Answer:
It is pure science
Explanation:
A basic knowledge for the discovery of unknown laws based on well controlled experiments and deductions from demonstrated facts or truths.
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.
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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
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 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]
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.
5. A body falls freely from rest. It covers as much distance in the last second of its
motion as covered in the first three seconds. The body has fallen for a time of:
B) 5s
C) 7s
D) 9s
A) 35
Answer:
B 5s
Explanation:
Because of the Displacement in the nth second of the free fall is
Snth=21g(t12−t22)
Given that (tn−tn−1)=1
Displacement in 3 seconds of the free fall
S=21gt2
S=21×10×32
S=45m
Given that: Snth=45
On solving that we get:
t1=5sec
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
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]
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
local unit of measurement of length : confined to a particular place cise Choose the best answer from the given alterna MKS system stands for ........ i. mass, kilogram and second ii. metre, kilogram and second iii. metre, kilometre and second iv. metre, kilogram and standard
MKS stands for metre , kilogram and second
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
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
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
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.
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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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;
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how can you relazie a perfect balck body in pratice
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
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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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!! :)
Please let me know if you have any questions
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:
Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth has proven much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate the difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic meter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about 2.0×1092.0×109 pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the magnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second.
Use two significant figures in your answer, expressed in kilometers.
Answer:
26 km
Explanation:
Let's say our "cable" has a cross section of 1 m²
Then each meter of cable would weight 7900(9.8) = 77420 N
A Pascal is a Newton per square meter
2 x 10⁹ / 77420 = 25840 m or about 26 km or about 16 miles
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]
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a. The molecules of a magnet are independent...
Answer:
variable
Explanation:
what is the area of velocity time graph
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
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.
help me I m stuck on this question
Answer:
Answer is in the picture.
Explanation:
Answer is in the picture.
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
