Answer:
Explanation:
We shall represent the point of destination and point of new starting point in vector form .
i and j will represent east and north .
point of destination = 123 j
starting point = 112 i
distance between the two
D = 123 j - 112 i
magnitude of D = √ ( 123² + 112² )
= 166.35 km
Direction of D
Tanθ = - 123 / 112
= - 1.09
θ = 132.54 degree from east in positive angle .
what type of image is formed by a lens if m = -0.87
A. The image is larger than the object and the image is real
B. The image is smaller than the object and the object is real
C. The image is larger than the object and is virtual
D. The image is smaller than the object and is virtual
Answer:
B.
Explanation:
as there is a '-' sign before the magnification value, it forms a real image.
so the last two options get cancelled.
as the value of magnification is 0.87 ie. lesser than 1, we can say that the image is smaller.
The type of image is formed by a lens if m = -0.87 would be smaller than the object and the object is real
What is a lens?A lens is a transmissive optical tool that employs refraction to focus or disperse a light beam. The power of the lens is expressed in the dioptre which is the reciprocal of the focal length of the lens.
The image formation through the lens is calculated with the help of the lens formula given as follows
1/f = 1/v - 1/u
where f is the focal length of the lens
u is the distance of the object from the lens
v is the distance of the image formed from the
magnification (m) = size of image /size of the object
As given in the problem image is formed by a lens if m = -0.87
Thus, the negative sign represents that the image formed by the object is real, and the magnitude of the magnification is less than 1 this represents that size of the image is smaller than the size of the object.
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Astronomers can now report that active star formation was going on at a time when the universe was only 20% as old as it is today. When astronomers make such a statement, how can they know what was happening inside galaxies way back then
Answer:
First, as you may know, the light travels at a given velocity.
In vaccum, this velocity is c = 3x10^8 m/s.
And we know that:
distance = velocity*time
Now, if some object (like a star ) is really far away, the light that comes from that star may take years to reach the Earth.
This means that the images that the astronomers see today, actually happened years and years ago (So the night sky is like a picture of the "past" of the universe)
Also, for example, if an astronomer sees some particular thing, he can apply a model (a "simplification" of some phenomena that is used to simplify it an explain it) and with the model, the scientist can infer the information of the given thing some time before it was seen.
The astronomers could know what was happening inside galaxies way back then by the fact that;
they examine the spectra of galaxies (or the overall colors of galaxies) with the highest redshifts they can find
Astronomers Measure the wavelength of the light that is stretched, so the light is seen as 'shifted' towards the red part of the spectrum by using spectroscopy. This measure is also called redshift.
This invokes a ray of light through a triangular prism that splits the light into various components known as spectrum.
The way the astronomers could use this concept to know what was happening in the galaxies before is by examining the spectra of galaxies that have the highest redshifts.
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SHOW ADEQUATE WORKINGS IN THIS SECTION
12. Wale and Lekan of 55kg and 60kg respectively ran a race of 200m.
i. Calculate their work done in KJ
ii. If wale finished the race in 25secs while Lekan finished in 30secs, calculate their
power and who is more powerful out of these two. ( g = 10m/secs)
13. A machine has a velocity ratio of 5 and is 800
/0 efficient. If the machine is carrying a load
of 200kg, what will be effort applied?
Please help me answer anyone that you understand
Answer:
12 i. The work done by Wale = 107.910 kJ
The work done by Lekan = 117.720 kJ
Total work done = 225.36 kJ
ii. Wale's power = 4.3164 kW
Lekan's power = 3.924 kW
Wale has more power and is more powerful than Lekan
13. 313.92 N
Explanation:
i. The work done, W = Force, F × Distance moved by the force, D
The given parameters are
The mass of Wale = 55 kg
The mass of Lekan = 60 kg
The acceleration due to gravity, g =9.81 m/s²
The motion force of Wale and Lekan are;
Motion force of Wale = 9.81 × 55 = 539.55 N
Motion force of Lekan = 9.81 × 60 = 588.6 N
The work done by Wale = 539.55 × 200 = 107910 J = 107.910 kJ
The work done by Lekan= 588.6 × 200 = 117720 J = 117.720 kJ
107910 + 117720 =225630 J = 225.36 kJ
ii. Power = Work done/time
Wale finished the race in 25 s, therefore, his power = 107910/25 = 4316.4 W
Lekan finished the race in 30 s, therefore, his power = 117720/30 = 3924 W
Wale has more power and is more powerful than Lekan
13. The velocity ratio = 5
V. R. = Distance moved by effort/(Distance moved by load)
Efficiency = 80%
Work done by effort = x
Work done by machine = Efficiency × Work done by effort = 0.8 × x
Distance moved by effort, E = V. R. × Distance moved by load, D = 5 × D
Work done by effort = Force × Distance moved = 200×9.81× E
Work done by effort = 1962×E = 1962×E = 1962×5×D
Work done by machine = 1962 × D, when D = 1, we have;
0.8 × 1962×1 = 1569.6 J
Work done by effort = Force × Distance moved
Work done by effort = Force × 5×D = Force × 5 (D = 1)
From the principle of conservation of energy, we have;
Energy is neither created nor destroyed
Therefore
Work done by effort = Force × 5 = 1569.6 J
Force = 1569.6 /5 = 313.92 N.
what is a hypothesis reffered to as after being verified by a large number or independent experiments
Answer:
The hypothesis may or may not be true and needs to be tested. It might be the answer to the problem. Hence, it must be tested thoroughly. When these predictions are tested again and again in independent scientific experiments and gets verified, the hypothesis is converted into a scientific theory.
A 0.149 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s . It has a head-on collision with a 0.308 kg glider that is moving to the left with a speed of 2.27 m/s . Suppose the collision is elastic.1. Find the magnitude of the final velocity of the 0.157kg glider.
2. Find the magnitude of the final velocity of the 0.306kg glider.
Answer:
v1 = −2.201946 m/s ( to the left)
v2 = 0.7780534 m/s ( to the right)
Explanation:
Given the following :
Mass of first glider (m1) = 0.149kg
Initial Speed of first glider (u1) = 0.710 m/s
Mass of second glider (m2) = 0.308kg
Initial Speed of second glider (u2) = 2.27m/s
For elastic collision:
m1u1 + mu2u2 = m1v1 + m2v2
Where V1 and v2 = final velocities if the body after collision.
Taking right as positive ; left as negative
u1 = 0.710m/s ; u2 = - 2.27m/s
u1 - u2 = - (v1 - v2)
0.710 - - 2.27 = - v1 + v2
v2 - v1 = 2.98 - - - - (1)
From:
m1u1 + mu2u2 = m1v1 + m2v2
(0.149 * 0.710) + ( 0.308 * - 2.27) = (0.149 * v1) + (0.308 * v2)
0.10579 + (-0.69916) = 0.149 v1 + 0.308v2
−0.59337 = 0.149 v1 + 0.308v2
Dividing both sides by 0.149
v1 + 2.067v2 = −0.59337 - - - - - (2)
From (1)
v2 = 2.98 + v1
v1 + 2.067(2.98 + v1) = −0.59337
v1 + 6.16 + 2.067v1 = −0.59337
3.067v1 = −0.59337 - 6.16
3.067v1 = −6.75337
v1 = −6.75337 / 3.067
v1 = −2.201946 m/s ( to the left)
From v2 = 2.98 + v1
v2 = 2.98 + (-2.201946)
v2 = 0.7780534 m/s ( to the right)
a 2-n force is applied to a spring, and there is displacement of 0.4 m. how much would the spring be displaced if a 5-n force was applied?
Answer:1m
Explanation:
2n=0.4m
5n=?
5n×0.4/2n=1m
what is space in detail?
Answer:
Space in the astronomy and cosmology space is 3 dimensional region and earth atmosphere end is called space.
Explanation:
Space is the everything of the top earth atmosphere moon, starts, milky way, black hole and GPS satellites an distant,space also called between stars,moon,planet.
Space is all the extends far in all directions, space is finite unbound space surface of the earth has finite are no beginning or not end.
Space contains there are three dimensions is called 3 D space,an space is to refer an interval during signal transmitted also used by the character, bytes, words and octets in digital signal.
space is that term can refer to various in science, communications and mathematics,and maintain orbits for responsible time, space is usually to begin at the lowest attitude satellites can maintain orbits.
space coordinates are uniquely define the location of any particular point and that continuum requires more than coordinates,and the number of dimensions and conventional space or digital communications during the signal represents logic is 0 words in a digital signal.
If the spring constant is 10 N/m and the spring is stretched 1 m, what is the Force?
Answer:
10N
Explanation:
Applying the Hooke law:
F = kx
F: Force
k: stiffness coefficient
x: stretched distance
F = 10N/m x 1m = 10N
The marginal cost curve
(a) Lies below the ATC curve when the ATC curve slopes upward.
(b) Intersects the AFC and ATC curves at their respective minimum points.
(c) Lies above the AVC curve when the AVC curve slopes downward.
(d) Intersects the AFC and AVC curves at their respective minimum points.
(e) Intersects the AVC and ATC curves at their respective minimum points
Answer:
c
Explanation:
The marginal cost curve image has been attached from which we can clearly, indicate that
ATC = average total cost
AFC = average fixed cost
AVC = average variable cost.
From the graph we can indicate that the marginal cost curve
(c) Lies above the AVC curve when the AVC curve slopes downward.
A car is driving at 99 km/h, calculate the distance it travels in 70 minutes.
Give your answer in correct SI units rounded to 0 decimal places.
Answer:
The distance the car travels is 115500 m in S.I units
Explanation:
Distance d = vt where v = speed of the car and t = time taken to travel
Now v = 99 km/h. We now convert it to S.I units. So
v = 99 km/h = 99 × 1000 m/(1 × 3600 s)
v = 99000 m/3600 s
v = 27.5 m/s
The speed of the car is 27.5 m/s in S.I units
We now convert the time t = 70 minutes to seconds by multiplying it by 60.
So, t = 70 min = 70 × 60 s = 4200 s
The time taken to travel is 4200 s in S.I units
Now the distance, d = vt
d = 27.5 m/s × 4200 s
d = 115500 m
So, the distance the car travels is 115500 m in S.I units
Compare diffusion of chlorine gas into air and into vaccuum. Explain your answer
Answer:
Diffusion depends among many other things also upon the concentration gradient of the diffusing substance.For example if there are two boxes with given particles first isolated from each other and if they are bought in contact, then there is a net diffusion of particles from a box with higher concentration of particles to box with lower concentration. And also diffusion rate depends at any instant directly on the concentration difference between them at that instant.Now the vaccum is equivalent to an empty box which means with this one leads to a maximum diffusion rate when bought in contact with a box with particles because there is maximum concentration difference.That is vaccum is empty ( except for energy perturbations and a small concentration of particles which will be zero for our purpose) and any bunch of particles will find least resistance to diffuse as just outside this bunch there is ‘0' concentration of the particles.
In a Young's double-slit experiment, a set of parallel slits with a separation of 0.102 mm is illuminated by light having a wavelength of 575 nm and the interference pattern observed on a screen 3.50 m from the slits.(a) What is the difference in path lengths from the two slits to the location of a second order bright fringe on the screen?(b) What is the difference in path lengths from the two slits to the location of the second dark fringe on the screen, away from the center of the pattern?
Answer:
Rounded to three significant figures:
(a) [tex]2 \times 575\; \rm nm = 1150\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].
(b) [tex]\displaystyle \left(1 + \frac{1}{2}\right) \times (575\;\rm nm) \approx 863\; \rm nm = 8.63\times 10^{-7}\; \rm m[/tex].
Explanation:
Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.
Let [tex]k[/tex] denote a natural number ([tex]k \in \left\lbrace0,\, 1,\, 2,\, \dots\right\rbrace[/tex].) In a double-split experiment of a monochromatic light:
A maximum (a bright fringe) is produced when light from the two slits arrive while they were in-phase. That happens when the path difference is an integer multiple of wavelength. That is: [tex]\text{Path difference} = k\, \lambda[/tex].Similarly, a minimum (a dark fringe) is produced when light from the two slits arrive out of phase by exactly one-half of the cycle. For example, The first wave would be at peak while the second would be at a crest when they arrive at the screen. That happens when the path difference is an integer multiple of wavelength plus one-half of the wavelength: [tex]\displaystyle \text{Path difference} = \left(k + \frac{1}{2}\right)\cdot \lambda[/tex].MaximaThe path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where [tex]k = 0[/tex] in [tex]\text{Path difference} = 0[/tex].
The path difference increases while moving on the screen away from the center. The first order maximum is at [tex]k = 1[/tex] where [tex]\text{Path difference} = \lambda[/tex].
Similarly, the second order maximum is at [tex]k = 2[/tex] where [tex]\text{Path difference} = 2\, \lambda[/tex]. For the light in this question, at the second order maximum: [tex]\text{Path difference} = 2\, \lambda = 2 \times 575\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].
Central maximum: [tex]k = 0[/tex], such that [tex]\text{Path difference} = 0[/tex].First maximum: [tex]k = 1[/tex], such that [tex]\text{Path difference} = \lambda[/tex].Second maximum: [tex]k = 2[/tex], such that [tex]\text{Path difference} = 2\, \lambda[/tex].MinimaThe dark fringe closest to the center of the screen is the first minimum. [tex]\displaystyle \text{Path difference} = \left(0 + \frac{1}{2}\right)\cdot \lambda = \frac{1}{2}\, \lambda[/tex] at that point.
Add one wavelength to that path difference gives another dark fringe- the second minimum. [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex] at that point.
First minimum: [tex]k =0[/tex], such that [tex]\displaystyle \text{Path difference} = \frac{1}{2}\, \lambda[/tex].Second minimum: [tex]k =1[/tex], such that [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex].For the light in this question, at the second order minimum: [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda = \left(1 + \frac{1}{2}\right)\times (575\; \rm nm) \approx 8.63\times 10^{-7}\; \rm m[/tex].
A dog is 10 m from a cat, whose speeds are 6 and 5 m / s, respectively. What time does the dog require to catch the cat?
Answer:
10 seconds
Explanation:
because the cat is moving one m/s slower than the dog, the dog has a relative speed of 1 m/s. 10 meters would take 10 seconds for the dog to cover
A car is moving on straight highway with a speed of 108 km/h.
Answer:
5.3333 sec
Explanation:
initial speed: u = 108km/hr or 30 m/s
final speed: v = 0m/s
distance travelled: s = 80m
time the car took to stop: = t sec
[tex]v^{2} - u^{2}[/tex] = 2as,
a = ([tex]v^{2} - u^{2}[/tex])/2s
a = (0-900)/160
a = -5.625 [tex]ms^{-2}[/tex]
v = u + at,
t = (v - u)/a
t= (0 - 30)/(-5.625)
t = 5.3333 sec
You have a cup with 50cm filled with water. How much pressure will the water act on the bottom of the cup? The density of water is 1000kg/m^3 and g = 10N/kg
Answer:
5000 N/m².
Explanation:
The following data were obtained from the question:
Density (d) = 1000 kg/m³
Acceleration due to gravity (g) = 10 N/kg
Height (h) = 50 cm = 50/100 = 0.5 m
Pressure (P) =.?
Pressure is related to density and height by the following equation:
P = dgh
Where
P is the pressure.
d is the density.
g is the acceleration due to gravity.
h is the height.
With the above formula, we can obtain the pressure at the bottom of the cup as follow:
P = dgh
P = 1000 x 10 x 0.5
P = 5000 N/m²
Therefore, the pressure at bottom of the cup is 5000 N/m².
Can someone explain how the weight of the block is 10.26N, with reference to an appropriate law of motion?
this process is called parellelogram method of resolving vectors.
Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of the block as it passes through the equilibrium position is 62 cm/s. What is the angular frequency of the block's motion
Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
A hammer is used to hit a nail into a board. Which statement is correct about the forces at play between the nail and the hammer? The nail exerts an equal force on the hammer in the same direction. The nail exerts a much smaller force on the hammer in the opposite direction. The nail exerts an equal force on the hammer in the opposite direction. The nail exerts a much smaller force on the hammer in the same direction.
Answer:
The nail exerts an equal force on the hammer in the opposite direction.
Explanation:
The Newtons third law states that there is an equal an opposite reaction for every action. When hammer pushes the nail, the nail will push the hammer back in opposite direction. When the hammer hits a nail then nail will exert the equal and opposite force to the hammer. These both objects will exert force on each other in opposite directions.
Hello, I am BrotherEye
Answer:
Answers are
1. "The nail exerts an equal force on the hammer in the opposite direction."
2. "500 N"
3. "The iron piece exerts a force of 1 N on the magnet in the opposite direction."
4. "When mass moves closer to the point of rotation, rotational inertia decreases."
5. "The skater spins slower because his rotational inertia has increased."
Explanation:
64. A heart pacemaker fires 72 times a minute, each time a 25.0-nF capacitor is charged (by a battery in series with a resistor) to 0.632 of its full voltage. What is the value of the resistance
Answer:
=33 .3×10^6Ω
=33.3M Ω
Explanation:
We were told to calculate the Resistance value,
Given the heart pacemaker fires as 72 times a minute, which is the time constant
Then we can convert the pacemaker fires of 72 times a minute to seconds for unit consistency.
1 minutes= 60secs
Then ,Time constant τ=60secs/72=0.8333 seconds
Time constant τ can be calculated using the formula below
τ= RC
Where R= resistance
C = capacitance
Then making RESISTANCE subject of formula we have
R=τ/C
But Capacitance=25.0-nF = 25×10^25F
Substitute the values we have
R=0.833/25×10^25
=33 .3×10^6 Ω
But can still be converted to M Ω= 33.3M Ω
Therefore, the resistance is 33 .3×10^6 Ω or 33.3M Ω
NOTE: 1M= 10^6
The value of the resistance will be "3.33×10⁷ Ω".
Resistance based problem:According to the question,
Capacitor, C = 25.0 nF
60 sec - 72 fires
now,
Time for 1 fire,
[tex]t = \frac{60}{72}[/tex]
[tex]= \frac{5}{6} \ sec[/tex]
Now,
⇒ [tex]V = V_0 (1-e^{-\frac{t}{RC} })[/tex]
[tex]0.632 V_0=V_0(1-e^{\frac{-\frac{5}{6} }{R.25 n C} })[/tex]
[tex]e^{-\frac{1}{R\times 30\times 10^{-9}} } = 1-0.632 = 0.368[/tex]
[tex]= 2.72[/tex]
By taking "log" both sides,
⇒ [tex]ln \ e^{\frac{1}{R\times 30\times 10^{-9}} } = ln \ 2.72[/tex]
hence,
The Resistance be:
⇒ [tex]R = \frac{10^9}{30}[/tex]
[tex]= 3.33\times 10^7 \ \Omega[/tex]
Thus the above approach is right.
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What did Bohr's model of the atom include that Rutherford's model did not have?
a nucleus
energy levels
electron clouds
smaller particles
Answer:
The correct option is energy levels
Explanation:
Rutherford's model of an atom suggests that an atom has a tiny positively charged central mass (now called the nucleus) which is surrounded by electrons (negatively charged) in a cloud-like manner.
Bohr's model went a bit further than the Rutherford's model in describing an atom by suggesting that the electrons which surrounds in the nucleus travel in fixed circular orbits. This description by Bohr was able to describe the energy levels of orbitals which assumes that smallest orbitals have the lowest energy while the largest orbitals have the highest energy.
Answer:
energy levels
hope this helped!
Explanation:
what is SI unit System ? why has SI system been developed ? Give reasons
Explanation:
SI is the international system of units
It was developed to express magnitudes and quantities
PLS HELP ME Define Derived Quantities ?
Derived Quantities
Explanation: Those physical quantities which are derived from fundamental quantities are called derived quantities and their units are called derived units. e.g., velocity, acceleration, force, work etc.
Answer:
These are quantities calculated from two or more measurements
Explanation:
They can't me measured directly.
They can only be computed.
They are calculated in PHYSICAL SCIENCE.
hope it helps.
Will mark as BRAINLIEST.......
The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation 4x³+3x²-5x+2 , where x is in meters and t is in sec.
a)Find velocity of particle at i) t=2 sec ii) t=4 sec.
b) Find the acceleration of the particle at t=3 sec.
Explanation:
It is given that,
The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:
[tex]x=4t^3+3t^2-5t+2[/tex]
Where,
x is in meters and t is in sec
We know that,
Velocity,
[tex]v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5[/tex]
(a) i. t = 2 s
[tex]v=12(2)^2+6(2)-5=55\ m/s[/tex]
At t = 4 s
[tex]v=12(4)^2+6(4)-5=211\ m/s[/tex]
(b) Acceleration,
[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6[/tex]
Pu t = 3 s in above equation
So,
[tex]a=24(3)+6\\\\a=78\ m/s^2[/tex]
Hence, (a) (i) v = 55 m/s (ii) v = 211 m/s and (b) 78 m/s²
A small cylinder is rolled along a ruler and completes two revolutions. The circumference is the distance around the outside of a circle. What is the circumference of the cylinder? A 4.4 cm B 5.2 cm C 8.8 cm D 10.2 cm
Answer; 4.4cm
Explanation: There is a ruler upon which the cylinder start from 1.4cm and reaches 10.2cm
distance traveled =10.2-1.4=8.8
since this cylinder is small so the linear distance can be approximately taked as rotational distance(as in case of point charge) so
2x2πxr =8.8
so the circumference will be 2πr=4.4cm
A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She throws the softball with a velocity of 23.5 m/s at an angle of 39.5∘ above the horizontal. When the softball leaves her hand, it is 11.5 m above the water. How far does the softball travel horizontally before it hits the water? Neglect any effects of air resistance when calculating the answer.
Answer:
66.86m
Explanation:
Velocity of ball thrown, u = 23.5 m/s
Initial height of the ball above the water, H = 11.5 m
Angle of projection, θ = 39.5°
Vertical components of veloclty = usinθ
Horizontal components of veloclty = ucosθ
The soft ball hits the water after time 't'
Considering the second equation of motion
S = ut + 1/2at^2........ 1
But since the ball went through motion under gravity ( free fall ) rather than linear motion, then equation 1 can be rewritten as:
H = ut +/- 1/2gt^2
H = - 11.5m
U = usinθ
θ = 39.5°
a = -g = -9.8m/s^2
- 11.5m = 23.5(sin39.5°)t + 1/2(-9.8)t^2
-11.5m = 23.5(0.6360)t - 4.9t^2
-11.5m = 14.946t - 4.9t^2
4.9t^2 -14.946t-11.5m = 0
Since the ball drifted horizontally
D = (Ucosθ)t
Where θ = 39.5°
U = 23.5m/s t=
Alternatively,
horizontal component of the velocity is 23.5 cos 39.5º = 18.1331 m/s
now how long does it take the ball to raise to a peak and fall to the water.
vertical component of velocity = 23.5 sin 39.5º = 14.947m/s
time to reach peak t = v/g = 11.947/9.8 = 1.5252 sec
peak reached above cliff top is
h = ½gt² = ½(9.8)(1.5252)²
= ½×22.797
= 11.3985m
now the ball has to fall 11.3985+ 11.5 = 22.8985m
time to fall from that height is
t = √(2h/g) = √(2• 22.8986/9.8) = 2.1617 sec
add up the two times to get time it is in the air, 2.1617 + 1.5252 = 3.6869
now haw far does the ball travel horizontally in that time
d = vt = 18.1331 ×3.6869= 66.856m
= 66.86m
One glass microscope slide is placed on top of another with their left edges in con- tact and a human hair under the right edge of the upper slide. As a result, a wedge of air exists between the slides. An interference pattern results when monochromatic light is incident on the wedge. What is observed at the left edge of the slides? a. A dark fringe b. A bright fringe c. Impossible to determine
Answer:
A dark fringe
Is there a way for us to control motion
Answer:
They are:
1) change position
2) distract yourself
3) Get fresh air
4) Face the direction you are going.
5) Drink water.
6) Play music.
7) Put your eyes on horizon.
Explanation:
Hope it helps.
Because the neutron has no charge, its mass must be found in some way other than by using a mass spectrometer. When a neutron and a proton meet (assume both to be almost stationary), they combine and form a deuteron, emitting a gamma ray whose energy is 2.2233 MeV. The masses of the proton and the deuteron are 1.007 276 467 u and 2.013 553 212 u, respectively. Based on this data, what is the mass of the neutron
Answer:
Explanation:
Energy of gamma ray = 2.2233 MeV
Let mass of neutron be n amu
mass defect of deuteron = 2.013553212 - ( 1.007 276 467 + n ) u .
in terms of energy this mass defect will be equal to energy of gamma ray
1 amu = 931 MeV
931 [ 2.013553212 - ( 1.007 276 467 + n ) ] = 2.2233
( 1.007 276 467 + n ) - 2.013553212 = .00238807733
n = 1.008664822 amu
so mass of neutron = 1.008664822 amu
A block is attached to the end of a spring. The block is then displaced from its equilibrium position and released. Subsequently, the block moves back and forth on a frictionless surface without any losses due to friction. Which one of the following statements concerning the total mechanical energy of the block-spring system this situation is true?
1. The total mechanical energy is dependent on the maximum displacement during the motion.
2. The total mechanical energy is at its maximum when the block is at its equilibrium position
3. The total mechanical energy is constant as the block moves back and forth.
4. The total mechanical energy is only dependent on the spring constant and the mass of the block.
Answer:
The correct option is;
3. The total mechanical energy is constant as the block moves back and forth
Explanation:
The total mechanical energy is the sum of the potential and kinetic energies of the system
For a system that is isolated from the effects of external forces, but being acted upon by the internal conservative forces within the system, the total mechanical energy is constant
For a black and spring system, we have total mechanical energy, E = 1/2×K×A².
Where;
K = Constant
A = The amplitude of motion
Therefore, where there is no loss to friction, with A, remaining constant, the total mechanical energy will be constant.
Light travels at a speed of 2.998*108 m/s. Light takes approximately 3.25 minutes to travel from the Sun to reach a planet. Calculate the distance from the Sun to this planet in meters. Give your answer to 0 decimal places.
Answer:
585×10⁸ m
Explanation:
Distance = rate × time
d = (2.998×10⁸ m/s) (3.25 min) (60 s/min)
d = 585×10⁸ m
