A tuning fork is held over a resonance tube, and resonance occurs when the surface of the water is 12 cm below the top of the tube. Resonance occurs again when the water is 34 cm below the top of the tube. If the air temperature is 23 degrees Celsius, find the frequency of the tuning fork.

Answer:

Explanation:

12.)

A. Opposite poles attract

B. Same poles repel

13.)

IDK

Answer:

1.92 Watt lost

Explanation:

Power rating of each toaster = 600 Watts

Current that flows = 5 Amperes

Wasted power = 1 Watt

Voltage of toaster can be gotten from P = [tex]I^{2}[/tex]R

where I = current

and R = Resistance

600 = [tex]5^{2}[/tex] x R

R = 600/25 = 24 Ohms.

According to joules loss due to heating of wire

Power loss P ∝ [tex]I^{2}[/tex]R

imputing values,

1 ∝ [tex]5^{2}[/tex] x 24

1 ∝ 600

to remove the proportionality sign, we introduce a constant k

1 = 600k

k = 1/600 = 0.00167

For the case where the current is doubled to 10 ampere, as the power doubles to 1200 W.

The resistance across the wire becomes

1200 = [tex]10^{2}[/tex]R

R = 1200/100 = 12 Ohms

power loss P = k x [tex]I^{2}[/tex]R

P = 0.0016 x [tex]10^{2}[/tex] x 12

P = 1.92 Watt lost

This question involves the concepts of power, current, and resistance.

The power wasted by the wires in the home for two units will be "4 watt".

The power wasted by the wires can be given in terms of current and resistance by the following formula:

[tex]P=I^2R\\\\\frac{P}{I^2}=R=Constant\\\\\frac{P_1}{I_1^2}=\frac{P_2}{I_2^2}[/tex]

where,

Therefore,

[tex]\frac{1\ watt}{(5\ A)^2}=\frac{P_2}{(10\ A)^2}\\\\P_2=\frac{(1\ watt)(100\ A^2)}{25\ A^2}[/tex]

P₂ = 4 watt

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Answer:

the answer is opposite.

plz mark brainliest

Explanation:

Answer:

  (possibly) Box D

Explanation:

The one that has balanced forces will not accelerate. The forces are unbalanced in figures A, B, C. We cannot tell about figure D, because the downward force is not shown. If that force is 20 N, box D will not accelerate.

Answer:

13.9 kJ

Explanation:

Given that

Length of the side, l = 1.8 m

Drag coefficient, C(d) = 1.4

Distance of run, d = 200 m

Velocity of run, v = 5 m/s

Density, ρ = 1.23

Using the Aerodynamics Drag Force formula. We have

F(d) = 1/2.ρ.A.C(d).v²

The Area, A needed is 1.8 * 1.8 = 3.24 m². So that,

F(d) = 1/2 * 1.23 * 3.24 * 1.4 * 5²

F(d) = 139.482/2

F(d) = 69.74

recall that, energy =

W = F * d

W = 69.74 * 200

W= 13948

W = 13.9kJ

Therefore, the thermal energy added to the air by the drag force is 13.9kJ

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it =  λ dx where  λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x  / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

=  k λ dx  .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position  on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫  k λ  .15  / (x² + .15² )³/² dx

=  k λ  x L / .15 √( L² / 4 + .15² )

b) The length of the rod:

[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )[/tex]

Given:

d = 1.5 mλ = 2.5 nC/m

Let the plastic rod extends from - L to + L .Consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .It will create a field at point P on y -axis.

Distance of point P =[tex]\sqrt{x^2 + 0.15^2}[/tex]

E.F at P due to small charged length[tex]dE = \frac{ k \lambda x.dx}{(x^2 + .15^2 )}[/tex]

Its component along Y - axis = dE cosθ where θ is angle between direction of field dE and y axis

[tex]= \frac{dE x .15 }{\sqrt{x^2 + .15^2} }\\\\= \frac{k \lambda dx .15}{(x^2 + .15^2 )^{1/2}}[/tex]

If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . We can say that the component of field in perpendicular to y axis will cancel out each other.

Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )}[/tex]

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Answer:

  More than 48%

Explanation:

If the interest is computed monthly on the outstanding balance, it has an effective annual rate of ...

  (1 +4%)^12 -1 = 60.1% . . . .  more than 48%

The effective annual or yearly interest rate would be=30.56% which is Between 4% and 48%

The formula used to calculate annual Interest rate =

[tex](1+ \frac{i}{n} ) {}^{n} - 1[/tex]

where i= nominal interest rate = 4%

n= number of periods= 12 months

Annual Interest rate=

[tex](1 + \frac{4\%}{12} ) {}^{12} - 1[/tex]

= (1+0.333)^12 -1

= (1.333)^12-1

= 31.56 - 1

= 30.56%

Therefore, the effective annual or yearly interest rate would be= 30.56%

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Answer:

1. The Universe is Expanding

2. It’s temperature is it’s uniform

3. Cosmic background radiation

4. I will give a hint for this one, since I don’t know, the hint is the universe is cooling.

5. It provides evidence of universe expansion.

6. Sorry I don’t know the rest

Explanation:

The universe is the collection of every item in space and time as well as the contents of those items

The correct options are as follows;

1. The universe is expanding

2. Its temperature is uniform

3. Composition of matter in the universe, cosmic background radiation

4. The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses

5. It provides evidence of universe expansion

6. Nuclear fission involves one large atom splitting into two smaller atoms

Nuclear fission takes place in the nucleus of an atom

7. A distant object travels rapidly towards an observer

8. The fuel in nuclear fusion is often uranium

Nuclear fusion is used to generate electricity at nuclear power plants

9. The universe began as a very high density singularity

Dark matter makes up majority of the universe

10.  Acceleration expansion — Dark energy

Structure forming in the early universe — Dark matter

11. It causes the expansion of the universe to accelerate

The reasons for selecting the above options are as follows;

1. The universe is expanding

The redshift of light from galaxies indicates that that are moving away

2. Its temperature is uniform

The uniform temperature of the microwave background suggest a common source

3. Composition of matter in the universe, cosmic background radiation

The matter present in the universe are characteristically similar in their origins

The cosmic background provides evidence of the existence of a singularity

4. The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses

Based on the Big Bang Theory, the temperature of the universe is reducing as the universe expands, compared to the initial temperature

5. It provides evidence of universe expansion

The background radiation coming from a single source as the rest of the universe is expected to spread throughout the universe

6. Nuclear fission involves one large atom splitting into two smaller atoms

Nuclear fission takes place in the nucleus of an atom

Nuclear fusion involves the joining of small atoms to form a larger atom

7. A distant object travels rapidly towards an observer

The redshift is the opposite, indicating that the object is moving further away

8. The fuel in nuclear fusion is often uranium

Nuclear fusion is used to generate electricity at nuclear power plants

Nuclear fusion usually consists of joining small atoms together. It has not been used for commercial energy production

9. The universe began as a very high density singularity

According to the Big Bang Theory, the universe started from the dense, high temperature singularity

Dark matter makes up majority of the universe

10.  Acceleration expansion — Dark energy

Dark energy causes expansion

Structure forming in the early universe — Dark matter

Dark matter is instrumental to the formation of structures in the universe

11. It causes the expansion of the universe to accelerate

Dark energy is seen as the cause of the accelerating expansion of the universe

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Answer:

d = 80 m

its vertical displacement (in m) after 4 s is 80 m

Explanation:

From the equation of motion;

d = vt + 0.5at^2 ......1

Where;

d = displacement

v = initial velocity = 0 (dropped with no initial speed)

t = time of flight = 4s

a = g = acceleration due to gravity = 10 m/s^2

Substituting the given values into equation 1;

d = 0(4) + 0.5(10 × 4^2)

d = 0.5(10×16)

d = 80 m

its vertical displacement (in m) after 4 s is 80 m

Answer:

Explanation:

The power rating of the battery isn't provided. But let us assume that it is one of the common batteries with ratings of 12 V and 50 A.h

Potential energy possessed by water at that height = mgh

m = mass of the water = ρV

ρ = density of water = 1000 kg/m³

V = volume of water = ?

g = acceleration due to gravity = 9.8 m/s²

h = height of water = 50 cm = 0.5 m

Potential energy = ρVgh = 1000 × V × 9.8 × 0.5 = (4900V) J

Energy of the battery = qV

q = 50 A.h = 50 × 3600 = 180,000 C

V = 12 V

qV = 180,000 × 12 = 2,160,000 J

Energy = 2,160,000 J

At a 100% conversion rate, the energy of the water totally powers the battery

(4900V) = (2,160,000)

4900V = 2,160,000

V = (2,160,000/4900)

V = 440.82 m³

Hence, with our assumed power ratings for the battery (12 V and 50 A.h), 440.82 m³ of water at the given height of 50 cm would power the battery.

Incase the power ratings of the battery in the complete question is different, this solution provides you with how to obtain the correct answer, given any battery power rating.

Hope this Helps!!!

The average acceleration [tex]\vec a_{\rm ave}[/tex] over some time interval [tex][t_1,t_2][/tex] is equal to the ratio of the change in velocity [tex]\vec v_2-\vec v_1[/tex] over the duration of the interval [tex]t_2-t_1[/tex], or

[tex]\vec a_{\rm ave}=\dfrac{\Delta\vec v}{\Delta t}=\dfrac{\vec v_2-\vec v_1}{t_2-t_1}[/tex]

which can be split into the [tex]x[/tex] and [tex]y[/tex] components as

[tex]a_{\rm{ave},x}=\dfrac{v_{2,x}-v_{1,x}}{t_2-t_1}=\dfrac{-170\frac{\rm m}{\rm s}-90\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-8.67\dfrac{\rm m}{\mathrm s^2}[/tex]

[tex]a_{\rm{ave},y}=\dfrac{v_{2,y}-v_{1,y}}{t_2-t_1}=\dfrac{40\frac{\rm m}{\rm s}-110\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-2.33\dfrac{\rm m}{\mathrm s^2}[/tex]

The magnitude of this average acceleration is

[tex]\left\|\vec a_{\rm ave}\right\|=\sqrt{{a_{\rm{ave},x}}^2+{a_{\rm{ave},y}}^2}\approx8.98\dfrac{\rm m}{\mathrm s^2}[/tex]

and its direction is [tex]\theta[/tex] such that

[tex]\tan\theta=\dfrac{a_{\rm{ave},y}}{a_{\rm{ave},x}}\implies\theta\approx-164.9^\circ[/tex]

which corresponds to a direction of about 15.1º South of West.

Answer:

none of the answers is correct, the time  is the same  t₁ = t₂ = 0.600 s

Explanation:

This is a kinematics exercise, analyze the situation a bit. The vertical speed in both cases is the same is zero, the horizontal speed in the second case is double (vₓ₂ = 2 vₓ₁)

let's find the time to hit the ground

     y = y₀ + I go t - ½ g t²

     0 = y₀ - ½ g t²

     t = √ 2y₀ / g

with the data from the first launch

     y₀i = ½ g t²

     y₀ = ½  9.8  0.6²

     y₀ = 1,764 m

with this is the same height the time to descend in the second case is the same

    t₂ = 0.600 s

this is because the horizontal velocity change changes the offset on the x axis, but does not affect the offset on the y axis

Therefore, none of the answers is correct, the time  is the same

t₁ = t₂ = 0.600 s

Answer:

Explanation:

Using one of the general gas equation to find the final volume of the R-134a.

According to pressure law; The volume of a given mas of gas is directly proportional to its temperature provided that the pressure remains constant.

VαT

V = kT

k = V/T

V1/T1 = V2/T2 = k

Given V1 = 0.14-m³ at T1 = –26.4°C = –26.4° + 273 = 246.6K

V2 = ? at T = 100°C = 100+273 = 373K

On substituting this values for T2;

0.14/246.6 = V2/373

373*0.14 = 246.6V2

V2 = 373*0.14 /246.6

V2 = 0.212m³

The final volume of R-134a is 0.212m³

Answer:

a)   E = 1.58 10²¹ J , b) Oil = 4,236 107 liter ,  e)   T = 54.3 C

Explanation:

a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface

     I = P / A

     A = 4π r²

in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m

     I = P / A

     I = P / 4π r²

let's calculate

     I = 3,828 10²⁵/4 pi (1.5 10¹¹)²

     I = 1.3539 10²W / m² = 135.4 W / m2

the energy that reaches the disk of the Earth is

    E = I A

the area of ​​a disc

    A = π r²

    E = I π r²

where r is the radius of the Earth 6.37 10⁶ m

     E = 135.4 π(6.37 10⁶)

     E = 1,726 10¹⁶ W

This is the energy per unit of time that reaches Earth

    t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s

    E = 1,826 10¹⁶ 86400

     E = 1.58 10²¹ J

b) for this part we can use a direct proportions rule

      Oil = 1.58 10²¹ (1 / 37.3 10⁶)

      Oil = 4,236 10⁷ liter

c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law

       P = σ A e T⁴

       T = [tex]\sqrt[4]{P/Ae}[/tex]

nos indicate the refect, therefore the amount of absorbencies

       P_absorbed = 0.7 P

let's calculate

       T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)

       T = RER (8,676 106)

       T = 54.3 C

b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere

Answer:

To solve this problem we just need to graph two forces with same direction, pointing to different sides with intensities of 40 N and 60 N.

The image attached shows these forces.

Notice that the vectors are parallel, that's because they have the same direction, but they point to different sides, and their magnitudes have a difference of 20 N.

Answer:

31.4 m/s

44.4°

Explanation:

Momentum is conserved in the horizontal direction:

pₓᵢ = pₓ

m vᵢ₂ + 2m vᵢ₁ cos θ = (m + 2m) vₓ

vᵢ₂ + 2 vᵢ₁ cos θ = 3 vₓ

17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vₓ

vₓ = 22.5 m/s

Momentum is conserved in the vertical direction:

pᵧᵢ = pᵧ

2m vᵢ₁ sin θ = (m + 2m) vᵧ

2 vᵢ₁ sin θ = 3 vᵧ

2 (41.5 m/s) (sin -52.7°) = 3 vᵧ

vᵧ = -22.0 m/s

The speed is:

v = √(vₓ² + vᵧ²)

v = √((22.5 m/s)² + (-22.0 m/s)²)

v = 31.4 m/s

The direction is:

θ = atan(vᵧ / vₓ)

θ = atan(-22.0 m/s / 22.5 m/s)

θ = -44.4°

The speed of the eagle at that instant is 31.4 m/s while it moves off in the direction of 44.4°.

Since momentum is conserved horizontally;

17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vx

vx = 17.1 m/s + 2 (41.5 m/s) (cos -52.7°)/3

vx =  22.5 m/s

Also, momentum is conserved vertically hence;

2 (41.5 m/s) (sin -52.7°) = 3 vy

vy = 2 (41.5 m/s) (sin -52.7°) /3

vy =  -22.0 m/s

The effective speed therefore, is;

v = √((22.5 m/s)² + (-22.0 m/s)²)

v = 31.4 m/s

The direction of this effective speed is;

θ = tan-1(22.0 m/s / 22.5 m/s)

θ = 44.4°

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Answer:

U = 218 nJ

Explanation:

We are given;

Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m

Voltage across the capacitor; V = 96 V

Dimension of the square plates is 7.2cm x 7.2cm.

So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²

Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²

From relative permeability table;

Dielectric constant of Pyrex; k1 = 5.6

Dielectric constant of polystyrene; k2 = 2.56

Now, formula for capacitance of a capacitor with Dielectric is;

C = kC_o

Where, C_o = ε_o(A/d)

Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)

Since we have 2 capacitor, thus ;

C1 = k1*ε_o*(A/d)

C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C1 = 1.51 × 10^(-10) F

Similarly;

C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C2 = 0.691 × 10^(-10) F

For capacitors in series, formula for total capacitance(Cs) is;

1/Cs = (1/C1) + (1/C2)

Simplifying this, we have;

Cs = (C1*C2)/(C1 + C2)

Plugging in the relevant values ;

Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))

Cs = 0.474 × 10^(-10) F

The formula for energy stored in a capacitor with 2 Dielectrics is given as;

U = ½Cs*V²

So,

U = ½ × 0.474 × 10^(-10) × 96²

U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

[tex]mv_{1i}+Mv_{2i}=(m+M)v[/tex]  (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

[tex]v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}[/tex]

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

Answer:

a.  4 V

b. 0.697 A

Explanation:

Magnetic field strength B =  0.732 T

length of rod l = 0.362 m

velocity of rod v = 15.1 m/s

a.  EMF can be calculated as

E = Blv = 0.732 x 0.362 x 15.1 = 4 V

b. If the rod is connected to a conducting rail, with resistance R = 5.74Ω

current I = V/R = 4/5.74 = 0.697 A

the current flows in a clockwise direction

Explanation:

We have, a pendulum on a planet, oscillates with a time period 5 sec. The formula used to find the time period is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is length of the pendulum

g is acceleration due to gravity on which it is placed

It is clear that, the time period of pendulum is independent of the mass. Hence, if the mass is increased six times, its time period remains the same.

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

[tex]v_y^2=v_{oy}^2+2gh[/tex]   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

[tex]v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}[/tex]

vx is calculated by using the information about the horizontal range of the ball:

[tex]R=v_o\sqrt{\frac{2h}{g}}[/tex]    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

[tex]v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}[/tex]

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

[tex]v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}[/tex]

The velocity of the ball just before it touches the ground is 46.99 m/s

Answer:

E = 389 MeV

Explanation:

The total energy of particle A, will be equal to the sum of rest mass energy and relative energy of particle A. Therefore,

Total Energy of A = E = Rest Mass Energy + Relative Energy

Using Einstein's Equation: E = mc²

E = m₀c² + mc²

From Einstein's Special Theory of Relativity, we know that:

m = m₀/[√(1-v²/c²)]

Therefore,

E = m₀c² + m₀c²/[√(1-v²/c²)]

E = m₀c²[1 + 1/√(1-v²/c²)]

where,

m₀c² = rest mass energy = 140 MeV

v = relative speed = 0.827 c

Therefore,

E = (140 MeV)[1 + 1/√(1 - (0.827c)²/c²)]

E = (140 MeV)(2.78)

E = 389 MeV

Answer:[tex]8.062\ m/s[/tex]

Explanation:

Given

masss of football player [tex]M=110\ kg[/tex]

Velocity of football player [tex]u_1=8\ m/s[/tex]

mass of football [tex]m=0.41\ kg[/tex]

velocity of football [tex]u_2=25\ m/s[/tex]

Final velocity will be given by applying conservation of linear momentum

After catching the ball Player and ball moves with same velocity

[tex]\Rightarrow Mu_1+mu_2=(M+m)v[/tex]

[tex]\Rightarrow 110\times 8+0.41\times 25=(110+0.41)v[/tex]

[tex]\Rightarrow 880+10.25=110.41\times v[/tex]

[tex]\Rightarrow v=\frac{890.25}{110.41}=8.063\ m/s[/tex]

So, final velocity will be [tex]8.062\ m/s[/tex]

t\12 and the parties are spreading ever

Explanation:

my point is that you can get sick if

you sont wash your ha

nds or be

save

Answer:

Vf = 0.0017 m³ = 1.7 L

Explanation:

The work done by the system on the surrounding at constant pressure is given by the following formula:

W = PΔV

W = P(Vf - Vi)

where,

W = Work done = 288 J

P = Constant Pressure = (2 atm)(101325 Pa/atm) = 202650 Pa

Vf = Final Volume f Cylinder = ?

Vi = Initial Volume of Cylinder = (0.25 L)(0.001 m³/ 1 L) = 0.00025 m³

Therefore,

288 J = (202650 Pa)(Vf - 0.00025 m³)

Vf = 288 J/202650 Pa + 0.00025 m³

Vf = 0.0017 m³ = 1.7 L

Sinkholes formation is one feature of karst topography. The top of a cave falls if it develops large enough and its top extends near enough to the surface.

Karst topography is a type of natural environment formed mostly by chemical weathering by water, resulting in caves, sinkholes, cliffs, and steep-sided hills known as towers.

The top of a cave falls if it develops large enough and its top extends near enough to the surface. Sinkholes are formed as a result of this, and they are one of the most distinguishing aspects of karst terrain.

When water absorbs carbon dioxide from the atmosphere and ground, it becomes carbonic acid.

Hence, sinkholes  formation is one feature of karst topography

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Answer: A) Caves

Explanation:

Answer:

The description of that same situation has been listed throughout the explanation segment below.

Explanation:

So that the above seems to be the right answer.

Answer:

it represents a fundamental difference. (more info below)

Explanation:

Production is incessantly developing and expanding in socialist countries, and employment is guaranteed for the entire productive population. Consequently, the relative overpopulation problem has been eliminated. This represents the fundamental difference between socialism's demographic law and capitalism's law.

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Starting from rest, the wheel attains an angular velocity of 25 rad/s in a matter of 10 s, which means the angular acceleration [tex]\alpha[/tex] is

[tex]25\dfrac{\rm rad}{\rm s}=\alpha(10\,\mathrm s)\implies\alpha=2.5\dfrac{\rm rad}{\mathrm s^2}[/tex]

For the next 37 s, the wheel maintains a constant angular velocity of 25 rad/s, meaning the angular acceleration is zero for the duration. After this time, the wheel undergoes an angular acceleration of -1.5 rad/s/s until it stops, which would take time [tex]t[/tex],

[tex]0\dfrac{\rm rad}{\rm s}=25\dfrac{\rm rad}{\rm s}+\left(-1.5\dfrac{\rm rad}{\mathrm s^2}\right)t\implies t=16.666\ldots\,\mathrm s[/tex]

which makes B, approximately 17 s, the correct answer.

The time interval of angular deceleration is 16.667 seconds, whose closest integer is 17 seconds. (B. 17 s.)

Let suppose that the grinding wheel has uniform Acceleration and Deceleration. In this question we need to need to calculate the time taken by the grinding wheel to stop, which is found by means of the following Kinematic formula:

[tex]t = \frac{\omega - \omega_{o}}{\alpha}[/tex] (1)

Where:

[tex]\omega_{o}[/tex] - Initial angular velocity, in radians per second.

[tex]\omega[/tex] - Final angular velocity, in radians per second.

[tex]\alpha[/tex] - Angular acceleration, in radians per square second.

[tex]t[/tex] - Time, in seconds.

If we know that [tex]\omega = 0\,\frac{rad}{s}[/tex], [tex]\omega_{o} = 25\,\frac{rad}{s}[/tex] and [tex]\alpha = -1.5\,\frac{rad}{s^{2}}[/tex], then the time taken by the grinding wheel to stop:

[tex]t = \frac{0\,\frac{rad}{s}-25\,\frac{rad}{s}}{-1.5\,\frac{rad}{s^{2}} }[/tex]

[tex]t = 16.667\,s[/tex]

The time interval of angular deceleration is 16.667 seconds. (Answer: B)

Please this related question: https://brainly.com/question/10708862

Answer:

14,260

Explanation:

Relevant data provided for computing the wavelengths are in one pulse is here below:-

The number of wavelengths in Ls = [tex]4.6\times 10_1_4[/tex]

Therefore the Number of in time = Δt = [tex]3.1\times 10_-_1_1[/tex]

The number of wavelengths are in one pulse is shown below:-

[tex]Number\ of\ wavelengths = \triangle t\times f[/tex]

[tex]= 3.1\times 10_-_1_1\times 4.6\times 10_1_4[/tex]

= 14,260

Therefore for computing the number of wavelengths are in one pulse we simply applied the above formula.