An 8.0g bullet, moving at 400 m/s, goes through a stationary block of wood in 4.0 x 10^-4s, emerging at a speed of 100 m/s. (a) what average force did the wood exert on the bullet? (b) how thick is the wood?

Answer:

HELLO BELOW IS THE ATTACHED DIAGRAM USED TO ANSWER YOUR QUESTION AS IT WAS MISSING

A) 1270⁰c

B) 65%

C) 1320⁰c

D)  62%

Explanation:

Nickel alloy composition : 50 wt% Ni - 50 wt%

initial temperature = 1200⁰c = 2190⁰F

A) The temperature at which the first liquid phase form

from the attached diagram the temperature at which the first liquid if formed is 1270⁰c ( at point 2 )

B ) The composition of this liquid phase ( THE FIRST LIQUID )

the composition is found at point 3

wt % of Nickel = 35%, wt% of copper = 100 - 35 = 65%

C ) The temperature at which the alloy melts completely

from the attached diagram the temperature = 1320⁰c ( point 4 )

D)  The composition of the last solid remaining prior to complete melting

this can be gotten at point 5 and calculated as

wt % of Ni = 62%

wt % of Cu = 100 - 62 = 38%

Answer:

[tex]\Large \boxed{\mathrm{5.67 \ seconds }}[/tex]

Explanation:

[tex]\displaystyle \mathrm{acceleration \ = \ \frac{final \ velocity - initial \ velocity }{elapsed \ time}}[/tex]

[tex]\displaystyle A \ = \ \frac{V_f - V_i }{t}[/tex]

[tex]\displaystyle 3 \ = \ \frac{25 - 8 }{t}[/tex]

[tex]\displaystyle 3 \ = \ \frac{17 }{t}[/tex]

[tex]\displaystyle t \ = \ \frac{17 }{3} \approx 5.67[/tex]

Answer: React with the skin

Explanation:

lab safety test Safety goggles and an apron must be worn when handling chemicals labeled corrosive because they React with the skin

Answer:

lathe

perfectly symmetrical

Answer: A lathe is used to turn wood . It rotates and carves wood to produce a perfectly symmetrical form .

First one is lathe .

Second one is perfectly symmetrical .

Explanation: I just took the test on Plato and had 100%

This question is incomplete; here is the complete question:

Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the frequency as 3 hertz, which statement about the wave is accurate?

A. The wave has traveled 32.4 cm in 3 seconds.

B. The wave has traveled 32.4 cm in 9 seconds.

C. The wave has traveled 97.2 cm in 3 seconds.

D. The wave has traveled 97.2 cm in 1 second.

The answer to this question is D. The wave has traveled 97.2 cm in 1 second.

Explanation:

The frequency of a wave, which is in this case 3 hertz, represents the number of waves that go through a point during 1 second. According to this, if the frequency of the wave is 3 hertz this means in 1 second there were 3 waves. Moreover, if you multiply the wavelength (32.4cm) by the frequency (3) you will know the distance the wave traveled in 1 second: 32.4 x 3 =  97.2 cm. This makes option D the correct one as the distance in 1 second was 97.2 cm.

Answer:Is D!

Explanation:TEAT(Sorry) -_-*

Answer:

22.2 m/s or 80 km/h

Explanation:

Given that

Distance travelled by the car, d = 240 km

Time taken by the car, t = 3 hours.

Speed of the car, v = ? m/s

for easy calculations, we will be converting the units to meters and seconds respectively.

240 km to meters would be

240 * 1000 m = 240000 m

3 hrs to seconds would be

3 * 60 mins * 60 seconds = 10800 s

now, we have our distance and time to be

d = 240000 m

t = 10800 s

speed is defined as the ratio of distance with respect to time taken, effectively,

Speed = distance/time

speed, v= 240000 / 10800

v = 22.2 m/s

therefore, the speed of the car during the time is 22.2 m/s, or if the speed is needed in km/h, we can convert it

22.2 * 3600/1000 =

80 km/h

Answer:

a)0.51°

b)1.47×10^-4m

Explanation:

a)for a single slit experiment, the minima that has an angle of θ towards the centre needs to satisfy the expression below.

bsin(θ)= mλ.........................(*)

Where b= width of the slit

The distance on the screen from Central angle can be expressed as

Sin(θ)= y/d............. (**)

d and y is the horizontal distance between slit and screen

If we input eqn(**) into equation (*) we have

y= mλd/b................(z)

In order to find angle (θ) we have

(θ)= sin-(1.99×10^-2)/2.25

= 0.51°

Therefore, angle of diffraction θ of the second minimum is 0.51°

b)to find the width of the sloth using eqn(z) by substitute the values, we have

b= (2)(649×10^-9)(2.25)/1.99×10^-2

b= 1.47×10^-4m

Therefore, the width of the slit is 1.47×10^-4m

Answer:

kinetic and potential

Answer:

first part the skater goes down a constant slope ramp, initially he has Newton's second law

pply Newton's third law, the normal is the reaction to the support of the body on the surface

the ramp shoots off.  axis becomes zero and therefore with Newton's first law its speed

Explanation:

It is the description of this movement let's write Newton's laws.

* The first law that a body goes at constant speed or zero if the sum of the external forces is zero

* the second law is F = m a

* The third law states that the forces act in pairs of equal magnitude and opposite direction, one applied to each body.

Let's apply these laws to our case

In the first part the skater goes down a constant slope ramp, initially he has Newton's second law when he accelerates from the initial velocity of zero to a terminal velocity.

The expression for this is

             Wₓ - fr = ma

             W sin θ - μ W cos θ = m a

             W = mg

             g (sin θ - μ cos) = a

the value of the coefficient of kinetic friction depends on the condition of the surface, dry, wet or muddy

This is Newton's second law

On the Y axis, which is perpendicular to the ramp we have

            N- [tex]W_{y}[/tex] = 0

If we apply Newton's third law, the normal is the reaction to the support of the body on the surface, note that it can be different from the weight.

In the second part when he is on the ramp.

In the ramp the skater enters with a speed v, suppose that the ramp has an incline so that the skater can jump, in this case the angle is positive with respect to the axis x

In this case the analysis is similar to the previous one

Newton's second law gives the acceleration of the skater, who when he reaches the end of the ramp shoots off.

 At this point the force in the x (horizontal) axis becomes zero and therefore with Newton's first law its speed this axis remains constant and the force in the y axis is the force of gravity and has an acceleration that changes if velocity according to Newton's second law

Answer:look at explanations

Explanation:

Answer:

answer is B

Explanation:

Acc. to Ohm's Law:

V = I.R and therefore R=( V/I )

For line A, R = 6/2 = 3

line B, R = 4/2 = 2 ( for all the points plotted on this line the ratio of V and I is 2)

Line C = 2/2 = 1

Line D < 1

Hence, the correct answer is Line B :-)

Answer:

(II). The fractions in simplest form

[tex]\dfrac{31}{35}[/tex], [tex]\dfrac{91}{165}[/tex] and [tex]\dfrac{13}{5}[/tex]

(III). The fractions in ascending order

[tex]\dfrac{91}{165}<\dfrac{31}{35}<\dfrac{13}{5}[/tex]

Explanation:

Given that,

(I). Represent the fractions in pictorial form

(II). Write the fractions in simplest form.

(III). Arrange them in ascending order.

Suppose, The fractions in pictorial form

(a). [tex]\dfrac{3}{5}+\dfrac{2}{7}[/tex]

(b). [tex]\dfrac{9}{11}-\dfrac{4}{15}[/tex]

(c). [tex]\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}[/tex]

We need to write in simplest form

Using given fraction

(a). [tex]\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{3\times7}{5\times7}+\dfrac{2\times5}{7\times5}[/tex]

[tex]\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{21}{35}+\dfrac{10}{35}[/tex]

[tex]\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{21+10}{35}[/tex]

[tex]\dfrac{3}{5}+\dfrac{2}{7}=\dfrac{31}{35}[/tex]

(b). [tex]\dfrac{9}{11}-\dfrac{4}{15}=\dfrac{9\times15}{11\times15}-\dfrac{4\times11}{15\times11}[/tex]

[tex]\dfrac{9}{11}-\dfrac{4}{15}=\dfrac{135-44}{165}[/tex]

[tex]\dfrac{9}{11}-\dfrac{4}{15}=\dfrac{91}{165}[/tex]

(c). [tex]\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}=\dfrac{7}{10}+\dfrac{2\times2}{5\times2}+\dfrac{3\times5}{5\times2}[/tex]

[tex]\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}=\dfrac{7}{10}+\dfrac{4+15}{10}[/tex]

[tex]\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}=\dfrac{7+4+15}{10}[/tex]

[tex]\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}=\dfrac{26}{10}[/tex]

[tex]\dfrac{7}{10}+\dfrac{2}{5}+\dfrac{3}{2}=\dfrac{13}{5}[/tex]

We need to arrange them in ascending order

Using simplest form

The simplest fraction in ascending order

[tex]\dfrac{91}{165}<\dfrac{31}{35}<\dfrac{13}{5}[/tex]

Hence, This is required solution.

Answer:

I assume its c. Since its talking about testing.

Explanation:

Answer:

The answer is test of durability

Explanation:

Answer:

  2.4583 ± 0.0207 seconds

Explanation:

The time period of a pendulum is approximately given by the formula ...

  T = 2π√(L/g)

The maximum period will be achieved when length is longest and gravity is smallest:

  Tmax = 2π√(1.51/9.7) ≈ 2.47903 . . . seconds

The minimum period will be achieved for the opposite conditions: minimum length and maximum gravity:

  Tmin = 2π√(1.49/9.9) ≈ 2.43756 . . . seconds

If we want to express the uncertainty using a symmetrical range, we need to find half their sum and half their difference.

  T = (2.47903 +2.43756)/2 ± (2.47903 -2.43756)/2

  T ≈ 2.4583 ± 0.0207 . . . seconds

__

We have about 2+ significant digits in the given parameters, so the time might be rounded to 2.46±0.02 seconds.

Answer:

17.15 m

Explanation:

Assuming the well is empty (full of air instead of water), the speed of sound is v = 343 m/s.

The water well acts as a pipe closed at one end.  Therefore, the depth of the water well must be an odd multiple of a quarter of the resonant wavelength.

L = (2n − 1) λ/4, n = 1, 2, 3, etc.

v = λf, so λ = v/f.  Substituting:

L = (2n − 1) v/(4f)

Solving for frequency:

f = (2n − 1) v/(4L)

The difference between two successive resonant frequencies is therefore:

Δf = (2(n+1) − 1) v/(4L) − (2n − 1) v/(4L)

Δf = (2n + 1) v/(4L) − (2n − 1) v/(4L)

Δf = 2 v/(4L)

Δf = v/(2L)

Plugging in values:

50 Hz − 40 Hz = 343 m/s / (2L)

2L = 34.3 m

L = 17.15 m

Answer:

The first answer is W and Z, since they appear to be a period apart. Dont know the second question. I did what I could, hope someone can answer the second.

Answer: 19 meters.

Explanation:

We want to find the total displacement between t = 2s and t = 4s.

To do it, we can integrate our function, first write our velocity equation.

for t ≤ 3s, we have a linear equation, let's write it:

A linear relationship can be written as:

y = a*t + b

where a is the slope and b is the y-axis intercept.

For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:

a = (y2 - y1)/(x2 - x1).

Now we can see that our line passes through the points (1, 0) and (0, -2)

then the slope is:

a = (0 -(-2)/(1 - 0) = 2/1 = 2

and knowing that when t = 0s, v(0s) = -2m/s, then our equation is:

v(t) = (2m/s^2)*t - 2m/s for t ≤ 3s

now, for t ≥3s the equation is constant, v(t) = 4m/s.

then we have

v(t) = (2m/s^2)*t - 2m/s -------if t ≤ 3s

v(t) = 4m/s ----- if t ≥ 3s

Now we integrate over time to get the position:

for t ≤ 3s we have:

p(t) = (1/2)*(2m/s^2)*t^2 - 2m/s*t + C

where C is a constant of integration, as we are calculating the displacement  this constant actually does not matter, so we can use C = 0m

p(t) = (1m/s^2)*t^2 - 2m/s*t  for t ≤ 3s

and p(3s) =  (1m/s^2)*3s^2 - 2m/s*3s = 9m - 6m = 3m is the initial position of the other part of the function.

for t ≥ 3s we have:

p(t) = 4m/s*t + p(3s) = 4m/s*t + 3m

then the position equation is:

p(t) = (1m/s^2)*t^2 - 2m/s*t  ---- t ≤ 3s

p(t) = 4m/s*t + 3m --- if t ≥ 3s

Now the displacement will be:

p(4s) - p(2s) where for each time, you need to use the correct function:

p(4s) = 4m/s*4s + 3m = 16m + 3m = 19m

p(2s) =  (1m/s^2)*2s^2 - 2m/s*2s = 4m - 4m = 0m

p(4s) - p(2s) = 19m - 0m = 19m

The butterfly displacement x from t=2 to 4s is 19 meters.

The spacing between two specified points is represented by the one-dimensional quantity of displacement (symbolised as d or s), commonly known as length or distance.

The total displacement between t = 2s and t = 4s.

Integrate our function, the velocity equation.

for t ≤ 3s, we have a linear equation, let's write it:

A linear relationship can be written as:

y = a x t + b

where a is the slope and b is the y-axis intercept.

For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:

a = (y2 - y1)/(x2 - x1).

The line passes through the points (1, 0) and (0, -2)

The slope is:

a = (0 -(-2)/(1 - 0) = 2/1 = 2

When t = 0s, v(0s) = -2m/s, then our equation is:

v(t) = (2m/s²) x t - 2m/s for t ≤ 3s

now, for t ≥3s the equation is constant, v(t) = 4m/s.

v(t) = (2m/s²) x t - 2m/s -------if t ≤ 3s

v(t) = 4m/s ----- if t ≥ 3s

Now we integrate over time to get the position:

for t ≤ 3s we have:

p(t) = (1/2) x (2m/s²) x t^2 - 2m/s x t + C

where C is a constant of integration, to calculate the displacement  this constant actually does not matter,

p(t) = (1m/s²)*t^2 - 2m/s x t  for t ≤ 3s

and p(3s) =  (1m/s^2) x 3s² - 2m/s x 3s = 9m - 6m = 3m is the initial position of the other part of the function.

for t ≥ 3s we have:

p(t) = 4m/s x t + p(3s) = 4m/s x t + 3m

then the position equation is:

p(t) = (1m/s^2) x t² - 2m/s x t  ---- t ≤ 3s

p(t) = 4m/s x t + 3m --- if t ≥ 3s

Now the displacement will be:

p(4s) - p(2s) where for each time, you need to use the correct function:

p(4s) = 4m/s x 4s + 3m = 16m + 3m = 19m

p(2s) =  (1m/s²) x 2s²- 2m/s x 2s = 4m - 4m = 0m

p(4s) - p(2s) = 19m - 0m = 19m

Thus, the displacement is 19 m.

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Answer:

Water waves are an example of waves that involve a combination of both longitudinal and transverse motions. As a wave travels through the waver, the particles travel in clockwise circles. The radius of the circles decreases as the depth into the water increases.

Answer:

Height of pole = Height of giraffe

Explanation:

Given:

Height of giraffe = 20 ft

Height of pole = 240 inch

Find:

Which is taller

Computation:

Height of giraffe = 20 ft

We know that 1 ft = 12 inch

So,

Height of giraffe = 20 × 12 inch

Height of giraffe = 240 inch

and

Height of pole = 240 inch

Height of pole = Height of giraffe

Answer:

(I). The Schwarzschild radius is [tex]2.94\times10^{8}\ km[/tex]

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is [tex]1.1\times10^{-7}\ km[/tex]

(IV). The Schwarzschild radius is [tex]7.4\times10^{-29}\ km[/tex]

Explanation:

Given that,

Mass of black hole [tex]m= 1\times10^{8} M_{sun}[/tex]

(I). We need to calculate the Schwarzschild radius

Using formula of radius

[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}[/tex]

[tex]R_{g}=2.94\times10^{8}\ km[/tex]

(II). Mass of block hole [tex]m= 6 M_{sun}[/tex]

We need to calculate the Schwarzschild radius

Using formula of radius

[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]

Put the value into the formula

[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}[/tex]

[tex]R_{g}=17.7\ km[/tex]

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]

Put the value into the formula

[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}[/tex]

[tex]R_{g}=1.1\times10^{-7}\ km[/tex]

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]

Put the value into the formula

[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}[/tex]

[tex]R_{g}=7.4\times10^{-29}\ km[/tex]

Hence, (I). The Schwarzschild radius is [tex]2.94\times10^{8}\ km[/tex]

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is [tex]1.1\times10^{-7}\ km[/tex]

(IV). The Schwarzschild radius is [tex]7.4\times10^{-29}\ km[/tex]

The energy conservation allows to find the Schwarschild radius for several bodies of different masses are:

  1)  Black hole quasar is:  r = 2.9 10⁸ km

  2) Blsck hole supernove is:  r = 17.7 km

  3)  Mini black hole is:   r = 1.1 10⁻⁷ km

  4) Human body is:  r=  7 10⁻²⁹ km

The schwarschild radius is defined as the distance from a black hole center at radius  which the escape velocity is equal to the light speed, in some cases it is also called the event horizon.

Let's use Newton's second law where force is the universal law of attraction and acceleration is centripetal.

        F = ma

        F = [tex]G \frac{Mm}{r^2}[/tex]        

Where F is the force, M the mass of the black hole, m the handle of the body, r the radius and v the speed of the body.

The energy of the gravitational field is

        F = [tex]- \frac{dU}{dr }[/tex]  

         U = [tex]-G \frac{Mm}{r}[/tex]  

Let's use conservation of energy

        Em₀ = K + U = ½ m v² -  [tex]G \frac{Mm}{r}[/tex]  

In infinity the energy

        Em_f = 0

energy is conserved

       Em₀ = Em_f  

       ½ m v² - [tex]G \frac{Mm }{r}[/tex]  = 0

       r = [tex]\frac{2GM}{v^2}[/tex]

From the definition of the Schwarschild radius this speed is equal to the light speed

        v = c

        r = [tex]\frac{2GM}{c^2 }[/tex]  

They ask to calculate the radius for several cases of different mass, claculate the constant value

      V = [tex]\frac{2 \ 6.67 \ 10^{-11} }{(3 \ 10^8) ^2 }[/tex]  

      V =  1.482 10⁻²⁷

1) A black hole of mass M = 1 10⁸ [tex]M_{sum}[/tex]

The tabulated mass of the sun is [tex]M_{sum}[/tex] = 1.989 10³⁰ kg

Let's  substitute

        r =  1.482 10⁻²⁷   1 10⁸ 1.989 10³⁰

        r = 2.94 10⁸ km

With two significant figures

        r = 2.9 10⁸ km

2) A black hole of mass M = 6 [tex]M_{sum}[/tex]

        r =  1.482 10⁻²⁷ 6 1.989 10-30

        r = 17.7 km

3) a mini black hole with the mass of the moon

    Tabulated mass of the moon M = 7.35 10²² kg

       r =  1.482 10⁻²⁷  7.35 10²²    

       r = 1.1 10⁻⁷ km

4) A person of M = 50 kg

    r =  1.482 10⁻²⁷ 50

    r=  7 10-29 km

In conclusion using the conservation of energy we can find the Schwarschild radius for several bodies of different masses are:

  1)  Black hole quasar is:  r = 2.9 10⁸ km

  2) Blsck hole supernove is:  r = 17.7 km

  3)  Mini black hole is:   r = 1.1 10⁻⁷ km

  4) Human body is:  r=  7 10⁻²⁹ km

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Answer:

Explanation:

Average rate = gallons / minutes

If 12.5gallons in 5mins

Xgallons in 1mins

12.5/5 = 2.5gallons per minute

Explanation:

current = velocity/resistance

I = V/R

15/4

current = 3.75A

hope this helps...

Answer:

Answer:

A. It is always a positive force

Explanation:

Hooke's law describes the relation between an applied force and extension ability of an elastic material. The law states that provided the elastic limit, e, of a material is not exceeded, the force, F, applied is proportional to the extension, x, provided temperature is constant.

i.e F = - kx

where k is the constant of proportionality, and the minus sign implies that the force is a restoring force.

The applied force can either be compressing or stretching force.

Answer:

27.34 (no unit)

Explanation:

26.975*82.33%+29.018*17.67%

=27.34

Answer:

Average acceleration = Displacement / (time)^2

Explanation:

The unit for acceleration is

[tex]m {s}^{ - 2} [/tex]

Displacement = m

Time = s

Hence the units of displacement and time should be manipulated to get the unit of acceleration.

You can't. You can only find average velocity.

But if you also know that initial velocity is zero ... the object started from rest ... then

Avg acceleration =

2 x displacement / time-squared

Answer:

Protons

Neutrons

Electrons.

Protons + charge(positively charged)

Neurons no charge.

Electrons - ( negatively charged).

Explanation:

Atom is the smallest unit of a matter that constitutes chemical elements.

It is composed of three particles which are protons,neutrons and electrons.

Protons is positively charged,neuron has no charge and electrons is negatively charged.

Proton and neuron are found in the nucleus while electron is found outside the neuron.

Answer:

[tex]\boxed{0}[/tex]

Explanation:

Momentum is the measure of mass in motion.

[tex]\sf momentum = mass \times velocity[/tex]

An object at rest has a velocity of 0.

[tex]p=mv[/tex]

[tex]p = 25 \times 0[/tex]

[tex]p=0[/tex]

The momentum of an object at rest is always 0.

Answer:

1) The density of the object = 2500 kg/m³

2) The density of the oil = 1250 kg/m³

Explanation:

1) The information relating to the question are;

The mass of the object in air = 0.250 kgf

The mass of the object in water = 0.150 kgf

The mass of the object in the oil  = 0.125 kgf

By Archimedes's principle, we have;

The upthrust on the object in water = Mass in air - mass in water = The weight of the water displaced

The upthrust on the object in water = 0.250 - 0.150 = 0.1 kgf

∴ The weight of the water displaced = 0.1 kgf

Given that the object is completely immersed in the water, we have;

The volume of the water displaced = The volume of the object

The volume of 0.1 kg of water water displaced = Mass of the water/(Density of water)

The volume of 0.1 kg of water = 0.1/1000 = 0.0001 m³

The density of the object = (Mass in air)/ volume = 0.250/0.0001 = 2500 kg/m³

The density of the object = 2500 kg/m³

2) Whereby the mass of the object in the oil = 0.125 kgf

The upthrust of the oil = The weight of the oil displaced

The upthrust of the oil on the object = Mass of the object in air - mass of the object in the oil

The upthrust of the oil on the object = 0.250 - 0.125 = 0.125 kgf

The weight of the oil displaced = The upthrust of the oil

Given that the volume of the oil displaced = The volume of the oil, we have;

The volume of the oil displaced = 0.0001 m³

The mass of the 0.0001 m³ = 0.125 kg

Therefore the density of the oil = 0.125/0.0001 = 1250 kg/m³.

The density of the oil = 1250 kg/m³.

Answer:

Well if they playing a game like that

Answer:

1. direction of propagation of sound

2.medium through which sound trsnsmitted