What is the difference between muscular endurance and muscular strength? A B or C

Answer:

a)  [tex]d_g=1.95m[/tex]

b) [tex]d_g'=2.3m[/tex]

Explanation:

From the question we are told that:

Depth [tex]d=2.60[/tex]

a)

Generally the equation for distance to ground is mathematically given by

[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]

Where

[tex]n=\frac{4}{3}[/tex]

Therefore

[tex]d_g=\frac{d}{n}[/tex]

[tex]d_g=\frac{2.6}{4/3}[/tex]

[tex]d_g=1.95m[/tex]

b)

For when The pool is filled halfway with water

[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]

[tex]d_g'=\frac{1.3}{4/3}[/tex]

[tex]d_g'=0.98m[/tex]

Therefore

[tex]d_g'=(1.3+0.98)[/tex]

[tex]d_g'=2.3m[/tex]

Answer:

24 meters

Explanation:

Find the final velocity. 12m/s

d=[final-initial]/2×time

D=(6m/s)×4=24 m/s

At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then

n / (19.2 L) = (1 mole) / (22.4 L)   ==>   n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol

If the sample has a mass of 12.0 g, then its molecular weight is

(12.0 g) / n ≈ 14.0 g/mol

Answer:

[tex]\rho_{avg}=4000kg/m^3[/tex]

Explanation:

From the question we are told that:

Density of Material 1 [tex]\rho_1=2000kg/m^3[/tex]

Density of Material 2 [tex]\rho_2=6000kg/m^3[/tex]

Generally the equation for Average density is mathematically given by

[tex]\rho_{avg}=frac{\rho _1+rho _2}{2}[/tex]

[tex]\rho_{avg}=\frac{2000+6000}{2}[/tex]

[tex]\rho_{avg}=4000kg/m^3[/tex]

Answer:

4.926 L Y 7.389 L

Explanation:

first you calculate the tank volume

V = π[tex](14 cm)^{2}[/tex](10 cm = [tex]12315 cm^{3}[/tex]

then you convert to liters

[tex]12315 cm^{3}[/tex] = 12.315 l

then you calculate the liters of water

2/5(12.35 l) = 4.926 l

finally we calculate the amount without water

12.315 l - 4.926 l = 7.389 l

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Answer:

Xc = (0.467 - 0.427j)R

Explanation:

Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is

Z = √[R² + (XL - XC)²]

Since the inductive reactance XL equals the resistance R, we have that

Z = √[R² + (XL - XC)²]

Z = √[R² + (R - XC)²]

Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]

Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus

Z' = √[R² + (R - XC')²]

Z' = √[R² + (R - 2XC)²]

The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]

Since the current doubles, I' = 2I.

V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]

1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]

√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]

squaring both sides, we have

[R² + (R - XC)²] = 4[R² + (R - 2XC)²]

expanding the brackets, we have

[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]

[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]

2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²

collecting like terms, we have

16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²

14RXC - 15XC² = 6R²

15XC² - 14RXC + 6R² = 0

Using the quadratic formula to find XC, we have

[tex]XC = \frac{-(-14R) +/- \sqrt{(-14R)^{2} - 4 X 15 X 6R^{2} } }{2 X 15}\\= \frac{-(-14R) +/- \sqrt{196R^{2} - 360R^{2} } }{30}\\ \\= \frac{14R +/- \sqrt{- 164R^{2} } }{30}\\ \\= \frac{14R +/- 12.81Ri }{30}\\\\= 0.467R +/- 0.427Ri[/tex]

Since it is capacitive, we take the negative part.

So, Xc = (0.467 - 0.427j)R

Answer:

[tex]H=1.35m[/tex]

Explanation:

From the question we are told that:

Height [tex]h=1.8m[/tex]

Depth of Water [tex]d=3.0m[/tex]

Where

Refraction index of water [tex]n = 4 / 3[/tex]

Generally the equation for Apparent height of the diver  is mathematically given by

[tex]H=h\frac{h}{n}[/tex]

[tex]H=\frac{1.8}{\frac{4}{3}}[/tex]

[tex]H=1.35m[/tex]

Answer:

The right answer is "[tex]\frac{KQ}{r_b_2}[/tex]".

Explanation:

As the outer spherical shell is conducting, so there is no electric field in side from

⇒ [tex]r_b_1 < r < r_b_2[/tex].

So the electric potential at all points inside the conducting shell that from

⇒ [tex]r_b_1<r<r_b_2[/tex]  

and will be similar as well as equivalent to the potential on the outer surface of the shell that will be:

⇒ [tex]v=\frac{KQ}{r_b_2}[/tex]

Thus the above is the right solution.

We have that the angle is

[tex]\theta=32.53[/tex]

From the Question we are told that

E⃗ =5.00×105ı^N/C

Generally the equation for Tension   is mathematically given

[tex]W=Tcos\theta[/tex]

Where

[tex]tan\theta=\frac{2.5*10^{-9}(5*10{5})}{2*10^{-3}(9.8)}[/tex]

[tex]\theta=32.53[/tex]

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Answer:

(a) the velocity of the shirt is 2.14 m/s

(b) the velocity of the shirt is 5.3 m/s

Explanation:

Given;

initial velocity of the shirt, u = 5.3 m/s

height of the platform above the ground, h = 4.00 m

(a) When the shirt is caught by someone whose hand is 5.20 m from the ground level, the height traveled by the shirt = 5.2 m - 4.0 m = 1.2 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 1.2)\\\\v^2 = 4.57\\\\v= \sqrt{4.57} \\\\v = 2.14 \ m/s[/tex]

(b) When the shirt is caught by someone whose hand is 4.00 m from the ground level, the height traveled by the shirt = 4.00 m - 4.00 m = 0 m

The velocity at this position is calculated as;

[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 0)\\\\v^2 = 28.09\\\\v= \sqrt{28.09} \\\\v = 5.3 \ m/s[/tex]

[tex]Hello[/tex] [tex]There![/tex]

[tex]AnimeVines[/tex] [tex]is[/tex] [tex]here![/tex]

The answer is...

C. Work and time.

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[tex]AnimeVines[/tex]

ANSWER; KE=5mv^2 so it is proportional to v^2.

Explanation:So if you triple the velocity you are replacing v with 3v. Then you get (3v)^2=9v^2.

Answer:

v₂ = 15.24 m / s

Explanation:

This is an exercise in fluid mechanics

Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

They indicate the pressure in the factory P₁ = 140000 Pa, the velocity

v₁ = 5.5 m / s and the initial height is zero y₁ = 0

the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m

          P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²

let's calculate

         140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²

         138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²

         v₂² = 2 (138000 /ρ - 58.8 + 15.125)

         v₂ = [tex]\sqrt{\frac{276000}{\rho } - 43.675 }[/tex]

In the exercise they do not indicate what type of liquid is being used, suppose it is water with

           ρ = 1000 kg / m³

           v₂ = [tex]\sqrt{\frac{276000}{1000} - 43.675}[/tex]

           v₂ = 15.24 m / s

Answer:

Explanation:

From the question we are told that:

Number of turns [tex]N=10[/tex]

Area [tex]a=0.23m^2[/tex]

Magnetic field [tex]B=0.947T[/tex]

Generally the equation for maximum flux is mathematically given by

 [tex]\phi=NBa[/tex]

 [tex]\phi=10*0.047*0.23[/tex]

 [tex]\phi=0.1081wbi[/tex]

Therefore induced emf

 [tex]e= \frac{d\phi}{dt}[/tex]

Since

 [tex]t=0[/tex]

Therefore

 [tex]e=0[/tex]

Answer:

[tex]\mathbf{y(t) =0.408e^{-6.20t }-0.408e^{-25.80t}}}[/tex]

Explanation:

Given that:

The weight of ball = 4 pounds

The spring stretch  x = 1/15 feet

Using the relation of weight on an object:

W = mg

m = W/g

m = 4 / 32

m = 1/8

Now, from Hooke's law:

F = kx

4 =k(1/5)

k = 20 lb/ft

However, since the air resistance is 4 times the velocity;

Then, we can say:

C = 4

Now, for the damped vibration in the spring-mass system, we have:

[tex]m\dfrac{d^2 y}{dx^2}+ c\dfrac{dy}{dt}+ky = 0[/tex]

[tex](\dfrac{1}{8})\dfrac{d^2 y}{dx^2}+ 4\dfrac{dy}{dt}+20y = 0[/tex]

[tex]\dfrac{d^2 y}{dx^2}+ 32\dfrac{dy}{dt}+160y = 0[/tex]

Solving the differential equation:

m² + 32m + 160 = 0

Solving the equation:

m = -25.80 or m = -6.20

So, the general solution for the equation is:

[tex]y (t)= c_1 e^{-6.20t}+c_2e^{-25.80t}[/tex]

[tex]y '(t)=-6.20 c_1 e^{-6.20t}-25.80c_2e^{-25.80t}[/tex]

y(0) = 0   ;  y'(0) = 8

[tex]y (0)= c_1 e^{-6.20(0)}+c_2e^{-25.80(0)}[/tex]

[tex]c_1 +c_2 = 0 ---(1)[/tex]

At y'(0) = 8

[tex]y '(0)=-6.20 c_1 e^{-6.20(0)}-25.80c_2e^{-25.80(0)} \\ \\ 8=-6.20 c_1 e^{0}-25.80c_2e^{0} \\ \\ 8=-6.20 c_1 -25.80c_2--- (2)[/tex]

From (1), let [tex]c_1 = -c_2[/tex], then replace the value of c_1 into equation (2)

[tex]8=-6.20 (-c_2)-25.80c_2[/tex]

[tex]8=6.20c_2-25.80c_2[/tex]

[tex]8=-19.60c_2[/tex]

[tex]c_2=\dfrac{ 8}{-19.60}[/tex]

[tex]c_2 = -0.408[/tex]

From [tex]c_1 = -c_2[/tex]

[tex]c_1 = -(-0.408)[/tex]

[tex]c_1 = 0.408[/tex]

∴

The required solution in terms of t is:

[tex]\mathbf{y(t) =0.408e^{-6.20t }-0.408e^{-25.80t}}}[/tex]

When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.

Now, given that:

Assumptions:

∴

For a motion under constant acceleration, we can apply the kinematic equation:

[tex]\mathsf{v^2 = u^2 + 2as}[/tex]

where;

From the above equation, making acceleration (a) the subject of the formula:

[tex]\mathsf{v^2 - u^2 =2as }[/tex]

[tex]\mathsf{a = \dfrac{v^2 - u^2 }{2s}}[/tex]

The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.

since 1 km/h = 0.2778 m/s

100 km/h = 27.78 m/s

[tex]\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}[/tex]

[tex]\mathsf{a = \dfrac{- 771.7284 }{2}}[/tex]

a = - 385.86 m/s²

Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration, and velocity is;

v = u + at

where;

v = 0

-u = at

[tex]\mathsf{t = \dfrac{-u}{a}}[/tex]

[tex]\mathsf{t = \dfrac{-27.78}{-385.86}}[/tex]

t = 0.07 seconds

An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.

Thus, we can conclude that in order to estimate how fast the airbag must inflate to effectively protect the driver, the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.

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Answer:

Weight, acceleration when it falls vertically, are not same as that of earth.

Explanation:

Weight of the object is given by the product of mass of the object and the acceleration due to gravity of the planet.

So, the weight of object is not same as that on earth.

The mass is defined as the amount of matter contained in the object.

So, the mass of the object is same as that of earth.  

The volume of the object is defined as the space occupied by the object.

So, the volume of the object is same as that of earth.  

The density is defined as the ratio of mass of the object to its volume.

So, the density of the object is same as that of earth.  

The acceleration due to gravity on a planet depends on the mass of planet and radius of planet.

So, the acceleration is not same as that of earth.

The color of the object is its characteristic.

It is same as that of earth.

Answer:

60 deg from Y-axis is 150 deg from X-axis

x = L cos theta = 23,2 cos 150 = -.867 * 23.2 = -20.1

Answer:

Final kinetic energies of both astronauts will be the same.

Explanation:

If we ignore all the friction present between the ramp and the person. Then essentially there is no loss of energy in the system. Hence the initial potential energies of the astronauts must be equal to their final kinetic energies.

Now the potential energy depends upon mass, height and acceleration due to gravity. All these parameters for both the astronauts. Therefore,both astronauts have same initial potential energies.

Similarly, the final kinetic energies of astronauts will also be the same.

Answer:

The answer is "[tex]143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}[/tex]"

Explanation:

For point a:

Energy balance equation:

[tex]\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\[/tex]

[tex]W=0\\\\Q=0\\\\m_e=0[/tex]

From the above equation:

[tex]\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\[/tex]

because the rate of air entering the tank that is [tex]h_i[/tex] constant.

[tex]\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\[/tex]

Since the tank was initially empty and the inlet is constant hence, [tex]m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\[/tex]

Interpolate the enthalpy between [tex]T = 300 \ K \ and\ T=295\ K[/tex]. The surrounding air  

temperature:

[tex]T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}[/tex]

Substituting the value from ideal gas:

[tex]\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}[/tex]

Follow the ideal gas table.

The [tex]u_2= 298.33\ \frac{kJ}{kg}[/tex] and between temperature [tex]T =410 \ K \ and\ T=240\ K.[/tex]

Interpolate

[tex]\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}[/tex]

Substitute values from the table.

 [tex]\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\[/tex]

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. [tex]\bar{R} \ is\ \frac{R}{M}[/tex] (M is the molar mass of the  gas that is [tex]28.97 \ \frac{kg}{mol}[/tex] and R is gas constant), and T is the temperature.

[tex]n=\frac{pV}{TR}\\\\[/tex]

[tex]=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\[/tex]

For point c:

 Entropy is given by the following formula:

[tex]\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}[/tex]

Answer:

30 m/s

Explanation:

Applying,

v = u+at................ Equation 1

Where v = final speed of the ball, u = initial speed of the ball, a = acceleration, t = time.

From the question,

Given: u = 0 m/s (stationary), a = 600 m/s², t = 0.05 s

Substitute these values into equation 5

v = 0+(600×0.05)

v = 30 m/s

Hence the speed at which the ball leaves the player's boot is 30 m/s

Answer:

Hence the square root of the wavelength. What is the wavelength of 27 MHz radio waves  11 m; option A is the correct answer.

Hence, as the frequency of the radio waves increases the penetration depth decreases.

Option C is the correct answer.

Hence the electric field amplitude has decreased by a factor of √3.

Hence the correct option is D.

Explanation:

[tex]c=n\lambda[/tex], where the symbols have their usual meanings.

[tex]\lambda =c/n\\= 3\times10^{8}/27\times10^{6}\\\\=11.11 m[/tex]

Hence, wavelength = 11 m; option A is the correct answer.

As given that, the penetration depth is proportional to the square root of wavelength, and we know that,

wavelength and frequency are inversely proportional to each other.

Hence, as the frequency of the radio waves increases the penetration depth decreases.

Option C is the correct answer.

Electric field amplitude is directly proportional to the square root of the intensity. Therefore, the ratio between the electric field amplitude is,

[tex]\frac{E\sigma_{1} }{E\sigma_{2} } =\sqrt{\frac{I_{1} }{I_{2} } }[/tex]

Here to find the ratio of intensities,

[tex]{\frac{I_{1} }{I_{2} } } = 3[/tex]

Therefore,

[tex]\frac{E\sigma_{1} }{E\sigma_{2} } =\sqrt{3 }[/tex]

Hence the correct option is D.

Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

[tex]m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C[/tex]

Therefore, the resulting temperature is 23.37 ⁰C

Answer:

d. The hammer falls with a constant acceleration

Explanation:

Since gravity is the only thing that is acting on the hammer as it falls and gravity is a form of acceleration then acceleration of 9.81m/s² which is gravity is the correct answer.

Answer:

The total number of maxima that can be seen is 11

Explanation:

Given the data in the question

wavelength λ = 500 nm = 5 × 10⁻⁷ m

if the third order maximum is 32

i.e m = 3 and θ = 32°

Now, we know that condition for diffraction maximum is as follows;

d × sinθ = m ×  λ

so we substitute in our given values

d × sin( 32° ) = 3 ×  5 × 10⁻⁷ m

d × sin( 32° ) = 1.5  × 10⁻⁶ m

d = [ 1.5  × 10⁻⁶ m ] / sin( 32° )

d = 2.83 × 10⁻⁶ m

Now, maxima n when θ = 90° will be;

sin( 90° ) = nλ / d

1 =  nλ / d

d =  nλ

n = d / λ

we substitute

n = [ 2.83 × 10⁻⁶ m ]  / [ 5 × 10⁻⁷ m ]

n = 5.66

so 5 is the max value

hence, total maxima value is;

⇒ 2n + 1 = 2( 5 ) + 1 = 10 + 1 = 11

Therefore, total number of maxima that can be seen is 11

Answer:

7.38 × 10-5

Explanation:

All numbers in scientific notation or standard form are written in the form m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.

To convert 0.0000738 into scientific notation, follow these steps:

Move the decimal 5 times to the right in the number so that the resulting number, m = 7.38, is greater than or equal to 1 but less than 10

Since we moved the decimal to the right the exponent n is negative

n = -5

Write in the scientific notation form, m × 10n

= 7.38 × 10-5

Therefore, the decimal number 0.0000738 written in scientific notation is 7.38 × 10-5 and it has 3 significant figures.

Answer = 7.38 × 10-5

Answer:

The density at 300 C is [tex]2641.68 kg/m^3[/tex].

Explanation:

density at 0 degree, d = 2700 kg/m3

coefficient of linear expansion, = 24 x 10^{-6} /K

Let the density at 300 C is d'.

Use the formula of the cubical expansion

[tex]d'= d (1-3\alpha \Delta T)\\\\d'= 2700(1- 3 \times 24\times 10^{-6}\times 300)\\\\d' = 2641.68 kg/m^3[/tex]

Answer:

I think Refraction

Explanation:

I think refraction is light which passes through a boundary between 2 transparent media.

Answer:

The electric potential energy is 6.72 x 10^-11 J.

Explanation:

Potential difference, V = 4.2 x 10^8 V

charge of electron, q = - 1.6 x 10^-19 C

Let the potential energy is U.

U = q V

U = 1.6 x 10^-19 x 4.2 x 10^8

U = 6.72 x 10^-11 J