Oil is optically denser than water. When sound/light goes from optically denser medium to optically rarer medium, their velocity increase and they moves away for normal.
Appropriate Answer:The sound wave speeds up and bends
[tex] \Large{ \underline{ \boxed{ \pink{ \bf{Option \: (D)}}}}}[/tex]
As, In optics we learnt that light undergoes refraction when travels from medium of different densities. Similarly, Sound also follows the law of refraction.
It is due to the change of speed of water in different mediums, This makes it speed up or down depending upon the medium and their densities.━━━━━━━━━━━━━━━━━━━━
Answer:
the sound wave slows down and bends
Explanation:
a p e x
If the mass of a ball is 50g on a height of 8m. Calculate the kinetic energy when it has a velocity of 3m/s.
Answer:
kinetic energy is 1/2mv^2.
which is 1/2×0.05×3^2
1/2×0.05×9.
1/2×0.45=
0.45÷2=0.225~0.23J
ASAP TWENTY POINTS What type of image is formed by a mirror if m = -0.4?
Answer:
OPTION (C)
Explanation:
m(magnification) = -0.4 means a real, inverted and diminished image is formed in front of the mirror.
Can you solve this question please help me with this
Answer:
Explanation:
The velocity ratio of a wheel and axle is the ratio of the radius (R) of the wheel to the radius (r) of the axle. It is expressed as;
VR = R/r
Since radius = diameter/2
VR = (D/2)/(d/2)
VR = D/d
D is the diameter of the wheel and 'd' is the diameter of the axle.
Given VR = 3 and d = 5cm
3 = D/5
D = 15 cm
If the diameter of the wheel is 15cm, the radius of the wheel will be 15/2 = 7.5cm.
b) Workdone by the load = Load * distance moved by load
Given load = 60kg
Distance moved by load = 2π*radius of axle
Distance moved by load = 2π(0.025) = 0.157
workdone by load = 60* 0.157 = 9.42J
Effort = Workdone by load/distance moved by the wheel
Effort = 9.42/2π(0.075)
Effort = 9.42/0.471
Effort = 20kg
Hence the effort applied is 20kg
c) MA = Load/Effort
MA = 60/20
MA = 3
d) Efficiency = MA/VR * 100%
Efficiency = 3/3 * 100%
Efficiency = 100%
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
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
plz help
I need it fast
Answer:
perpendicular to
Explanation:
it means perpendicular to .....should u come across something like this / / , this one means parallel to .....
Answer:
perpendicular
Explanation:
Some of the most popular symbols are:
Heart symbol: this represents love, compassion and health.
Dove symbol: this represents peace, love, and calm.
Raven symbol: this represents death and doom.
Tree symbol: this represents growth, nature, stability, and eternal life.
Owl symbol: this represents wisdom and intelligence.
Determine the gradient and the co-ordinates of the x and y intercept of line whose equation is 2y + 3x = 1
Answer:
The x - intercept is 1/3
The y - intercept is 1/2
The gradient is -3/2
Explanation:
To find the x - intercept of the equation 2y + 3x = 1, we find the value of x when y = 0. So,
2y + 3x = 1
2(0) + 3x = 1
0 + 3x = 1
3x = 1
x = 1/3
So, the x - intercept is 1/3
To find the y - intercept of the equation 2y + 3x = 1, we find the value of y when x = 0. So,
2y + 3x = 1
2y + 3(0) = 1
2y + 0 = 1
2y = 1
y = 1/2
So, the y - intercept is 1/2
To find the gradient of the equation 2y + 3x = 1, we re-write it in gradient intercept form by making y subject of the formula.
So, 2y + 3x = 1
2y = -3x + 1
y = -3x/2 + 1/2
The coefficient of x which equals -3/2 is the gradient.
The gradient is -3/2
A 26-foot ladder is placed against a wall. If the top of the ladder is sliding down the wall at -2 feet per second (note that the rate is negative because the height is decreasing). At what rate is the bottom of the ladder moving away from the wall when the bottom of the ladder is 10 feet away from the wall?
Answer:
Dx/dt = 4,8 f/s
Explanation:
The ladder placed against a wall, and the ground formed a right triangle
with x and h the legs and L the hypothenuse
Then
L² = x² + h² (1)
L = 26 f
Taking differentials on both sides of the equation we get
0 = 2x Dx/dt + 2h Dh/dt (1)
In this equation
x = 10 distance between the bottom of the ladder and the when we need to find, the rate of the ladder moving away from the wall
Dx/dt is the rate we are looking for
h = ? The height of the ladder when x = 10
As L² = x² + h²
h² = L² - x²
h² = (26)² - (10)²
h² = 676 - 100
h² = 576
h = 24 f
Then equation (1)
0 = 2x Dx/dt + 2h Dh/dt
2xDx/dt = - 2h Dh/dt
10 Dx/dt = - 24 ( -2 ) ( Note the movement of the ladder is downwards)
Dx/dt = 48/10
Dx/dt = 4,8 f/s
You have a reservoir held at a constant temperature of –30°C. You add 400 J of heat to the reservoir. If you have another reservoir at 0°C, how much heat must you add to it so that both reservoirs have the same increase in entropy?
Answer:
449.38 J
Explanation:
ΔS = ΔQ/T
Where ΔS = entropy change
Q = quantity of heat
T = temperature
First reservoir :
T = –30°C = - 30 + 273 = 243K
Q = 400 J
Second reservoir :
T = 0°C = 273K
Q =?
To have same increase in entropy for both reservoirs :
Q/T of first reservoir = Q/T of second reservoir
400/243 = Q/273
243 * Q = 400 * 273
Q = (400 * 273) / 243
Q = 109,200 / 243
Q = 449.38271
Q = 449.38 J
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].
Need help finding the average speed.
Explanation:
To find the average of these numbers, we just have to add the three numbers together and divide by 3.
2.07 + 0. 74 + 1.33 = 4.14. 4.14 / 3 = 1.381.09 + 1.40 + 0.31 = 2.8. 2.8 / 3 ≈ 9.3333333/ 9 1/30.95 + 1.61 + 0.56 = 3.12 / 3 = 1.040.81 + 1.89 + 1.08 = 3.78 / 3 = 1.26The acceleration due to gravity near Earth ... Select one: a. varies inversely with the distance from the center of Earth. by. varies inversely with the square of the distance from the center of Earth. c. is a constant that is independent of altitude d. varies directly with the distance from the center of Earth.
Answer:
b. varies inversely with the square of the distance from the center of Earth.
Explanation:
Comparing the Newton's law of universal gravitation and second law of motion;
from Newton's second law of motion,
F = ma ............. 1
from New ton's law of universal gravitation,
F = [tex]\frac{GMm}{r^{2} }[/tex] ........... 2
Equating 1 and 2, we have;
mg = [tex]\frac{GMm}{r^{2} }[/tex]
g = [tex]\frac{GM}{r^{2} }[/tex]
Therefore, the acceleration due to gravity near Earth, g, is inversely proportional to the square of the distance from the center of Earth.
Atoms of the same element will always have the same number of Question Blank but will have different numbers of Question Blank if their mass numbers are different.
Answer:
proton and neutron respectively.
Explanation:
Atoms of the same element will always have the same number of proton but will have different numbers of neutron if their mass numbers are different.
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
A missile is moving 1350 m/s at a 25° angle it needs to hit a target 23,500 m away in a 55° direction in 10.2 seconds what is the magnitude of its final velocity
Answer:
3504 m/s
Explanation:
Let x be the horizontal component of distance
y - vertical component of distance
t-time
ax- horizontal component of acceleration
ay-Vertical component of acceleration
Vx-horizontal component of velocity
Vy-Vertical component of velocity
horizontally: x = V_x ×t + ½×a_x×t²
plugging the values we get
23500× cos 55º = 1350×cos25.0º × 10.20 + ½×a_x× (10.20)²
⇒ax = 19.2 m/s²
Moreover,
V'x = V_x + a_x×t = 1350×cos25.0º + 19.2×10.20= 1419 m/s
similarly in vertical direction:
y = V_y×t + ½×a_y×t²
23500×sin55º = 1350×sin25.0º×10.20s + ½×a_y×(10.20)²
⇒a_y = 258 m/s²
Also,
V'y = V_y + a_y×t = 1350×sin25.0º + 258×10.20 = 3204 m/s
Therefore
V = √(V'x² + V'y²) = 3504 m/s
therefore, magnitude of final velocity of missile=3504 m/s
THANKS
3
5.0
(7 votes)
Not the answer you're looking for?
FIND MORE
Previous
Next
Manetho
Manetho
Rank: Ambitious
29830/500
0/5
Your influence
1
14h : 28m
Answering Streak
Answer a question in time to keep your streak going and the unlocks coming.
Fun fact
An apple, potato, and onion all taste the same if you eat them with your nose plugged
1
2
2
Get +4
bonus
2
3
3
Get +6
bonus
3
Last 7 days
Overall
Your stats from the last 7 days compared with the previous 7 days
Popularity
881
People learning from your answers.
+4%
Brainliest
A car accelerate uniformly from rest at 5m/s2 . Determine it's speed after 10s
Answer:
50m/s.
Explanation:
Let's take acceleration as A and speed as S:
A = 5m/s²
S = A × 10s = 5 × 10 = 50m/s
The answer is 50m/s.
Answer:
50m/s
Explanation:
Given:
a=5m/s^2
t=10s
Required:
v=?
Formula:
a=v/t
Solution:
a=v/t
v=a*t
v=5m/s^2*10s
v=50m/s
Hope this helps ;)❤❤❤
What is the value of work done on an object when a 0.1x102–newton force moves it 30 meters and the angle between the force and the displacement is 25°?
Answer: A. 2.7 x 10^2 joules
Explanation: I’m sorry for the guy above me!
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.
Why does it take a longer time for a kilogram of water than a kilogram of copper to reach the same temperature?
Answer:
Since water has a higher specific heat than copper.
Explanation:
Dimensionally speaking, the specific heat of a material ([tex]c[/tex]) is represented by:
[tex][c] = \frac{[Energy]}{[Mass]\cdot [Temperature]}[/tex]
The specific heats of water and copper are [tex]4186\,\frac{J}{kg\cdot ^{\circ}C}[/tex] and [tex]390\,\frac{J}{kg\cdot ^{\circ}C}[/tex], respectively. Let suppose that temperature change and masses of water and copper are the same. Then, a kilogram of water takes a longer time than a kilogram of copper since the first has a higher specific heat.
A box with mass of 2 kg is pushed directly horizontally over a horizontal surface (with friction) at a constant speed of 10 m/s. The force of the push is 60 N. How much thermal energy is generated pushing the box a distance of 15 m
Answer:
E= 600 W
Explanation:
Given that
m = 2 kg
Speed , v= 10 m/s
Force , F= 60 N
Given that box is moving with constant velocity, it means that friction force will be 60 N.
f = 60 N
Therefore total energy generated
E= f x v
E= 60 x 10 = 600 W
E= 600 W
Thus the answer will be 600 W.
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.
Will mark as BRAINLIEST..... A balloon is ascending at the rate of 4.9 m/s. A packet is dropped from from the balloon when situated at a height of 100m. How long does it take the packet to reach the ground ? What is it's final velocity ?
Answer:
PFA
:-)
Explanation:
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.
Read more at; https://brainly.com/question/15995216
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
A body of mass 20 kh changes its velocity from 5m/s to 17 m/s in 3 seconds. find the force applied on this body .
please answer fast
Answer:
80 N.
Explanation:
The following data were obtained from the question:
Mass (m) = 20 Kg
Initial velocity (u) = 5 m/s
Final velocity (v) = 17 m/s
Time (t) = 3 secs
Force (F) =.?
Next, we shall determine the acceleration of the body.
Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:
Acceleration = change of velocity /time
Acceleration (a) = (final velocity (v) – initial velocity (u)) / time (t)
a = (v – u) /t
With the above formula, we can obtain the acceleration of the body as follow:
Initial velocity (u) = 5 m/s
Final velocity (v) = 17 m/s
Time (t) = 3 secs
Acceleration (a) =?
a = (v – u) /t
a = (17 – 5) /3
a = 12/3
a = 4 m/s²
Finally, we shall determine the force applied to the body as follow:
Force (F) = mass (m) x acceleration (a)
F = ma
Mass (m) = 20 Kg
Acceleration (a) = 4 m/s²
Force (F) =.?
F = ma
F = 20 x 4
F = 80 M.
Therefore, the force applied on the body is 80 N.
[tex]r=s^2/t^2[/tex] 1. If s is tripled and t stays constant, r is multiplied by... 2. If t is doubled, and s stays constant, r is multiplies by...
Answer:
9 and 4
Explanation:
The relation is:
● r = s^2 / t^2
Triplind s means multiplying it by 3. Since it's an equation we should multiply both sides by the same number
Let k be the number we should multiply by r
●k* r = (3s)^2 / t^2
●k* r = 9s^2 /t^2
We have multiplied s^2 by 9 so we should do the same for r.
k = 9
■■■■■■■■■■■■■■■■■■■■■■■■■■
Doubling t means multiplying it by 2.
Let x be the number we shoukd multiply by r.
● x* r = s^2/(2t)^2
● x*r = s^2/ 4t^2
We have multiplied t^2 by 4 so we should do the same for r.
x= 4
A drone has been designed that can do monitoring and surveillance at considerable heights due to its light weight of 0.800 kg. For this purpose, they are doing a test to determine its maximum height and they make it start in a vertical direction, using its thrusters it manages to achieve a thrust of 35.0 N during the first 6.00 s. until the battery runs out. What was the maximum height that the drone reached?
Answer:
2660 m
Explanation:
Sum of the forces in the first 6.00 s:
∑F = ma
F − mg = ma
35.0 N − (0.800 kg) (10 m/s²) = (0.800 kg) a
a = 33.75 m/s²
The height it reaches during this time is:
Δy = v₀ t + ½ at²
Δy = (0 m/s) (6.00 s) + ½ (33.75 m/s²) (6.00 s)²
Δy = 607.5 m
The velocity it reaches is:
v = at + v₀
v = (33.75 m/s²) (6.00 s) + 0 m/s
v = 202.5 m/s
After the battery runs out, the drone is in free fall. At the highest point, the velocity is 0. The height at this point is:
v² = v₀² + 2aΔy
(0 m/s)² = (202.5 m/s)² + 2 (-10 m/s²) (h − 607.5)
h ≈ 2660 m
A spring of initial length 35 cm acquires a length of 55 cm when we hang from it a mass of 3.5 kg. Calculate:
a) The elastic constant of the spring.
b) The length of the spring when we hang a mass of 5 Kg.
Answer:
the elastic constant of the spring=1.715
the length of the spring=0.28
Explanation:
we know that according to hooks law
F=-k x
F= force
k= elastic constant
x= extension or compression
given
length change from 35cm to 55 cm so delta x = L2-L1= 55-35=20 cm
now to find k we need F and F =ma
M for part a is 3.5 kg
so F=3.5 kg *9.8=34.3
now k=F/x
k=34.3/20=1.715 N/cm=171.5 N/m
now to find length given mass is 5 kg so
F= ma
F=5*9.8=49 N
so x =F/k
x=49/171.5
x=0.28
A storm is moving east towards your house at an average speed of 35 km / hr. If the storm is currently 80 km from your house, how much time do you expect it to arrive
Answer:
The expected time is 2.28 hours.
Explanation:
The speed of storm = 35 km/hr
The distance between the house and the storm = 80 km.
Now, we have to find the time taken by storm to arrive at the house. Here, we can determine the time by dividing the distance with speed.
The time, taken by storm = Distance/speed
The time, taken by storm = 80 / 35
The time, taken by storm = 2.28 hours.
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
