Please help! Average speed. Show work!

Answer:

k = 0.6 N/m

Explanation:

The time period of a spring mass oscillation system is given by the following formula:

T = 2π√(m/k)

where,

T = Time Period of Oscillation = 10 s

m = Mass attached to the spring = 1.5 kg

k = spring constant = ?

Therefore,

10 s = 2π√(1.5 kg/k)

squaring on both sides we get:

100 s² = 4π²(1.5 kg/k)

k = 6π² kg/100 s²

k = 0.6 N/m

Answer:

Explanation:

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Answer:

Dead lifting uses tho muscle fundamentals

Explanation:

Answer:

Fast twitch

Explanation:

Edmentum

Answer:

The kinetic energy at a displacement of half the amplitude is 37.5 J

Explanation:

Given;

total energy on the spring, E = 50 J

When the displacement is half the amplitude, the total energy in the spring is sum of the kinetic energy and elastic potential energy.

E = K + U

Where;

K is the kinetic energy

U is the elastic potential energy

K = E - U

K = E - ¹/₂KA²

When the displacement is half = ¹/₂(A) = A/₂

K = E - ¹/₂K(A/₂)²

K = E - ¹/₂K(A²/₄)

K = E - ¹₄(¹/₂KA²)

Recall, E = ¹/₂KA²

K = ¹/₂KA² - ¹₄(¹/₂KA²)     (recall from simple arithmetic, 1 - ¹/₄ = ³/₄)

K = 1(¹/₂KA²) - ¹₄(¹/₂KA²)  = ³/₄(¹/₂KA²)

K = ³/₄(¹/₂KA²)

But E = ¹/₂KA² = 50J

K = ³/₄ (50J)

K = 37.5 J

Therefore, the kinetic energy at a displacement of half the amplitude is 37.5 J

The kinetic energy when the displacement is half the amplitude

Given the following data:

To find the kinetic energy when the displacement is half the amplitude:

The total energy of the system of a block and a spring is the sum of the spring's elastic potential energy and kinetic energy of the block and it's proportional to the square of the amplitude.

Mathematically, the total energy of the system of a block and a spring is given by the formula:

[tex]T.E = U + K.E[/tex]   .....equation 1.

[tex]T.E = \frac{1}{2} kA^2[/tex]

Where:

Making K.E the subject of formula, we have:

[tex]K.E = T.E - U[/tex]   .....equation 2.

But, [tex]U = \frac{1}{2} kx^2[/tex]    ....equation 3.

Where:

Substituting the eqn 3 into eqn 2, we have:

[tex]K.E = T.E - \frac{1}{2} kx^2[/tex]

[tex]K.E = T.E - \frac{1}{2} k(\frac{A}{2})^2\\\\K.E = T.E - \frac{1}{2} k(\frac{A^2}{4})\\\\K.E = T.E - \frac{1}{4} (\frac{1}{2} kA^2)\\\\K.E = T.E - \frac{1}{4} (T.E)\\\\K.E = 50 - \frac{1}{4} (50)\\\\K.E = 50 - 12.5[/tex]

K.E = 37.5 Joules.

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Answer:

E = 0.645J

Explanation:

In order to calculate the total mechanical energy of the system, you take into account that if the zero of energy is at the equilibrium position, then the total mechanical energy is only the elastic potential energy of the spring.

You use the following formula:

[tex]E=U_e=\frac{1}{2}kA^2[/tex]         (1)

k: spring constant = ?

A: amplitude of the oscillation = 7.50cm = 0.075m

The spring constant is given by:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

[tex]k=4\pi^2f^2m[/tex]         (2)

f: frequency of the oscillation = 1.95Hz

m: mass of the piece of iron = 1.53kg

You replace the expression (1) into the equation (2) and replace the values of all parameters:

[tex]E=\frac{1}{2}(4\pi^2f^2m)A^2=2\pi^2f^2mA^2\\\\E=2\pi^2(1.95Hz)^2(1.53kg)(0.075m)^2=0.645J[/tex]

The totoal mechanical energy of the system is 0.645J

Answer:

a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Explanation:

a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:

First arrow

[tex]U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{x,1}[/tex], [tex]K_{x,2}[/tex] - Initial and final horizontal translational kinetic energy, measured in joules.

[tex]K_{y,1}[/tex], [tex]K_{y,2}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.

Now, the system is expanded and simplified:

[tex]m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0[/tex]

[tex]g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})[/tex]

[tex]y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}[/tex]

Where:

[tex]y_{1}[/tex]. [tex]y_{2}[/tex] - Initial and final height of the arrow, measured in meters.

[tex]v_{y,1}[/tex], [tex]v_{y,2}[/tex] - Initial and final vertical speed of the arrow, measured in meters.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

The initial vertical speed of the arrow is:

[tex]v_{y,1} = v_{1}\cdot \sin \theta[/tex]

Where:

[tex]v_{1}[/tex] - Magnitude of the initial velocity, measured in meters per second.

[tex]\theta[/tex] - Initial angle, measured in sexagesimal degrees.

If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the initial vertical speed is:

[tex]v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}[/tex]

[tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex]

If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex] and [tex]v_{y,2} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:

[tex]y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]y_{2} - y_{1} = 56.712\,m[/tex]

Second arrow

[tex]U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,3}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{y,1}[/tex], [tex]K_{y,3}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.

[tex]m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0[/tex]

[tex]g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})[/tex]

[tex]y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}[/tex]

If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} = 82\,\frac{m}{s}[/tex] and [tex]v_{y,3} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:

[tex]y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]y_{3} - y_{1} = 342.816\,m[/tex]

The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.

b) The total energy of each system is determined hereafter:

First arrow

The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:

[tex]E = U + K_{x}[/tex]

The expression is now expanded:

[tex]E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}[/tex]

Where [tex]v_{x}[/tex] is the horizontal speed of the arrow, measured in meters per second.

[tex]v_{x} = v_{1}\cdot \cos \theta[/tex]

If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the horizontal speed is:

[tex]v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}[/tex]

[tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex]

If [tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{max} = 56.712\,m[/tex] and [tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex], the total mechanical energy is:

[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}[/tex]

[tex]E = 201.720\,J[/tex]

Second arrow:

The total mechanical energy is equal to the potential gravitational energy. That is:

[tex]E = m\cdot g \cdot y_{max}[/tex]

[tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]y_{max} = 342.816\,m[/tex]

[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)[/tex]

[tex]E = 201.720\,J[/tex]

Both arrows have a total mechanical energy at their maximum height of 201.720 joules.

Answer:

r = 0.22m

Explanation:

To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.

Then, you have:

[tex]F_c=F_e=ma_c[/tex]      (1)

m: mass of the particle = 20g = 20*10-3 kg

ac: centripetal acceleration = ?

q: charge of the particle = 5*10^-6C

Fe: electric force between the charges

The electric force is given by:

[tex]F_e=k\frac{qq'}{r^2}[/tex]             (2)

r: radius of the orbit

q': charge of the particle at the center of the orbit = -5*10^-6C

Furthermore, the centripetal acceleration is:

[tex]a_c=\frac{v^2}{r}[/tex]                 (3)

v: speed of the particle = 7m/s

You replace the expressions (2) and (3) in the equation (1) and solve for r:

[tex]k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}[/tex]

Finally, you replace the values of all parameters in the previous expression:

[tex]r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m[/tex]

The radius of the circular trajectory is 0.22m

Answer:

1.06 atm

Explanation:

On the open end of the tube, the pressure will be the sum of atmospheric pressure and the pressure due to the height of water

The pressure due to a height of water = ρgh

where ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the height of the water column

The height of water column on the open end = 100 cm = 1 m

pressure on this end = ρgh = 1000 x 9.81 x 1 = 9810 Pa

Atmospheric pressure = 101325 Pa

The total pressure on the open end =  101325 Pa + 9810 Pa = 111135 Pa

The pressure due to the water column on the closed end = ρgh

The height of the water in the closed end = 40 cm = 0.4 m

The pressure due to this column of water = 1000 x 9.81 x 0.4 = 3924 Pa

The resultant pressure on the water on the top of the closed end of the tube = 111135 Pa - 3924 Pa = 107211 Pa

In atm unit, this pressure = 107211/101325 = 1.06 atm

Answer:

8.92 rpm

Explanation:

Given that

Radius of the merry go round, r = 2.2 m

Initial moment of inertia, I1 = 245 kgm²

Initial speed of rotation, w1 = 11 rpm

Mass of the child, m = 26 kg

To solve the problem, we use the law conservation of momentum

I1w1 = I2w2

I2 = mr² + I1

I2 = 245 + 26 * 2.2

I2 = 245 + 57.2

I2 = 302.2 kgm²

Now, applying the formula, we have

I1w1 = I2w2

245 * 11 = 302.2 * w2

w2 = 2695 / 302.2

w2 = 8.92 rpm

Thus, the new angular speed of the merry go round is 8.92 rpm

Answer:

B : It is hard to stop using.

Explanation:

just took the quiz ! hope this helps with anyone who needs it !

Due to the dependency on natural gas as a fuel, it is hard to stop using.

Natural gas is a fossil fuel which is obtained from the ground in association with petroleum.

Natural gas consists mainly of petroleum.

It is a non-renewable energy source.

Natural gas use contributes to global warming

However, due to the dependency on natural gas as a fuel, it is hard to stop using.

Learn more about natural gas at: https://brainly.com/question/815922

Answer:

Momentum is conserved when there are no outside forced present and it has an equal and opposite reaction, also momentum is conserved the ball's momentum is transferred to the ground. This first instance is the case of a Closed system.

The second case where momentum is not conserved is when there is a variation or difference in the moment of the ball because of influence of external forces

Answer:

Explanation:

Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.

Answer:

4.78 second

Explanation:

given data

vertical cliff = 41 m

height = 112 m

solution

we know here time taken to fall vertically from the cliff =  time taken to move horizontally   ..........................1

so we use here vertical component of ball

and that is accelerated motion with initial velocity = 0

so we can solve for it as

height = 0.5 ×  g ×  t²     ........................2

put here value

112 = 0.5 ×  9.8 ×  t²    

solve it we get

t²   = 22.857

t = 4.78 second

ball thrown horizontally from the top of the cliff in 4.78 second

Answer:

14.79 kgm/s

Explanation:

Data provided in the question

Let us assume the mass of baseball =  m = 0.145 kg

The Initial velocity of pitched ball = [tex]v_i[/tex] = 47 m/s

Final velocity of batted ball in the opposite direction = [tex]v_f[/tex]= -55m/s

Based on the above information, the change in momentum is

[tex]\Delta P = m(v_f -v_i)[/tex]

[tex]= 0.145 kg(-55m/s - 47m/s)[/tex]    

= 14.79 kgm/s

Hence, the magnitude of the change in momentum of the ball is 14.79 kg m/s

Answer:

C - Arteries grow smaller

Explanation:

The option choices are:

A. Faster post-exercise recovery time

B. Lungs expand more easily

C. Arteries grow smaller

D. Diaphragm grows stronger

Explanation:

There are many advantages of cardiorespiratory fitness. It can decrease the risk of heart disease, lung cancer, type 2 diabetes, stroke, and other diseases. Cardiorespiratory health helps develop lung and heart conditions and enhances feelings of well-being.

Answer:

The bat and the ball were in contact for 1.8 x 10⁻³ s

Explanation:

Given;

mass of baseball, m = 0.14 kg

initial velocity of the baseball, u = 27.1 m/s

applied force in opposite direction, F = -5000 N

final velocity in opposite direction, v = -37 m/s

Note: The applied force and final velocity are negative because they act in opposite direction to the initial velocity.

impulse received by the body = change in momentum of the  body

Ft = Δmv

Ft = mv - mu

Ft = m(v-u)

t = m(v-u) / F

[tex]t = \frac{0.14(-37-27.1)}{-5000} \\\\t = \frac{0.14(-64.1)}{-5000} \\\\t = \frac{-8.974}{-5000} \\\\t = 0.0018 \ s\\\\t = 1.8*10^{-3} \ s[/tex]

Therefore, the bat and the ball were in contact for 1.8 x 10⁻³ s

Answer:

Explanation:

speed of light = 3 x 10⁸ m /s .

distance between moon and the earth = 3.77 x 10⁵  x 10³m .

Time taken by light to cover the distance

= distance / speed

= 3.77 x 10⁸ / 3 x 10⁸

= 1.256  s

Answer:

The speed of the arrow after passing through the target is 30.1 meters per second.

Explanation:

The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:

[tex]m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}[/tex]

Where:

[tex]m_{a}[/tex], [tex]m_{t}[/tex] - Masses of arrow and target, measured in kilograms.

[tex]v_{a,o}[/tex], [tex]v_{a,f}[/tex] - Initial and final speeds of the arrow, measured in meters per second.

[tex]v_{t,o}[/tex], [tex]v_{t,f}[/tex] - Initial and final speeds of the target, measured in meters per second.

The final speed of the arrow is now cleared:

[tex]m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})[/tex]

[tex]v_{a,f} = v_{a,o} + \frac{m_{t}}{m_{a}} \cdot (v_{t,o}-v_{t,f})[/tex]

If [tex]v_{a,o} = 2.1\,\frac{m}{s}[/tex], [tex]m_{t} = 0.3\,kg[/tex], [tex]m_{a} = 0.0225\,kg[/tex], [tex]v_{t,o} = 2.10\,\frac{m}{s}[/tex] and [tex]v_{t,f} = 0\,\frac{m}{s}[/tex], the speed of the arrow after passing through the target is:

[tex]v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )[/tex]

[tex]v_{a,f} = 30.1\,\frac{m}{s}[/tex]

The speed of the arrow after passing through the target is 30.1 meters per second.

Answer:

I₂/I₁ = 0.53

Explanation:

During the motion the angular momentum of the skater remains conserved. Therefore:

Angular Momentum of Skater Before Pulling Arms = Angular Momentum of Skater After Pulling Arms

L₁ = L₂

but, the formula for angular momentum is:

L = Iω

Therefore,

I₁ω₁ = I₂ω₂

I₂/I₁ = ω₁/ω₂

where,

I₁ = Initial Moment of Inertia

I₂ = Final Moment of Inertia

ω₁ = Initial Angular Velocity = 3.14 rad/s

ω₂ = Final Angular velocity = 5.94 rad/s

Therefore,

I₂/I₁ = (3.14 rad/s)/(5.94 rad/s)

I₂/I₁ = 0.53

Answer:

The  diameter is  [tex]d = 6.5 *10^{-4} \ m[/tex]

Explanation:

From the question we are told that

   The length of the cylinder is  [tex]l = 120 \ m[/tex]

     The resistance is  [tex]\ 6.0\ \Omega[/tex]

     The  resistivity of the metal is [tex]\rho = 1.68 *10^{-8} \ \Omega \cdot m[/tex]

Generally the resistance of the cylindrical wire is  mathematically represented as

         [tex]R = \rho \frac{l}{A }[/tex]

The cross-sectional area of the cylindrical wire is  

        [tex]A = \frac{\pi d^2}{4}[/tex]

Where  d is the diameter, so

         [tex]R = \rho \frac{l}{\frac{\pi d^2}{4 } }[/tex]

=>     [tex]d = \sqrt{ \rho* \frac{4 * l }{\pi * R } }[/tex]

       [tex]d = \sqrt{ 1.68 *10 ^{-8}* \frac{4 * 120 }{3.142 * 6 } }[/tex]

       [tex]d = 6.5 *10^{-4} \ m[/tex]

Answer:

  p₁ = - p₂

the moment value of the two particles is the same, but its direction is opposite

Explanation:

When a nucleus divides spontaneously, the moment of the nucleic must be conserved, for this we form a system formed by the initial nucleus and the two fragments of the fission, in this case the forces during the division are internal and the moment is conserved

initial instant. Before fission

               p₀ = 0

since they indicate that the nucleus is at rest

final moment. After fission

             [tex]p_{f}[/tex] = m₁ v₁ + m₂ v₂

             p₀ = p_{f}

             0 = m₁ v₁ + m₂v₂

             m₁ v₁ = -m₂ v₂

              p₁ = - p₂

this indicates that the moment value of the two particles is the same, but its direction is opposite

Answer:

71nC is the total charge of the rod

Explanation:

See attached file

The total charge on the rod is equal to 3.3 × 10⁻⁸ C.

The force on the charge in a uniform electric field E is given by:

F = qE    where q is charge in coulombs

The electric field due to the charge associated with the rod is given by:

[tex]E =\frac{kQ}{r\sqrt{r+\frac{L^2}{4} } }[/tex]

Where r is the distance between the bead and the rod, L is the length of the glass rod and Q is the charge on the rod.

The force experienced by the bead charged is,

[tex]F =\frac{kqQ}{r\sqrt{r+\frac{L^2}{4} } }[/tex]

From the above equation, we can find the value of Q as:

[tex]Q =\frac{Fr\sqrt{r+\frac{L^2}{4} } }{kq}[/tex]

Given, the value of force, F = 940μN = 940 ×10⁻⁹ C

The length of glass rod, L = 10cm = 0.1 m and r = 4cm = 0.04 m

[tex]Q =\frac{(940\times 10^{-6}N)(0.04m)\sqrt{(0.04m)+\frac{(0.1m)^2}{4} } }{(8.99\times 10^9 N.m^2/C^2)(8\times 10^{-9}C)}[/tex]

[tex]Q= 0.5228\times 10^{-6}\times\sqrt{0.0041}[/tex]

[tex]Q = 3.3\times 10^{-8} C[/tex]

Therefore, the total charge on the rod is  3.3 × 10⁻⁸ C.

Learn more about the force on a charge, here:

https://brainly.com/question/22042360

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Answer:

55.80 g/s

Explanation:

From the question,

Flow rate = density×Area×velocity.

φ = ρ×A×V................... Equation 1

Where φ = flow rate of blood, ρ = density of blood, A = cross sectional area of blood, V = velocity of blood.

Given: ρ = 0.846 g/cm³, A = 1.36 cm², V = 48.5 cm/s.

Substitute these values into equation 1

φ = 0.846×1.36×48.5

φ = 55.80 g/s

Hence, the flow rate of  the blood = 55.80 g/s

Answer:

Only if it starts moving

Explanation:

Work done is defines as [tex]W=F.d=Fdcos\Theta[/tex]

In two case work done will be zero

First case is that when force and displacement are perpendicular to each other

What's more, other case is that when there is no displacement

So for work to be done there must have displacement,  in the event that there is no displacement then there is no work done

Answer:

Length = 0.4882 m

Explanation:

given data

fundamental frequency = 175 Hz

air temperature = 18.0°C

solution

we will apply here fundamental frequency formula that is

F = [tex]\frac{v}{4L}[/tex]      ....................1

here v =   [tex]331 \sqrt{1+\frac{T}{273}}[/tex]  

here 331 m/s is speed of sound in air

so v =  [tex]331 \sqrt{1+\frac{18}{273}}[/tex]   = 341.74 m/s

now put value in equation 1 we get

F = [tex]\frac{v}{4L}[/tex]

[tex]175 = \frac{341.74}{4L}[/tex]  

Length = 0.4882 m

Answer:

2. Dot Product

Explanation:

The calculation of the electric flux gives an scalar result.

When we tray to calculate how much electric field passes trough a surface, we are calculating a scalar value. Furthermore, the concept of flux requires the calculation of a scalar value.

Also it is necessary to take into account that the magnitude of the flux trough a surface depends of the inclination of the surface respect to the direction of the electric field. This is taken into account sufficiently by a dot product.

Then, the answer is:

2. Dot Product

Answer:  

Kepler's Third Law:  The square of the period of any planet about the sun is proportional to cube of its mean distance from the sun.

Mathematically:  T^2 = K R^3

So  (TA / TB)^2 = (RA / RB)^3

TB^2 = TA^2 * (RB / RA)^3

TB^2 = 10^2 * 16^3

TB = (409600)^1/2 = 640 days

Answer:

14.1 m/s

Explanation:

From the question,

μk = a/g...................... Equation 1

Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.

make a the subject of the equation

a = μk(g).................. Equation 2

Given: μk = 0.160, g = 9.8 m/s²

Substitute into equation 2

a = 0.16(9.8)

a = 1.568 m/s²

Using,

F = ma

Where F = force, m = mass.

Make m the subject of the equation

m = F/a................... Equation 3

m = 160/1.568

m = 102.04 kg.

Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.

Assuming the initial velocity of the skier to be zero

Fd+mgμ = 1/2mv²........................Equation 4

Where v = speed of the skier after crossing the patch, d = distance/width of the patch.

v = √2(Fd+mgμ)/m)................ Equation 5

Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16

Substitute these values into equation 5

v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]

v = √197.57

v = 14.1 m/s

v = 9.86 m/s

Answer:

a = 0.43 m/s²

Explanation:

First, we need to find the velocity of the satellite:

Velocity = V = Distance Covered/Time Taken

here,

Distance = 1 revolution = 2π(1.5 x 10⁷ m) = 9.43 x 10⁷ m

Time = (8.65 hours)(3600 s/1 hour) = 31140 s

Therefore,

V = (9.43 x 10⁷ m)/(31140 s)

V = 3028.26 m/s

Now, the acceleration of the satellite will be equal to the centripetal acceleration, with the center of circular motion as the center of earth:

a = V²/R

where,

R = 1.5 x 10⁷ m + 0.638 x 10⁷ m

R = 2.138 x 10⁷ m

Therefore,

a = (3028.26 m/s)²/(2.138 x 10⁷ m)

a = 0.43 m/s²

By  the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "

An experiment would be a technique used to confirm or deny a hypothesis, as well as assess the likelihood or effectiveness of something that has never been tried before.

Hydrochloric acid is a kind of compound in which hydrogen and chlorine element is present.

Maintain a safe distance between your hands and your body, mouth, eyes, as well as a face when utilizing lab supplies and chemicals.

By  the experiment "By  the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "

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