Which of the following emissions is associated with burning coal? a. sulfur dioxide b. carbon dioxide c. nitrous oxides d. all of the above

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:

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:  

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:

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:

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.

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Answer:

11.7°

Explanation:

See attached file

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:

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:

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:

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:

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:

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:

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.

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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Answer:

15amps

Explanation:

V=IR

I=V/R

I = 120/8

I = 15 amps

Answer:

Explanation:

check this out and rate me

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:

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:

Cost = $ 1.48

Cost = $ 0.005

Cost = $ 0.38

Explanation:

given data

electrical transmission varies = $0.070/kWh to $0.258/kWh

average value = $0.110/kWh

solution

when leaving a 40-W porch light on for two weeks while you are on vacation so cost will be

first we get here energy consumed that is express as

E = Pt    .................1

here E is Energy Consumed and Power Delivered is P and t is time

so power is here 0.04 KW and t = 2 week = 336 hour

so

put value in 1 we get

E = 0.04 × 336

E = 13.44 KWh

so cost will be as

Cost = E × Unit Price    .............2

put here value and we get

Cost = 13.44 × 0.11

Cost = $ 1.48

and

when you making a piece of dark toast in 3.00 min with a 970-W toaster

so energy consumed will be by equation 1 we get

E = Pt

power is = 0.97 KW and time = 3 min = 0.05 hour

put value in equation 1 for energy consume

E = 0.97 × 0.05 h

E = 0.0485 KWh

and we get cost by w\put value in equation 2 that will be

cost =  E × Unit Price

cost = 0.0485 × 0.11

Cost = $ 0.005

and

when drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer

from equation 1 we get energy consume

E = Pt

Power Delivered = 5.203 KW and time = 40 min = 0.67 hour

E = 5.203 × 0.67

E = 3.47 KWh

and

cost will by put value in equation 2

Cost = E × Unit Price

Cost = 3.47 × 0.11

Cost = $ 0.38

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is  [tex]v = 3.79 *10^{5} \ m/s[/tex]  

Explanation:

From the question we are told that

    The  magnitude of the electric field is  [tex]E = 144 \ kV /m = 144*10^{3} \ V/m[/tex]

     The magnetic field is  [tex]B = 0.38 \ T[/tex]

The force due to the electric field is mathematically represented as

      [tex]F_e = E * q[/tex]

and

The force due to the magnetic field is mathematically represented as

    [tex]F_b = q * v * B * sin(\theta )[/tex]

Now given that it is perpendicular ,  [tex]\theta = 90[/tex]

=>   [tex]F_b = q * v * B * sin(90)[/tex]

=>   [tex]F_b = q * v * B[/tex]

Now  given that it is not deflected it means that

        [tex]F_ e = F_b[/tex]

=>    [tex]q * E = q * v * B[/tex]

=>   [tex]v = \frac{E}{B }[/tex]

 substituting values

     [tex]v = \frac{ 144 *10^{3}}{0.38 }[/tex]

     [tex]v = 3.79 *10^{5} \ m/s[/tex]

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:

The sun will be located near the Gemini constellation at sunset

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:

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:

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:

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:

a. for destructive interference

λmax= 240m

b. for constructive interference

λmax = 120m

Explanation:

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:

the correct answer is

All four known fundamental interactions are non-contact forces: Gravity, the force of attraction that exists among all bodies that have mass. ... Examples of this force include: electricity, magnetism, radio waves, microwaves, infrared, visible light, X-rays and gamma rays.

Explanation:

hope this helps you!!!!