Explanation:
Recall that
[tex]K = \dfrac{v_0^2\sin2\theta}{g}\:\:\:\:\:\:\:\:\:(1)[/tex]
and
[tex]Q = \dfrac{v_0^2\sin^2\theta}{2g}\:\:\:\:\:\:\:\:\:(2)[/tex]
From Eqn(2), we can write
[tex]\sin\theta = \sqrt{\dfrac{2gQ}{v_0^2}}\:\:\:\:\:\:\:\:\:(3)[/tex]
Using the identity [tex]\sin\theta = 2\sin\theta \cos\theta[/tex], we can rewrite Eqn(1) as
[tex]\dfrac{gK}{2v_0^2} = \sin\theta \cos\theta[/tex]
Squaring the above equation, we get
[tex]\dfrac{g^2K^2}{4v_0^4} = \sin^2\theta \cos^2\theta[/tex]
[tex]\:\:\:\:\:\:\:\:\:=\sin^2\theta(1 - \sin^2\theta)\:\:\:\:\:\:\:(4)[/tex]
Use Eqn(3) on Eqn(4) and we will get the following:
[tex]\dfrac{g^2K^2}{4v_0^4} = \dfrac{2gQ}{v_0^2}(1 - \dfrac{2gQ}{v_0^2})[/tex]
This simplifies to
[tex]\dfrac{gK^2}{8v_0^2Q} = 1 - \dfrac{2gQ}{v_0^2}[/tex]
Rearranging this further, we get
[tex]1 = \dfrac{2gQ}{v_0^2} + \dfrac{gK^2}{8v_0^2Q}[/tex]
Putting [tex]v_0^2[/tex] to the left side, we get
[tex]v_0^2 = 2qQ + \dfrac{gK^2}{8Q}[/tex]
Finally, taking the square root of the equation above, we get the expression for the muzzle velocity [tex]v_0[/tex] as
[tex]v_0 = \sqrt{2gQ + \dfrac{gK^2}{8Q}}[/tex]
How is centripetal force affected if an object increases its speed?
A. Decreases
B. Increases
C. Cut in half
D. No effect
Answer:
B. Increases.
Explanation:
[tex]{ \bf{F = \frac{m {v}^{2} }{r} }} \\ { \bf{F \: \alpha \: {v}^{2} }}[/tex]
Keeping mass, and radius constant, speed or velocity is directly proportioanal to centripetal force.
,In order to increase the speed of an object on a circular path, YOU have to increase the centripetal force acting on it.
need help with question after number 40
Answer:
4 device 3 e e and 4 divide 3 + 5 barabar 11 33 size to 46 size 35 and size 49 browser
What is the average velocity if the initial velocity of an object is 19 mph and the final velocity of 75 mph ?
Answer:
Hi I hope this is correct!
Explanation:
To find average velocity you can use the formula av = (v1 + v2) / 2
*I converted everything into m/s because that it usually the measurement for velocity*
v1 = initial velocity = 8.49376 m/s , v2 = final velocity = 33.528 m/s
av = 8.49376 + 33.528 / 2
= 21.01088 m/s
*If you were required to leave the final answer in mph here it is
av = 19 + 75 / 2
= 47 mph
Hope this helps! Best of luck <3
Explanation:
hope it helps you
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Which of the following measures is equal to 700 km?
Answer:
1km=1000m
700km=
700×1000=700000
=700000metres
hope this helps
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
Explain how blood circulation takes place in humans?
Blood comes into the right atrium from the body, moves into the right ventricle and is pushed into the pulmonary arteries in the lungs. After picking up oxygen, the blood travels back to the heart through the pulmonary veins into the left atrium, to the left ventricle and out to the body's tissues through the aorta.
Hope it helps you
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There is more than 1 answer,
The picture is down
Answer:
test her prototype and collect data about its flight
why is nut-cracker 2nd class lever?
2nd class leaver refers to such leaver in which load lies between effort and fulcrum.In a nut cracker too load is in between effort and fulcrum.Thus, nut cracker is a 2nd class leaver.......
OBJECTI
1. The motion of a liquid inside a U-tube is an
example of what type of motion?
a. Simple Harmonic c. Random
b.Rectilinear
d. Circular
Answer:
option A
Explanation:
simple harmonic motion
Answer:
random motion I think not sure
importance of SI system in points
Answer:
SI unit is an international system of measurements that are used universally in technical and scientific research to avoid the confusion with the units. Having a standard unit system is important because it helps the entire world to understand the measurements in one set of unit system.
Hi Friends!
please help me with these questions!
SUBJECT: Chemistry, Physics,Biology
Answer:
q.1 : Air near candle gets heated up and after this it rises by convection so the thermometer B will receive more heat than the thermometer A So, according to the given condition thermometer B will show a greater rise in temperature.
q.2 : x is the pure sample of compound . y is the pure sample of element . z is the mixture of different elements
q.3 : the saliva contains an enzyme salivary amylase (ptyalin) which converts starch in roti into maltose, isomaltose and small dextrins called a-dextrin.
SI units are used for the scientific works,why?
Answer:
SI is used in most places around the world, so our use of it allows scientists from disparate regions to use a single standard in communicating scientific data without vocabulary confusion
Using this information...
Determine the velocity of the pebble as it passes over the top of the tree.
[tex]19.2\:\text{m/s}[/tex]
Explanation:
At the top of the tree, the velocity of the pebble is purely horizontal so we can calculate it as
[tex]v_{y} = v_{0y} = v_0\cos 40° = (25\:\text{m/s})(0.766)[/tex]
[tex]\:\:\:\:\:= 19.2\:\text{m/s}[/tex]
The height of the tree is approximately 12.5 meters when velocity of the pebble as it passes over the top of the tree.
Let's calculate the height of the tree step by step:
Given:
Initial velocity (v0) = 25 m/s
Launch angle (θ) = 40° above the horizontal
Time after launch (t) = 2 seconds
Acceleration due to gravity (g) = -9.8 m/s² (negative because it acts downward)
Step 1: Calculate the vertical component of the initial velocity (Vy):
Vy = v0 * sin(θ)
Vy = 25 m/s * sin(40°)
Vy ≈ 25 m/s * 0.6428 ≈ 16.07 m/s (rounded off to two decimal places)
Step 2: Calculate the vertical displacement (change in height) of the pebble after 2 seconds:
d = vot + (1/2)at²
d = (16.07 m/s) * (2 s) + (1/2) * (-9.8 m/s²) * (2 s)²
d ≈ 32.14 m - 19.6 m
d ≈ 12.54 m (rounded off to two decimal places)
Step 3: The height of the tree is equal to the vertical displacement of the pebble:
Height of the tree ≈ 12.54 m ≈ 12.5 m (rounded off to one decimal place)
The height of the tree is approximately 12.5 meters.
To know more about Acceleration here
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A proposed communication satellite would revolve around the earth in a circular orbit in the equatorial plane at a height of 35880Km above the earth surface. Find the period of revolution of the satellite. (Take the mass of earth =5.98×10²⁴kg, the radius of the earth 6370km and G=6.6×10–¹¹Nm²/kg2)
Answer:
Period is 86811.5 seconds.
Explanation:
[tex]{ \boxed{ \bf{T {}^{2} = (\frac{4 {\pi}^{2} }{GM}) {r}^{3} }}}[/tex]
[tex]{ \tt{T {}^{2} = \frac{4 {(3.14)}^{2} }{(6.6 \times {10}^{ - 11} ) \times (5.98 \times {10}^{24} )} \times {((35880\times {10}^{3}) } + (6370 \times {10}^{3} )) {}^{3} }} \\ \\ { \tt{T {}^{2} = 7.54 \times {10}^{9} }} \\ { \tt{T = \sqrt{7.54 \times {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}[/tex]
Which of the following scientists won a Nobel Prize for pioneering work in the
study of the evolution of stars?
A. Christian Doppler
B. Warren Washington
C. Charles Kuen Kao
-D. Subrahmanyan Chandrasekhar
Answer:
Subrahmanyan Chandrasekhar
Answer:
D. Subrahmanyan Chandrasekhar
Explanation:
calculate the force of gravitation due to Earth on A ball of 1 kg mass lying on the ground
Using newtons second law
[tex]\\ \sf\longmapsto Force=Mass\times Acceleration[/tex]
[tex]\\ \sf\longmapsto Force=1(9.8)[/tex]
[tex]\\ \sf\longmapsto Force=9.8N[/tex]
Suppose you inflate your 2000 kg car's tires to the recommended pressure, as measured by a gauge. The resulting contact patch is 18 cm wide and 12 cm long. What does the gauge read
the molecule of magnet are independent _____________
Answer:
The first is the electric field, which describes the force acting on a stationary charge and gives the component of the force that is independent of motion. The magnetic field, in contrast, describes the component of the force that is proportional to both the speed and direction of charged particles.
For a transverse wave, what is a wavefront?
A a line joining all points on the same crest of a wave
B a line showing the displacement of a wave
C the energy content of a wave
D the first part of a wave to reach a point
wavefront is the long edge that moves, for example, the crest or the trough
a reagent is added to the blue solution to identify the copper 2 ions name the blue solution
Answer:
buriret i belive
Explanation:
.
Answer:
The blue solution is named copper sulfate
A 25g rock is rolling at a speed of 5 m/s. What is the kinetic energy of the rock?
Answer:
The answer is 312.5j
Explanation:
The kinetic energy (KE):
KE=1/2*m*v^2
M= mass of the object
v= velocity of the object
We have;
m=25g
v=5m/s
KE=1/2*25g*5^2m/s
KE =312.5j
3. Is it possible for a scientific theory to become a law? Why or why not?
A theory does not change into a scientific law with the accumulation of new or better evidence. A theory will always remain a theory; a law will always remain a law. Both theories and laws could potentially be falsified by countervailing evidence. Theories and laws are also distinct from hypotheses.
The power dissipated in a series RCL circuit is 68.3 W, and the current is 0.548 A. The circuit is at resonance. Determine the voltage of the generator. (The current given is the rms current.)
The generator is 124.6 volts RMS.
A rabbit is moving in the positive x-direction at 2.70 m/s when it spots a predator and accelerates to a velocity of 13.3 m/s along the positive y-axis, all in 2.40 s. Determine the x-component and the y-component of the rabbit's acceleration.
Answer:
the answer is nearly 5.655 [tex]ms^{-2}[/tex]
Explanation:
Given,
[tex]v_{x}=2.7 ms^{-1}[/tex]
[tex]v_{y}=13.3 ms^{-1}[/tex]
[tex]t=2.4 s[/tex]
[tex]a_{x}=\frac{2.7}{2.4}=1.125 ms^{-2}[/tex] (as [tex]a=\frac{v-u}{t}[/tex])
[tex]a_{y}=\frac{13.3}{2.4}=5.542 ms^{-2}[/tex]
[tex]a=\sqrt{a_{x}^{2}+a_{y}^{2} }[/tex]
[tex]=\sqrt{1.125^{2}+5.542^{2} }[/tex]
[tex]=5.655 ms^{-2}[/tex]
hope you have understood this...
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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
hope this helps
A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top of it as shown. The block w2 is attached to avertical wall by a string 6m long. If the coefficient of friction between all surface is 0.25 and the system is in equilibrium find the magnitude of the horizontal force applied to the lower block
The horizontal force applied to the block is approximately 1,420.84 N
The known parameters;
The mass of the block, w₁ = 400 kg
The orientation of the surface on which the block rest, w₁ = Horizontal
The mass of the block placed on top of the 400 kg block, w₂ = 100 kg
The length of the string to which the block w₂ is attached, l = 6 m
The coefficient of friction between the surface, μ = 0.25
The state of the system of blocks and applied force = Equilibrium
Strategy;
Calculate the forces acting on the blocks and string
The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N
The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N
Let T represent the tension in the string
The upward force from the string = T × sin(θ)
sin(θ) = √(6² - 5²)/6
Therefore;
The upward force from the string = T×√(6² - 5²)/6
The frictional force = (W₂ - The upward force from the string) × μ
The frictional force, [tex]F_{f2}[/tex] = (981 - T×√(6² - 5²)/6) × 0.25
The tension in the string, T = [tex]F_{f2}[/tex] × cos(θ)
∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6
Solving, we get;
[tex]T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27[/tex]
[tex]Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92[/tex]
The frictional force on the block W₂, [tex]F_{f2}[/tex] ≈ 219.92 N
Therefore;
The force acting the block w₁, due to w₂ [tex]F_{w2}[/tex] = 219.92/0.25 ≈ 879.68
The total normal force acting on the ground, N = W₁ + [tex]\mathbf{F_{w2}}[/tex]
The frictional force from the ground, [tex]\mathbf{F_{f1}}[/tex] = N×μ + [tex]\mathbf{F_{f2}}[/tex] = P
Where;
P = The horizontal force applied to the block
P = (W₁ + [tex]\mathbf{F_{w2}}[/tex]) × μ + [tex]\mathbf{F_{f2}}[/tex]
Therefore;
P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84
The horizontal force applied to the block, P ≈ 1,420.84 N
Learn more about friction force here;
https://brainly.com/question/18038995
a. The molecules of a magnet are independent...
Answer:
variable
Explanation:
help me I m stuck on this question
Answer:
Answer is in the picture.
Explanation:
Answer is in the picture.
You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction between the box and the inclined plane is 0.680. a) Determine the static frictional force which holds the box in place. b) You slowly raise one end of the track, slowly increasing the incline of the angle. Determine the maximum angle that the incline can make with the horizontal so that the box just remains at rest. Ms 680 u Fgsin 281 Ffg Mgm r 680 55 4 8
Answer:
[tex]\theta=34 \textdegree[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=55kg[/tex]
Angle [tex]\theta =28.0[/tex]
Coefficient of static friction [tex]\alpha =0.680[/tex]
Generally, the equation for Newtons second Law is mathematically given by
For
[tex]\sum_y=0[/tex]
[tex]N=mgcos \theta[/tex]
for
[tex]\sum_x=0[/tex]
[tex]F_{s}=mgsin\theta[/tex]
Where
[tex]F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta[/tex]
[tex]F_{s}=0.68*55*9.8*cos 28[/tex]
[tex]F_{s}=323.62N[/tex]
Therefore
[tex]\alpha mgcos \theta=mg sin \theta[/tex]
[tex]\theta=tan^{-1}(0.68)[/tex]
[tex]\theta=34 \textdegree[/tex]
(a) The static frictional force which holds the box in place is 323.62 N.
(b) The maximum angle that the incline can make with the horizontal is 34.2⁰.
Net forceThe net force applied to keep the box at rest must be zero in order for the box to remain in equilibrium position. Apply Newton's second law of motion to determine the net force.
∑F = 0
Static frictional forceThe static frictional force is calculated as follows;
Fs = μFncosθ
Fs = 0.68 x (55 x 9.8) x cos28
Fs = 323.62 N
Maximum angle the incline can makeFn(sinθ) - μFn(cosθ) = 0
mg(sinθ) - μmg(cosθ) = 0
μmg(cosθ) = mg(sinθ)
μ(cosθ) = (sinθ)
μ = sinθ/cosθ
μ = tanθ
θ = tan⁻¹(μ)
θ = tan⁻¹(0.68)
θ = 34.2⁰
Learn more about net force of inclined here: https://brainly.com/question/25784024
A man is driving a car at speed 25m/s. calculate the distance covered by it in one hour.
Answer:
6.94 km/hr
Explanation:
m/s to km/hr -> Multiply by 18/5
25/(18/5)
=> 25 x 5/18
=> 125/18 km/hr
=> 6.94 km/hr
Answer: 90,000 m = 90 km
Explanation:
Given information
Time = 1 hour
Speed = 25 m/s
Given expression deducted from the given information
Distance = speed × time
Convert units of time
1 hour = 60 minutes
1 minute = 60 seconds
1 hour = 60 × 60 = 3600 seconds
Substitute values into the expression
Distance = 25 × 3600
Simplify by multiplication
Distance = [tex]\boxed{90,000 m=90km}[/tex]
Hope this helps!! :)
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