax vf = vi + at x = 1/2(vi + vf)delta t
1. A car in front of the school goes from rest to 27 ms in 3.0 seconds. What is its acceleration (assuming it is constant)?
2. A rocket starts from rest and accelerates at a uniform rate of 190 m/s^2 for 2.4 seconds. What is its final velocity?
3. A car has a velocity of 15 m/s. It then accelerates at a uniform rate of 3.5 m/s^2 for the next 5.0 seconds. What distance does the car cover during this time?
4. What distance is required for a train to stop if its initial velocity is 23 m/s and its deceleration is 0.25 m/s^2? (Assume the train decelerates at a constant rate.
5. What distance will a car cover while uniformly accelerating from 12 m/s to 26 m/s in 14 seconds?
6. A person starts at rest and accelerates at a uniform rate of at 3.2 m/s^2 for 3.0 seconds. What distance does this person cover in that time?
7. What must be the acceleration of a train in order for it to stop from 12 m/s in a distance of 541 m? (Assume the train accelerates uniformly.)
Answer:
Explanation:
1) Acceleration is the change in velocity of a body with respect to time.
Acceleration a = vf - vi/t
vf is the final velocity
vi is the initial velocity.
t is the time taken
Since the body accelerates from rest, vi = 0m/s
a = 27-0/3
a = 27/3
a = 9m/s²
2) Given
u = 0m/s (accelerates from rest)
a = 190m/s²
t = 2.4seconds
v = ?
Using v = u+at
v = 0+190(2.4)
v = 190×2.4
v = 456m/s
The final velocity of the car is 456m/s
3) Given
u = 15m/s
a = 3.5m/s
t = 5seconds
Using the relationship
S = ut+1/2at²
S is the distance covered by the car.
S = 15(5)+1/2(3.5)×5²
S = 75+25×3.5/2
S = 75+43.75
S = 118.75m
4) Given
initial velocity u = 23m/s
Deceleration a = -0.25m/s²(negative acceleration)
Final velocity v = 0m/s
Using the relationship
V² = u²+2as
0² = 23²+2(-0.25)s
-23² = -0.5S
23² = 0.5S
S = 529/0.5
S = 1058m
The distance required for the train to stop is 1058m.
5) Given
initial velocity u = 12m/s
Final velocity v = 26m/s
time = 14sec
Acceleration a = v-u/t
a = 26-12/14
a= 14/14
a = 1m/s²
For Distance covered
v² = u²+2as
26² = 12²+2(1)S
676 = 144 +2S
2S = 676-144
2S = 532
S = 532/2
S = 266m
Distance that the car will cover is 266m
6) Given
Initial velocity u = 0m/s (person starts from rest)
acceleration a = 3.2m/s²
time t = 3.0s
To get the distance;
S = ut + 1/2at²
S = 0(3)+1/2(3.2)×3²
S = 0+1.6×9
S= 9×1.6
S = 14.4m
The distance that the person covered in 3.0s is 14.4m
7) initial velocity of train u = 12m/s
Distance covered S = 541m
Final velocity = 0m/s (on stopping)
acceleration a= ?
Acceleration will be negative since the train is coming to a stop (decelerating)
Using the formula v² = u²+2as
0² = 12² - 2a(541)
-12² = -1082a
144 = 1082a
a = 1082/144
a = 7.51m/s²
Hence the acceleration of a train in order for it to stop from 12 m/s in a distance of 541 m is 7.51m/s²
How much energy is stored in the capacitor before the dielectric is inserted?Express your answer in joules.
Answer:
0.004394Joules
Explanation:
The question is incomplete. Find the complete question in the attachment below;
Energy stored in a capacitor = 1/2CV² where;
C is the capacitance of the capacitor
V is the potential difference across the plates
Given parameters
C = 13.0μF = 13 * 10⁻⁶F
V = 26.0V
Required
Energy stored in the capacitor
Substituting the values into the equation given;
E = 1/2 * 13 * 10⁻⁶ * 26²
E = 1/2 * 13 * 10⁻⁶ * 676
E = 1/2* 8788 * 10⁻⁶
E = 4394 * 10⁻⁶
Hence the Energy stored in the capacitor in Joules is 0.004394Joules
A uniform seesaw is balanced at its center of mass, as seen below. The smaller boy on the right has a mass of 40.0 kg. What is the mass of his friend is (A) 40 kg (B) 20 kg (C) 80 kg (D) Cannot be determined
Answer:
C. 80kgExplanation:
The question lacks the appropriate diagram. Find the diagram attached below:
Let the mass of his friend be M.
Using the principle of moment which states that the sum of clockwise moment is equal to the sum of its anticlockwise moment to solve the problem,.
Since Moment = Force * perpendicular distance.
The smaller boy will move in the clockwise direction and his friend will move in the anti clockwise direction.
Clockwise moment = 40 * 4 = 160kgm
anticlockwise moment = M * 2 = 2Mkgm
Equating both moments to get the mass M of his friend
160 = 2M
Divide both sides by 2
2M/2 = 160/2
M = 80kg
Hence the mass of his friend that will keep the seesaw balanced horizontally is 80kg.