A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises to a height of 1.5 m. a) What is the ball's velocity just before it hits the floor? b) What is the ball's velocity just after it leaves the floor? c) If the ball is in contact with the floor for 0.02 seconds, what are the magnitude and direction of the ball's average acceleration while in contact with the floor?

Answer:

B. children have too much energy and play will rid them of that energy.

Explanation:

The surplus energy theory of play suggests that human being have excess of energy that must be released through active play.

This theory is given by Friedreich Schiller. Therefore, the correct answer is children have too much energy and play will rid them of that energy. rest of the option are examples of other theories, whereas Option B is the example of surplus energy theory.

Answer:

Zero

Explanation:

Electric flux is defined as the number of electric field lines which passes through any area in the direction of area vector.

The formula for the electric flux is given by

[tex]\phi =\overrightarrow{E}.\overrightarrow{dS}[/tex]

Here, E is the strength of electric field and dS be the area vector.

It is a scalar quantity.

According to the Gauss's theorem, the electric flux passing through any surface is equal to the [tex]\frac{1}{\epsilon _{0}}[/tex] times the total charge enclosed through the surface.

Here, the charge enclose is zero, so the total flux is also zero.  

The electric field strengths at various points around a charged object can be calculated using Coulomb's Law. They depend on the charge of the object and the distance from it. Fields produced by multiple charges need to be added vectorially.

You're asking about the electric field strengths at varying distances from a charged glass rod, located next to a plastic rod with a charge of equal magnitude, but of opposite sign. The electric field strength E at a point in the vicinity of a charged object can be derived using Coulomb's Law. This law states that the force F between two charges q1 and q2 is proportional to the product of the charges and inversely proportional to the square of the distance r between them.

Given this, the electric field E created by a charge q at a distance r from the charge is given by E = k|q| / r², where k is Coulomb's constant, approximately 9 × 10⁹ N•m²/C². The direction of the electric field is determined by the sign of the charge. When it's positive, the field lines are going outward, and when it's negative, they are directed inward.

To calculate the net electric field at a point on the line connecting the midpoints of the rods, we consider the electric fields produced by both rods. Since they're opposite in sign, the fields will have opposite directions, and they should be added vectorially. The magnitudes and directions of the fields will also depend on the specific distances (1.0 cm, 2.0 cm, and 3.0 cm) from the glass rod.

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Answer:

a) 1511 MW

b) 44%

Explanation:

The thermal power will be the electric power plus the heat taken away by the cooling water.

Qt = P + Qc

The heat taken away by the water will be:

Qc = G * Cp * (t1 - t0)

The Cp of water is 4180 J/(kg K)

The density of water is 1 kg/L

Then

G = 1.17 * 10^8 L/h * 1 kg/L * 1/3600 h/s = 32500 kg/s

Now we calculate Qc

Qc = 32500 * 4180 * (29.8 - 23.6) = 842*10^6 W = 842 MW

The total thermal power then is

Qt = 669 + 842 = 1511 MW

The efficiency is

η = P / Qt

η = 669 / 1511 = 44%

Answer:

The electric field intensity at the aircraft is [tex]7.18\times10^{5}\ N/C[/tex]

Explanation:

Given that,

Altitude = 2000 m

Charge = +40 C

At the position of the airplane, the electric field will be due to both the positive and negative charges and will be downwards due to both the charges.

The position is equidistant from both the charges.

We need to calculate the resultant electric field

The contribution due to the positive charge at 3000 m altitude

Using formula of electric field

[tex]E_{+}=k\dfrac{|q|}{r^2}[/tex]

Put the value into the formula

[tex]E_{+}=8.99\times10^{9}\times\dfrac{40}{(3000-2000)^2}[/tex]

[tex]E_{+}=3.59\times10^{5}\ N/C[/tex]

The contribution due to the negative charge at 1000 m altitude

Using formula of electric field

[tex]E_{-}=8.99\times10^{9}\times\dfrac{40}{(2000-1000)^2}[/tex]

[tex]E_{-}=3.59\times10^{5}\ N/C[/tex]

We need to calculate the electric field intensity at the aircraft

The resultant electric field is

[tex]E=E_{+}+E_{-}[/tex]

Put the value into the formula

[tex]E=3.59\times10^{5}+3.59\times10^{5}[/tex]

[tex]E=7.18\times10^{5}\ N/C[/tex]

Hence, The electric field intensity at the aircraft is [tex]7.18\times10^{5}\ N/C[/tex]

Answer:

Explanation:

Given

ambient Pressure =98.10 kPa

(a)gauge pressure 152 kPa

we know

Absolute pressure=gauge pressure+Vacuum  Pressure

[tex]P_{abs}[/tex]=152+98.10=250.1 kPa or 36.27 psi

(b)[tex]P_{gauge}[/tex]=67.5 Torr or 8.99 kpa

as 1 Torr is 0.133 kPa

[tex]P_{abs}[/tex]=8.99+98.10=107.09 kPa or 15.53 psi

(c)[tex]P_{vaccum}[/tex]=0.1 bar or 10 kPa

Thus absolute pressure=98.10-10=88.10 kPa or 12.77 psi

as 1 kPa is equal to 0.145 psi

(d)[tex]P_{vaccum}[/tex]=0.84 atm  or 85.113 kPa

as 1 atm is equal to 101.325 kPa

[tex]P_{abs}[/tex]=98.10-85.11=12.99 kPa or 1.88 psi

Answer:

Current through circuit will be [tex]0.2706\times 10^6A[/tex]

Explanation:

We have given source voltage v = 50 volt

Resistance [tex]R=25Mohm=25\times 10^6ohm[/tex]

Capacitance [tex]C=0.1\mu F=0.1\times 10^{-6}F[/tex]

Time t = 5 sec

Time constant of RC circuit is given by [tex]\tau =RC=25\times 10^6\times 0.1\times 10^{-6}=2.5sec[/tex]

We know that voltage across capacitor is given by [tex]v_c=v_s(1-e^{\frac{-t}{\tau }})[/tex]

[tex]v_c=50(1-e^{\frac{-5}{2.5 }})=43.2332v[/tex]

So current will be [tex]=\frac{v_s-v_c}{R}=\frac{50-43.2332}{25\times 10^{-6}}=0.2706\times 10^6A[/tex]

So current through circuit will be [tex]0.2706\times 10^6A[/tex]

Answer:

option B

Explanation:

the correct answer is option B

when negative charge is placed in electric field then direction will be accelerated in opposite direction of the field.

charged electric particle produce electric field and they exert force on other charged particle.

when a charged particle is positive it will accelerate in the direction of the electric field if the charge is negative then particle will accelerate in opposite direction of electric field.

The pitch of the helical path followed by an electron in a magnetic field is given by the formula p = v * cos(theta) * T, where v is the velocity of the electron, theta is the angle between the direction of the magnetic field and the electron's velocity, and T is the period of the electron's motion in the magnetic field. By substituting the given values into this formula, we can calculate the pitch.

The pitch of a helical path can be determined by considering how the charged particle reacts in a magnetic field. The helical path of an electron in a magnetic field is caused by the perpendicular and parallel components of the electron's velocity. In particular, the parallel component of the velocity, v(cos(theta)), is responsible for the linear movement of the electron along the field lines. This leads to the corkscrew-like helical path, and the pitch of this helix is equivalent to the linear distance moved by the electron in one revolution.

The pitch (p) can be calculated by the formula p = v * cos(theta) * T, where v is the velocity of the electron, theta is the angle between the direction of the magnetic field and the velocity of the electron, and T is the period of circular motion of the electron in the magnetic field. From the provided data, we can first calculate the velocity of the electron using its kinetic energy, 22.5 eV: v = sqrt((2 * KE) / m), where KE is the kinetic energy and m is the mass of the electron (9.11 * 10^-31 kg). Next, we calculate the period using the formula T = 2 * pi * m / (q * B), where q is the charge of the electron (-1.6 * 10^-19 C) and B is the magnetic field intensity (4.55 * 10^4 T).

Finally, we substitute these values into the pitch formula and calculate the pitch of the helical path taken by the electron in the magnetic field.

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The pitch of the helical path taken by the electron is approximately [tex]\( 9.20 \times 10^{-11} \)[/tex] meters.

To find the pitch of the helical path taken by the electron, we need to follow these steps:

1. Calculate the speed of the electron:

  The kinetic energy (K.E.) of the electron is given by:

  [tex]\[ \text{K.E.} = \frac{1}{2}mv^2 \][/tex]

  where [tex]\( m \)[/tex] is the mass of the electron [tex](\( 9.11 \times 10^{-31} \) kg)[/tex] and [tex]\( v \)[/tex] is the speed of the electron.

First, we need to convert the kinetic energy from electron volts (eV) to joules (J):

  [tex]\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \][/tex]

  [tex]\[ 22.5 \, \text{eV} = 22.5 \times 1.602 \times 10^{-19} \, \text{J} = 3.6045 \times 10^{-18} \, \text{J} \][/tex]

  Now, using the kinetic energy formula:

 [tex]\[ \text{K.E.} = \frac{1}{2}mv^[/tex]

To find the pitch of the helical path taken by the electron, we need to determine the components of the electron's velocity and the motion along the magnetic field.

1. Calculate the speed of the electron:

The kinetic energy (KE) of the electron is given by:

  [tex]\[ \text{KE} = \frac{1}{2} mv^2 \][/tex]

  where m is the mass of the electron [tex](\( 9.11 \times 10^{-31} \) kg)[/tex] and v is the speed of the electron.

  First, convert the kinetic energy from electron volts (eV) to joules (J):

 [tex]\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \][/tex]

 [tex]\[ 22.5 \, \text{eV} = 22.5 \times 1.602 \times 10^{-19} \, \text{J} = 3.6045 \times 10^{-18} \, \text{J} \][/tex]

  Now, solve for the speed \( v \) using the kinetic energy formula:

  [tex]\[ \frac{1}{2} mv^2 = 3.6045 \times 10^{-18} \, \text{J} \][/tex]

 [tex]\[ v^2 = \frac{2 \times 3.6045 \times 10^{-18}}{9.11 \times 10^{-31}} \][/tex]

  [tex]\[ v^2 = 7.9137 \times 10^{12} \][/tex]

 [tex]\[ v = \sqrt{7.9137 \times 10^{12}} \approx 2.81 \times 10^6 \, \text{m/s} \][/tex]

2. Determine the components of the velocity:

The angle between the velocity and the magnetic field is [tex]\( 65.5^\circ \)[/tex]. The component of the velocity parallel to the magnetic field [tex](\( v_{\parallel} \))[/tex] and the component perpendicular to the magnetic field[tex](\( v_{\perp} \))[/tex] can be found using trigonometry:

 [tex]\[ v_{\parallel} = v \cos(65.5^\circ) \][/tex]

  [tex]\[ v_{\perp} = v \sin(65.5^\circ) \][/tex]

  Calculate these components:

  [tex]\[ v_{\parallel} = 2.81 \times 10^6 \, \text{m/s} \times \cos(65.5^\circ) \approx 2.81 \times 10^6 \times 0.417 \approx 1.17 \times 10^6 \, \text{m/s} \][/tex]

 [tex]\[ v_{\perp} = 2.81 \times 10^6 \, \text{m/s} \times \sin(65.5^\circ) \approx 2.81 \times 10^6 \times 0.909 \approx 2.55 \times 10^6 \, \text{m/s} \][/tex]

3. Find the cyclotron frequency [tex](\( f \))[/tex] and the period [tex](\( T \))[/tex] of the electron's circular motion:

  The cyclotron frequency is given by:

  [tex]\[ f = \frac{qB}{2\pi m} \][/tex]

  where q is the charge of the electron [tex](\( 1.602 \times 10^{-19} \) C)[/tex], B is the magnetic field [tex](\( 4.55 \times 10^4 \) T)[/tex], and m is the mass of the electron.

[tex]\[ f = \frac{1.602 \times 10^{-19} \times 4.55 \times 10^4}{2\pi \times 9.11 \times 10^{-31}} \][/tex]

[tex]\[ f \approx \frac{7.29 \times 10^{-15}}{5.72 \times 10^{-31}} \approx 1.27 \times 10^{16} \, \text{Hz} \][/tex]

  The period [tex](\( T \))[/tex] is the inverse of the frequency:

  [tex]\[ T = \frac{1}{f} \approx \frac{1}{1.27 \times 10^{16}} \approx 7.87 \times 10^{-17} \, \text{s} \][/tex]

4. Calculate the pitch of the helical path:

The pitch is the distance the electron travels parallel to the magnetic field in one period:

 [tex]\[ \text{Pitch} = v_{\parallel} \times T \][/tex]

 [tex]\[ \text{Pitch} \approx 1.17 \times 10^6 \, \text{m/s} \times 7.87 \times 10^{-17} \, \text{s} \approx 9.20 \times 10^{-11} \, \text{m} \][/tex]

The temperature of the boiling chlorine is calculated using the resistance change of a platinum wire thermometer and the temperature coefficient of resistivity. After calculations, the boiling point of chlorine is found to be -34.6°C.

The temperature of the boiling chlorine can be deduced using the relation between temperature change and resistance change for a platinum wire thermometer. The equation we use is:

R = R0 (1 + αΔT)

Where R represents the resistance at the new temperature, R0 the resistance at a reference temperature (20°C in this case), α the temperature coefficient of resistivity, and ΔT the change in temperature. We can rearrange this equation to solve for the new temperature:

ΔT = ∂(T₂ - T₁)

Giving us:

(99.6 ohms / 125 ohms - 1) = 3.72 × 10-3°C-1 ∂T

Simplifying:

ΔT = (0.7968 - 1) / 3.72 × 10-3°C-1

ΔT = -0.2032 / 3.72 × 10-3°C-1

ΔT = -54.6°C

Thus, the boiling point of chlorine is:

20°C – 54.6°C = -34.6°C

Answer:

the height above the ground where  the board makes contact with the wall equals 5.297 meters above ground.

Explanation:

The wall and ladder are shown in the attached figure

In the figure we can see that

[tex]cos(28^{o})=\frac{H}{L}\\\\cos(28^{o})=\frac{H}{6}\\\\\therefore H=6.0\times cos(28^{o})==5.297meters[/tex]

Answer:

Current density [tex]j=1.44\times 10^7A/m^2[/tex]

Electron density [tex]=5.55\times 10^{18}electron/sec[/tex]

Explanation:

We have given power = 100 watt

Current = 0.89 A

Diameter d = 0.280 mm

So radius [tex]r=\frac{d}{2}=\frac{0.28}{2}=0.14mm=0.14\times 10^{-3}m[/tex]

Area [tex]A=\pi r^2=3.14\times (0.14\times 10^{-3})^2=0.016\times 10^{-6}m^2[/tex]

We know that current density [tex]J=\frac{I}{A}=\frac{0.89}{0.016\times 10^{-6}}=1.44\times 10^7A/m^2[/tex]

Now we have to calculate the electron density

We have current i = 0.89 A = 0.89 J/sec

Charge on 1 electron [tex]1.6\times 10^{-19}C/electron[/tex]

So electron density [tex]=\frac{0.89j/sec}{1.6\times 10^{-19}C/electron}=5.55\times 10^{18}electron/sec[/tex]

Final answer:

The current density in the filament is 14.4 x 10^6 A/m^2. The electron current in the filament is 1.42 x 10^-19 C/s.

Explanation:

The current density in a filament can be calculated by dividing the current in the lightbulb by the cross-sectional area of the filament. To find the current density, we first need to find the cross-sectional area of the filament. The diameter of the filament is given as 0.280 mm, so the radius is 0.140 mm (0.140 mm = 0.280 mm / 2).

Using the formula for the area of a circle (A = πr^2), we can find the cross-sectional area of the filament:

A = π(0.140 mm)^2 = 0.0616 mm^2

Now, we can calculate the current density by dividing the current (0.890 A) by the cross-sectional area (0.0616 mm^2), and converting mm to m:

Current density = 0.890 A / (0.0616 mm^2) * (1 m / 1000 mm)^2 = 14.4 x 10^6 A/m^2

The electron current in the filament can be calculated using the formula:
Electron current = charge * number of charges passing per second

The charge of an electron is approximately 1.6 x 10^-19 C, and the current is given as 0.890 A, so:

Electron current = (1.6 x 10^-19 C) * (0.890 A) = 1.42 x 10^-19 C/s

Answer:

6

Explanation:

Number of lines emanate from + 5 micro coulomb is 15 .

They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.

the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.

So the lines terminating at - 3 micro coulomb

                                    = [tex]\frac{3}{5}\times 15 = 9[/tex]

So the lines terminating at - 2 micro coulomb

                                    = [tex]\frac{2}{5}\times 15 = 6[/tex]

So, the number of filed lines terminates at - 2 micro Coulomb are 6.

The number of lines emanating from the negative charged particles are 9.09 and 6.06 respectively.

In an electric field, the number of lines emanating from a charged particle is directly proportional to the magnitude of the charge. Thus, this is given by this mathematical expression:

[tex]q \;\alpha \;L\\\\q=kL\\\\5=k15\\\\k=\frac{5}{15}[/tex]

k = 0.33.

For the -3.0 μC particle:

[tex]L=\frac{q}{k} \\\\L=\frac{3.0}{0.33}[/tex]

L = 9.09.

For the -2.0 μC particle:

[tex]L=\frac{q}{k} \\\\L=\frac{2.0}{0.33}[/tex]

L = 6.06.

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Answer:

The mass of the water on earth is [tex]5.4537\times10^{20}\ kg[/tex]

Explanation:

Given that,

Average depth h= 0.95 mi

[tex]h=0.95\times1.609[/tex]

[tex]h =1.528\ km[/tex]

[tex]h=1.528\times10^{3}\ m[/tex]

Radius of earth [tex]r= 6.37\times10^{6}\ m[/tex]

Density = 1000 kg/m³

We need to calculate the area of surface

Using formula of area

[tex]A =4\pi r^2[/tex]

Put the value into the formula

[tex]A=4\pi\times(6.37\times10^{6})^2[/tex]

[tex]A=5.099\times10^{14}\ m^2[/tex]

We need to calculate the volume of earth

[tex]V = Area\times height[/tex]

Put the value into the formula

[tex]V=5.099\times10^{14}\times1.528\times10^{3}[/tex]

[tex]V=7.791\times10^{17}\ m^3[/tex]

Now, 70 % volume of the total volume

[tex]V= 7.791\times10^{17}\times\dfrac{70}{100}[/tex]

[tex]V=5.4537\times10^{17}\ m^3[/tex]

We need to calculate the mass of the water on earth

Using formula of density

[tex]\rho = \dfrac{m}{V}[/tex]

[tex]m = \rho\times V[/tex]

Put the value into the formula

[tex]m=1000\times5.4537\times10^{17}[/tex]

[tex]m =5.4537\times10^{20}\ kg[/tex]

Hence, The mass of the water on earth is [tex]5.4537\times10^{20}\ kg[/tex]

Answer:

v =4.36 m/s

Explanation:

given,

mass of object A = 18.5 Kg

initial velocity of object A = 8.15 m/s in east

mass of object B = 30.5 kg

initial velocity of object B = 5 m/s

[tex]P = P_A+P_B[/tex]

[tex]P = m_Av_A\widehat{i} + m_B v_B\widehat{j}[/tex]

[tex]P = 18.5\times 8.15 \widehat{i} + 30.5\times 5\widehat{j}[/tex]

[tex]P = 150.775 \widehat{i} + 152.5 \widehat{j}[/tex]

[tex]P = \sqrt{150.775^2+152.5^2}[/tex]

P = 214. 45 N s

velocity after collision is equal to

[tex]v =\dfrac{214.45}{18.5+30.5}[/tex]

v =4.36 m/s

hence, velocity after collision is equal to 4.36 m/s

Answer:

The magnitude of the final velocity of the two-object system is [tex]v=4.37\frac{m}{s}[/tex]

Explanation:

As the Momentum is conserved, we can compare the instant before the collision, and the instant after. Also, we have to take in account the two components of the problem (x-direction and y-direction).

To do that, we put our 0 of coordinates where the collision takes place.

So, for the initial momentum we have that

[tex]p_{ix}=m_{a}v_{0a}+0[/tex]

[tex]p_{iy}=0+m_{b}v_{0b}[/tex]

Now, this is equal to the final momentum (in each coordinate)

[tex]p_{fx}=(m_{a}+m_{b}) v_{fx}[/tex]

[tex]p_{fy}=(m_{a}+m_{b}) v_{fy}[/tex]

So, we equalize each coordinate and get each final velocity

[tex]m_{a}v_{0a}=(m_{a}+m_{b}) v_{fx} \Leftrightarrow v_{fx}=\frac{m_{a}v_{0a}}{(m_{a}+m_{b})}[/tex]

[tex]m_{b}v_{0b}=(m_{a}+m_{b}) v_{fy} \Leftrightarrow v_{fy}=\frac{m_{b}v_{0b}}{(m_{a}+m_{b})}[/tex]

Finally, to calculate the magnitude of the final velocity, we need to calculate

[tex]v_{f}=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}[/tex]

which, replacing with the previous results, is

[tex]v_{f}=\sqrt{(v_{fx})^{2}+(v_{fy})^{2}}=(\sqrt{(\frac{18.5*8.15}{49})^{2}+(\frac{30.5*5.00}{49})^{2}})\frac{m}{s}[/tex]

Therefore, the outcome is

[tex]v_{f}=4.37\frac{m}{s}[/tex]

Answer:

They're going to increase the total resistance as [tex]R_{T} = \sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}[/tex]

Explanation:

If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then

[tex]I = I_1 + I_2 + ... + I_N[/tex]

where

[tex]I_i = \frac{V_i}{R_i}[/tex]

but

[tex]V_i = V_j = V[/tex] for [tex]i,j= 1, 2,..., N[/tex]

so

[tex]I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}[/tex]

where

[tex]R_T = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)^{-1} =\sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1} [/tex]

Answer:

[tex]t=6.96s[/tex]

Explanation:

From this exercise, our knowable variables are hight and initial velocity

[tex]v_{oy}=96ft/s[/tex]

[tex]y_{o}=112ft[/tex]

To find how much time does the ball strike the ground, we need to know that the final position of the ball is y=0ft

[tex]y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}[/tex]

[tex]0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}[/tex]

Solving for t using quadratic formula

[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]

[tex]a=-\frac{1}{2} (32.2)\\b=96\\c=112[/tex]

[tex]t=-0.999s[/tex] or [tex]t=6.96s[/tex]

Since time can't be negative the answer is t=6.96s

Answer:

Explanation:

The digits which are reliable and first uncertain digit is called significant figure.

In any measurement, if the significant figure is more than the accuracy of the measurement is also more.

If there are some numbers in a calculations and they having different numbers of significant figures, then in the final result the number of significant figures is the least number of significant figures which are in the individual number.

When performing calculations with measurements that have different numbers of significant figures, the final answer will be rounded to the same number of significant figures as the measurement with the fewest significant figures. This applies to both addition and subtraction, as well as multiplication and division.

For addition and subtraction, the result should be rounded to the same decimal place as the least precise measured value. For example, if one measurement has two decimal places and another measurement has three decimal places, the final answer should be rounded to two decimal places.

For multiplication and division, the final answer should have the same number of significant figures as the measurement with the fewest significant figures. For example, if one measurement has three significant figures and another measurement has four significant figures, the final answer should have three significant figures.

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Answer:371.564 mi

Explanation:

Given

Airplane flies northwest for 250 mi and then travels west 150 mi

That is first it travels 250cos45 in - ve x direction and simultaneously 250sin45 in y direction

after that it travels 150 mi in -ve x direction

So its position vector is given by

[tex]r=-250cos45\hat{i}-150\hat{i}+250sin45\hat{j}[/tex]

[tex]r=-\left ( 250cos45+150\right )\hat{i}+250sin45\hat{j}[/tex]

so magnitude of displacement is

[tex]|r|=\sqrt{\left ( \frac{250}{\sqrt{2}}+150\right )^2+\left ( \frac{250}{\sqrt{2}}\right )^2}[/tex]

|r|=371.564 mi

Answer:

The wavelength required is 102.9 nm.

Explanation:

The energy levels for the hydrogen atom are

[tex]E(n) = \frac{-13.6 \ eV}{n^2}[/tex]

So, for a transition from the first level to the third level we got

[tex]\Delta E = E(3) - E (1)[/tex]

[tex]\Delta E = \frac{-13.6 \ eV}{ 3 ^2} - \frac{-13.6 \ eV}{1^2}[/tex]

[tex]\Delta E = \frac{-13.6 \ eV}{ 9} - \frac{-13.6 \ eV}{1}[/tex]

[tex]\Delta E = \frac{8}{9} 13.6 \ eV[/tex]

[tex]\Delta E = 12.09 \ eV[/tex]

[tex]\Delta E = 12.09 \ eV * \frac{1.6 \ 10^-19 \ Joules}{1 \ eV}[/tex]

[tex]\Delta E = 1.93 \ 10^-18 \ Joules[/tex]

So we need a photon with this energy.

The energy of a photon its given by

[tex]E = h \nu = h \frac{c}{\lambda}[/tex]

So, the wavelength will be

[tex]\lambda = \frac{h c}{E}[/tex]

[tex]\lambda = \frac{6.62 \ 10^{-34} \ \frac{m^2 kg}{s} \ *  3.00 \ 10^8 \ \frac{m}{s}}{1.93 \ 10^-18 \ Joules}[/tex]

[tex]\lambda = 10.29 \ 10^{-8} m[/tex]

[tex]\lambda = 1.029 \ 10^{-7} m[/tex]

[tex]\lambda = 102.9 \ nm[/tex]

Answer:

Approximate linear dimension is 2 light years.

Explanation:

Radius of the spiral galaxy r = 62000 LY

Thickness of the galaxy h = 700 LY

Volume of the galaxy = πr²h

                                   = (3.14)(62000)²(700)

                                   = (3.14)(62)²(7)(10)⁸

                                   = 84568×10⁸

                                   = [tex]8.45\times 10^{12}[/tex] (LY)³

Since galaxy contains number of stars = 1078 billion stars ≈ [tex]1.078\times 10^{12}[/tex]

Now volume covered by each star of the galaxy = [tex]\frac{\text{Total volume of the galaxy}}{\text{Number of stars}}[/tex]

= [tex]\frac{8.45\times 10^{12} }{1.078\times 10^{12}}[/tex]

= 7.839 Light Years

Now the linear dimension across the volume

= [tex](\text{Average volume per star})^{\frac{1}{3}}[/tex]

= [tex](7.839)^{\frac{1}{3}}[/tex]

= 1.99 LY

≈ 2 Light Years

Therefore, approximate linear dimension is 2 light years.

Answer:

F=94.32*10⁻⁹N  , The force F is repusilve because both charges have the same sign (+)

Explanation:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F=K*q₁*q₂/d² Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)  

d: distance between the charges in meters(m)

Equivalence  

1nC= 10⁻⁹C

Data

K=8.99x10⁹N*m²/C²

q₁ = 7.94-nC= 7.94*10⁻⁹C

q₂= 4.14-nC=  4.14 *10⁻⁹C

d= 1.77 m

Magnitude of the electrostatic force that one charge exerts on the other

We apply formula (1):

[tex]F=8.99x10^{9} *\frac{7.94*10^{-9} *4.14 *10^{-9} }{1.77^{2} }[/tex]

F=94.32*10⁻⁹N  , The force F is repusilve because both charges have the same sign (+)

In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:

a) [tex]h*(1 - 1/2 g * h/v_0^2)[/tex]

b)[tex]h = v_0^2/ g[/tex]

c)[tex]h = v_0^2/ g[/tex]

So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:

[tex]y = y_0 + v_0*t + 1/2 * a * t^2\\v = v_0 + a * t[/tex]

where:

a) When the balls collide, h1 = h2. Then,

[tex]h_1 = h_2\\v_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\\v_0 * t = h\\t = h / v_0[/tex]

Replacing in the equation of the height of the first ball:

[tex]h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\\h_1 = h - 1/2 g * h^2/ v_0^2\\h_1 = h*(1 - 1/2 g * h/v_0^2)[/tex]

b)  that the balls collide at t = h/v0. Then:

[tex]h/ v_0 = -v_0/-g\\h = v_0^2/ g[/tex]

c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:

[tex]h = v_0^2/ g[/tex]

See more about velocity at brainly.com/question/862972

Answer:

(a). The average speed is 97.23 km/hr.

(b). The average velocity is 81.70 km/min.

Explanation:

Given that,

Distance in east = 14.5 km

Distance in south = 66.5 km

Time = 50 min = 0.833 hr

(a). We need to calculate the average speed

Using formula of average speed

[tex]v_{avg}=\dfrac{D}{T}[/tex]

Where, D = total distance

T = total time

Put the value into the formula

[tex]v_{avg}=\dfrac{14.5+66.5}{0.833}[/tex]

[tex]v_{avg}=97.23\ Km/hr[/tex]

(b). We need to calculate the displacement

Using Pythagorean theorem

[tex]d=\sqrt{(d_{e})^2+(d_{s})^2}[/tex]

Put the value into the formula

[tex]d=\sqrt{(14.5)^2+(66.5)^2}[/tex]

[tex]d=68.06\ km[/tex]

(b). We need to calculate the average velocity

Using formula of average velocity

[tex]v_{avg}=\dfrac{d}{t}[/tex]

Where, d = displacement

t = time

Put the value into the formula

[tex]v_{avg}=\dfrac{68.06}{0.833}[/tex]

[tex]v_{avg}=81.70\ km/min[/tex]

Here, We can not necessary to find the unit vectors for displacement because we need to the displacement for find the average velocity.

Hence, (a). The average speed is 97.23 km/hr.

(b). The average velocity is 81.70 km/hr.

Answer:

a) 0.76 m

b) 1.53 s

c) 0.39 s

d) 3.8 m/s

Explanation:

This is a problem in which we need to use the equations pertaining Uniformly Accelerated Motion, as the acceleration during all this process is constant: The gravitational pull on the ball, 9.8 m/s².

To make things easier, we can divide this process in two parts: The first one (A) is from the moment the ball is thrown, until the moment it reaches it highest point and momentarily stops. The second one (B) from the moment it starts descending until it hits the ground.

a) During part A, we use the formula Vf²=Vi² +2*a*x . Where Vf is the final velocity (0 m/s, as the ball stopped in midair), Vi is the initial velocity (15 m/s), a is the acceleration (-9.8 m/s², it has a minus sign, as it goes against the direction of the movement) and x is the distance; thus, we're left with:

0 m/s=(15.0 m/s)²+2*(-9.8 m/s²)*x

We solve for x

x = 0.76 m

b) The formula is Vf=Vi +a*t, where t is the time. We're left with:

0 m/s=15.0 m/s + (-9.8 m/s²)*t

We solve for t

t= 1.53 s

c) Now we focus on part B, and use the formula x=Vi * t + [tex]\frac{at^{2}}{2}[/tex] , with the difference that the Vi is 0 m/s. We already know the value of x in exercise a). Note that a does not have a negative sign, as the direction of movement is opposite to the direction of part A

0.76 m=0 m/s * t +[tex]\frac{9.8\frac{m}{s^{2} } *t^{2} }{2}[/tex]

Solve for t

0.76=4.9t²

t=0.39 s

d) Once again we use the formula Vf=Vi +a*t, using the value of t previously calculated in exercise c).

Vf=0 m/s + 9.8 m/s² * 0,39 s

Vf=3.8 m/s

The ball rises to a height of 11.48 m, takes 1.53 s to reach its highest point, the same amount of time to fall back down, and has a velocity of -15.0 m/s when it returns to the starting level.

The question involves concepts from physics, specifically the kinematics of one-dimensional motion under constant acceleration due to gravity. Here is how we can solve each part of the given problem:

(a) How high does it rise?

To find the maximum height reached by the ball, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:

v² = u² + 2as

At the highest point, the final velocity (v) is 0 m/s, the initial velocity (u) is 15.0 m/s (upward), the acceleration due to gravity (a) is -9.8 m/s² (downward), and s represents the height. Solving for s:

0 = (15.0)² + 2(-9.8)s

s = (15.0)² / (2 * 9.8)

s = 11.48 m

(b) How long does it take to reach its highest point?

To find the time (t) it takes for the ball to reach its highest point, we can use the equation:

v = u + at

Since the final velocity at the highest point is 0 m/s:

0 = 15.0 + (-9.8)t

t = 15.0 / 9.8

t = 1.53 s

(c) How long does the ball take to hit the ground after it reaches its highest point?

The time for the ball to fall back down is the same as the time taken to reach the highest point, so this is also 1.53 s.

(d) What is its velocity when it returns to the level from which it started?

The velocity on returning to the starting level will be the same magnitude as the initial velocity but in the opposite direction, so it will be -15.0 m/s, with the negative sign indicating the downward direction.

Answer:

56.7°

Explanation:

Imagine a rectangle triangle.

The triangle has 3 sides.

One side is the height of the tower, let's name it A.

Another side is the distance from the base of the tower to the point where the waire touches the ground. Let's name that B.

Sides A and B are perpendicular.

The other side is the length of the wire. Let's name it C.

From trigonometry we know that:

cos(a) = B / C

Where a is the angle between B anc C, between the wire and the ground.

Therefore

a = arccos(B/C)

a = arccos(552/1005) = 56.7°

[tex]56.7^\circ[/tex] angle the wire makes with the ground.

Given :

A guy wire 1005 feet long is attached to the top of a tower.

When pulled taut, it touches level ground 552 feet from the base of the tower.

Solution :

We know that,

[tex]\rm cos \theta = \dfrac{Base}{Hypotanuse}[/tex]

Given that,

base = 552 ft

hypotanuse = 1005 ft

Therefore,

[tex]\rm cos\theta = \dfrac{552}{1005}[/tex]

[tex]\rm \theta = cos^-^1 \dfrac{552}{1005}[/tex]

[tex]\theta = 56.7^\circ[/tex]

[tex]56.7^\circ[/tex] angle the wire makes with the ground.

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Answer:

a)[tex] C_p=35.42\ \rm J/mol.K[/tex]

b)[tex] C_v=27.1\ \rm J/mol.K[/tex]

Explanation:

Given:

Let R be the gas constant which has value [tex]R=8.31432\times10^3\ \rm N\ m\ kmol^{-1}K^{-1}[/tex]

Work done in the process

[tex]W=PdV\\W=0.947\times(104-56.7)\times10^{-3}\ \rm L\ atm\\W=44.8\times10^{-3}\times101.33\ \rm J\\W=4.43\ \rm J[/tex]

Now Using First Law of thermodynamics

[tex]Q=W+\Delta U\\19.3=4.43+\Delta U\\\Delta U=14.77\ \rm J[/tex]

Let[tex]C_v[/tex] be the molar specific heat of the gas at constant volume given by

[tex]\Delta U=nC_v\Delta T\\14.77=\dfrac{C_v}{R}{PdV}\\14.77=\dfrac{C_v}{R}(0.987\times10^5(104-56.7)\times10^-6})\\C_v=3.26R\\C_v=27.1\ \rm J/mol.K[/tex]

Also we know that

Let[tex]C_p[/tex] be the molar specific heat of the gas at constant pressure given by

[tex]C_p-C_v=R\\C_p=R+C_v\\C_p=4.26R\\C_p=35.42\ \rm J/mol.K[/tex]

Answer:

accelerate in the direction in which the electric field is pointing.

Explanation:

The positive charge feels a force in the same direction as the electric field

F=Eq  

F and E are vectors, q is a scalar

(if it were a negative charge the force would be in the opposite direction)

that force will produce an acceleration in the same direction, that acceleration will cause the particle to move in the same direction, ie the direction of the electric field.

A free positive charge in an electric field will accelerate in the direction in which the electric field is pointing due to electrostatic forces.

In the context of physics and electric fields, a free positive charge released in an electric field will accelerate in the direction in which the electric field is pointing. This is because electric field lines are drawn from positive to negative, indicating the direction that a positive test charge would move. Therefore, when a positive charge is placed in an electric field, it is repelled from the positive source charge and attracted to the negative source charge, causing it to accelerate along the electric field lines in the direction they are pointing.

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Answer:

-65 degC

Explanation:

ΔD=[tex]D_{0}[/tex]αΔT

1.869-1.871=1.871(1.2E-5)ΔT

ΔT≈-89 degC

24-89=-65 degC

Explanation:

The position of a particle along x - axis is given by :

[tex]x=2.08+3.06t-1t^2[/tex]

(a) Position at t = 3.3 s

[tex]x=2.08+3.06(3.3)-1(3.3)^2[/tex]

x = 1.288 m

(b) Velocity at t = 3.3 s

[tex]v=\dfrac{dx}{dt}[/tex]

[tex]v=\dfrac{d(2.08+3.06t-1t^2)}{dt}[/tex]  

[tex]v=3.06-2t[/tex]

at t = 3.3 s

[tex]v=3.06-2(3.3)[/tex]

v = -3.54 m/s

(c) Acceleration,

[tex]a=\dfrac{dv}{dt}[/tex]

[tex]a=\dfrac{d(3.06-2t)}{dt}[/tex]  

[tex]a=-2\ m/s^2[/tex]

Hence, this is the required solution.

Final answer:

At t = 3.30 s, the particle is at the position of 1.288 meters, moves with a velocity of -3.54 m/s, and is subjected to a constant acceleration of -2 m/s^2.

Explanation:

The student's question about a particle's motion along the x-axis can be answered using kinematic equations from physics, which relate to the position, velocity, and acceleration of the particle.

Position at t = 3.30 s

To find the position of the particle at t = 3.30 s, we substitute the time into the position function x = 2.08 + 3.06t - 1.00t2.

2)>

Calculating this gives x(3.30 s) = 2.08 + 10.098 - 10.89 = 1.288 meters. The particle is at 1.288 meters from the origin at 3.30 seconds.

Velocity at t = 3.30 s

To find the velocity, we differentiate the position function with respect to time: v(t) = 3.06 - 2.00t.

The particle's velocity at 3.30 seconds is -3.54 meters per second.

Acceleration at t = 3.30 s

The acceleration is the second derivative of the position, which in this case is a constant because the t2 term has a constant coefficient: a(t) = -2.00 m/s2.

Therefore, the acceleration of the particle at any time, including at t = 3.30 s, is -2 meters per second squared.