Select the only true statement:


A beam in bending experiences tensile stresses on one side and compressive stresses on the other side.


A beam in bending experiences tensile stresses along the beam center and compressive stresses along the beam’s edges.


A beam in bending experiences only compressive stresses.


A beam in bending experiences only tensile stresses.

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:

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:

I think Refraction

Explanation:

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

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.

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:

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:

(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]

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:

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:

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

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:

[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:

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

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:

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:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

[tex]v =f\lambda[/tex]

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

[tex]\mu = \frac{m}{l}[/tex]

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

[tex]\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m[/tex]

Now, for the tension we use the formula:

[tex]v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T[/tex]

T = 0.015 N

(b)

The mass to be hung is:

[tex]T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\[/tex]

M = 1.53 x 10⁻³ kg = 1.53 g

Answer:

120 km/s

Explanation:

Given data :

Radius of the star is r = 19 km

Rotational time period of the star is T = 1 s

Therefore, we know that the velocity of the star is given by :

[tex]$V=\frac{2\pi r}{T}$[/tex]

[tex]$V=\frac{2 \times 3.14 \times 19\times 10^3}{1}$[/tex]

V = 119380.52 m/s

Therefore, the velocity of the point on the equator of the star is = 120 km/s

A body of mass m feels a gravitational force due to the planet of

F = GmM/R ² = ma

where

• G = 6.67 × 10⁻¹¹ N•m²/kg² is the universal gravitational constant

• M is the mass of the planet

• R is the distance between the body and the planet's center

• a is the acceleration due to gravity

Solving for a gives

a = GM/R ²

Notice that 28,000 km is twice 14,000 km. The equation says that the acceleration varies inversely with the square of the distance. So if R is changed to 2R, we have a new acceleration of

GM/(2R)² = 1/4 × GM/R ² = a/4

so the acceleration of the body at 28,000 km from the planet's center would be (32 m/s²)/4 = 8 m/s².

Answer:

Force = mass × acceleration

[tex]F =(100 \times 1000) \times 10 \\ = 1 \times {10}^{6} \: newtons[/tex]

A. true

means decelerating

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shirleywashington

Ambitious

2.3K answers

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Explanation :

We know that the slope of velocity -time graph gives the acceleration. Acceleration of an object is defined as the rate of change of velocity i.e.



Suppose a driver suddenly applies brakes. In this case the initial velocity of his or her vehicle is more and the final velocity is less.

So, the acceleration is negative in this case i.e. the object is decelerating.

A negative slope on the velocity versus time graph indicates that an object is not accelerating. This statement is false as the object is decelerating.

Answer:

do this Q yourself because i havent read the chapter

The maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk is 63 rpm.

This is calculated using the coefficient of static friction between the penny and block, which is 0.495.

The maximum angular velocity of the disk is when the force of static friction is just sufficient to prevent the penny from sliding.

This force is equal to the mass of the penny multiplied by the acceleration due to gravity, multiplied by the coefficient of static friction.

The angular velocity of the disk is then calculated from this force and the radius of the disk.

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Answer:

The energy from food and then from plants and then from sun.

As sun is the ultimate source of energy.

Explanation:

Distance = 100 m, 1000m, marathon

As the distance is covered by the person, so the muscular energy is used and thus the energy comes form out food.

As we know that the energy can neither be created nor be destroyed it can transform from one form to another.

So, the energy form the food which we consume is converted into the kinetic energy as we run.

"Longitudinal wave" is the wave where the difference between the coils increases as well as decreases.

Thus the above answer is correct.

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Answer:

Njfjrhrjrkrirkehrbrhrrhrhehrhrhejejebrbrhrbrbbbrhje

Answer:

The answer is 286.2815.

Answer:

Spaceship speed is 36000 km/h

So, in 1 hour spaceship travel 36000 km

Or we can say that in 60×60 second spaceship travel 36000 km

Therefore in 1 sec spaceship travel

=

= 10 km/s

Answer:

Explanation:

360 km/hr(1000 m/km) / 3600 s/hr) = 100 m/s

Answer:

b

Explanation:

Answer:

Moves towards left

Explanation:

When the truck is moving towards right then there is pseudo force acting on the fog statue which is acting left wards.

A seen from the stationary observer, the dog statue moves towards left.

Answer:

[tex]\rho=2.28\times 10^5\ C/m^3[/tex]

Explanation:

Given that,

Charge, Q = 2.1 C

The radius of sphere, r = 1.3 cm = 0.013 m

We need to find the charge per unit volume for this situation. It can be calculated a follows:

[tex]\rho=\dfrac{Q}{\dfrac{4}{3}\pi r^3}\\\\\rho=\dfrac{2.1}{\dfrac{4}{3}\pi \times (0.013)^3}\\\\\rho=2.28\times 10^5\ C/m^3[/tex]

So, the charge per unit volume is [tex]2.28\times 10^5\ C/m^3[/tex].

Explanation:

pls write the full question