What happens when a sound wave passes from oil into air? A. the sound wave slows down and bounces back B. the sound wave slows down and bends C. the sound wave speeds up and bounces back D. the sound wave speeds up and bends

Answer:

OPTION (C)

Explanation:

m(magnification) = -0.4 means a real, inverted and diminished image is formed in front of the mirror.

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.

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.

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 ;)❤❤❤

Answer: A. 2.7 x 10^2 joules

Explanation: I’m sorry for the guy above me!

Answer:

PFA

:-)

Explanation:

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

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

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.

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.

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

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

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.

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%

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

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.

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  

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Brainliest

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

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

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

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

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

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.

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:

The 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].

The 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.

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].

Explanation:

To find the average of these numbers, we just have to add the three numbers together and divide by 3.

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.

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

Answer:1m

Explanation:

2n=0.4m

5n=?

5n×0.4/2n=1m

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.

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