a vector is 253m long and points in a 55.8degree direction. Find the y-component of the vector.

Answer:

The answer to your question is:

a) t = 3.64 s

b) vox = 31.59 m/s

c) vy = 35.71 m/s

d) v = 47.67 m/s

Explanation:

a) To calculate the time, we know that voy = 0 m/s so we this this formula

   h = voy + 1/2(gt²)

   65  = 0 + 1/2(9.81)(t²)

   65 = 4.905t²

   t² = 65/4.905 = 13.25

   t = 3.64 s

b) To calculate vox  we use this formula vox = d/t ; vox is constant

   vox = 115/3.64 = 31.59 m/s

c) To calculate voy we use the formula    vy = voy + gt

but voy = 0

               vy = gt = 9.81 x 3.64 = 35.71 m/s

d) To calculate v  we use the pythagorean theorem

            c2 = a2 + b2

            c2 = 31.59² + 35.71² = 997.92 + 1275.20

            c2 = 2273.12

              c = 47.67 m/s

Answer:

You're supposed to measure the distance from X to the end of vector 5 using the appropriate scale, and measure the angle (counterclockwise from X) using a protractor.  

Mathematically:  

x = [83cos90+55cos141+69cos229+41cos281+61co... cm = -27 cm  

y = [83sin90+55sin141+69sin229+41sin281+61si... cm = 55 cm  

so d = √(x² + y²) = 61 cm  

and Θ = arctan(55/-27) = -64º +180º (to get into QII) = 116º

Explanation:

To calculate the pinball's total displacement, you sum up the x and y components of each vector separately and find the resultant vector's magnitude and direction, which is approximately 96.60 cm at an angle of 88.7 degrees from the horizontal.

To find the total displacement of the pinball, we need to add the vectors given by their magnitude and direction. We first break down each vector into its horizontal (x) and vertical (y) components using trigonometric functions:

To find the resultant vector, sum the x and y components separately:

The magnitude of the resultant displacement vector (S) is calculated using the Pythagorean theorem:

S = √(x² + y²) = √(1.97² + 96.13²) = √(9330.56) = 96.60 cm approximately

The angle made with the horizontal (θ) is found using the tangent function:

θ = arctan(y/x) = arctan(96.13/1.97) ≈ 88.7 degrees

The work function of the metal is approximately 3.67 eV.

The work function, or binding energy, of a metal is the minimum energy required to eject an electron from the metal surface. In the photoelectric effect, the maximum kinetic energy of ejected electrons (photoelectrons) is given by the equation KE = hf - BE, where hf is the photon energy and BE is the work function. To find the work function W0 of the metal, we can use Equation 6.16, which states that the threshold wavelength for observing the photoelectric effect is given by λ = (hc)/(W0), where h is Planck's constant, c is the speed of light, and W0 is the work function.

Given that the maximum wavelength that can still eject electrons is 542 nm, we can rearrange the equation to solve for W0:

W0 = (hc)/λ

Plugging in the values, we have:

W0 = (6.626 × 10^-34 J·s)(3 × 10^8 m/s)/(542 × 10^-9 m)

Simplifying the calculation, we find that the work function W0 of the metal is approximately 3.67 eV.

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Final answer:

The maximum wavelength for ejecting electrons from the metal surface is 542 nm. Using the energy equation for photons, we calculate the work function (W0) to be approximately 2.29 electron volts.

Explanation:

The question relates to the photoelectric effect and the calculation of the work function of a metal when electromagnetic radiation of a certain wavelength is incident upon it. The work function (W0) is the minimum energy needed to eject an electron from the metal surface. According to the question, the maximum wavelength that can eject an electron is 542 nm.

To find the work function in electron volts, we can use the equation:

E(photon) = hc / λ

Where:

E(photon) is the energy of the photon

h is Planck's constant (4.135667696 × [tex]10^-^1^2[/tex] eV·s)

c is the speed of light (3 × 108 m/s)

λ is the wavelength of the photon in meters (542 nm = 542 × [tex]10^-^9[/tex] m)

First, convert the energy into electron volts:

E(photon) = (4.135667696 × [tex]10^-^1^2[/tex] eV·s × 3 × 108 m/s) / (542 × [tex]10^-^9[/tex] m)

After calculation, we find that E(photon) is approximately 2.29 eV. For a photon to eject an electron, its energy must be equal to or greater than the work function of the metal. Therefore, the work function of the metal is also 2.29 eV.

Answer:

Explanation:

[tex]\overrightarrow{A} = 3\widehat{i}+3\widehat{j}[/tex]

[tex]\overrightarrow{B} = \widehat{i}-4\widehat{j}[/tex]

[tex]\overrightarrow{C} = -2\widehat{i}+5\widehat{j}[/tex]

(a)

[tex]\overrightarrow{D} =\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}[/tex]

[tex]\overrightarrow{D} =\left ( 3+1-2 \right )\widehat{i} +\left ( 3-4+5 \right )\widehat{j}[/tex]

[tex]\overrightarrow{D} =\left 2\widehat{i} +4\widehat{j}[/tex]

Magnitude of [tex]\overrightarrow{D}[/tex] = [tex]\sqrt{2^{2}+4^{2}}[/tex]

                                                                     = 4.47 m

Let θ be the direction of vector D

[tex]tan\theta =\frac{4}{2}[/tex]

θ = 63.44°

(b)

[tex]\overrightarrow{E} =

- \overrightarrow{A}-\overrightarrow{B}+\overrightarrow{C}[/tex]

[tex]\overrightarrow{E} =\left ( - 3- 1 -2 \right )\widehat{i} +\left ( - 3 + 4+5 \right )\widehat{j}[/tex]

[tex]\overrightarrow{E} =- \left 6\widehat{i} +6\widehat{j}[/tex]

Magnitude of [tex]\overrightarrow{E}[/tex] = [tex]\sqrt{6^{2}+6^{2}}[/tex]

                                                                     = 8.485 m

Let θ be the direction of vector D

[tex]tan\theta =\frac{6}{-6}[/tex]

θ = 135°

a) For vector [tex]\( \mathbf{D} \)[/tex]: Magnitude: [tex]\( 4 \, \text{meters} \)[/tex] ,Direction: [tex]\( 90^\circ \)[/tex]

b) For vector [tex]\( \mathbf{E} \)[/tex]: Magnitude: [tex]\( 7.21 \, \text{meters} \)[/tex] ,Direction: [tex]\( 123.69^\circ \)[/tex]

To find the resultant vectors [tex]\( \mathbf{D} = \mathbf{A} + \mathbf{B} + \mathbf{C} \) and \( \mathbf{E} = -\mathbf{A} - \mathbf{B} + \mathbf{C} \)[/tex] using the component method, we need to break each vector into its [tex]\( i \)[/tex] and [tex]\( j \)[/tex] components, sum these components, and then find the magnitude and direction of the resulting vectors.

Given vectors:

[tex]\[ \mathbf{A} = 3\mathbf{i} + 3\mathbf{j} \][/tex]

[tex]\[ \mathbf{B} = -\mathbf{i} - 4\mathbf{j} \][/tex]

[tex]\[ \mathbf{C} = -2\mathbf{i} + 5\mathbf{j} \][/tex]

a) Vector [tex]\(\mathbf{D} = \mathbf{A} + \mathbf{B} + \mathbf{C}\)[/tex]

1. Sum the components:

[tex]\[ \mathbf{D} = (3\mathbf{i} + 3\mathbf{j}) + (-\mathbf{i} - 4\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) \][/tex]

Combine like terms:

 [tex]\[ D_i = 3 - 1 - 2 = 0 \][/tex]

  [tex]\[ D_j = 3 - 4 + 5 = 4 \][/tex]

  So, the components of [tex]\( \mathbf{D} \)[/tex] are:

  [tex]\[ \mathbf{D} = 0\mathbf{i} + 4\mathbf{j} \][/tex]

2. Magnitude of [tex]\( \mathbf{D} \)[/tex]:

 [tex]\[ |\mathbf{D}| = \sqrt{D_i^2 + D_j^2} = \sqrt{0^2 + 4^2} = 4 \, \text{meters} \][/tex]

3. Direction of [tex]\( \mathbf{D} \)[/tex]:

  The angle [tex]\( \theta_D \)[/tex] from the positive x-axis is:

 [tex]\[ \theta_D = \tan^{-1}\left(\frac{D_j}{D_i}\right) = \tan^{-1}\left(\frac{4}{0}\right) = 90^\circ \][/tex]

b) Vector [tex]\(\mathbf{E} = -\mathbf{A} - \mathbf{B} + \mathbf{C}\)[/tex]

1. Sum the components:

 [tex]\[ \mathbf{E} = - (3\mathbf{i} + 3\mathbf{j}) - (-\mathbf{i} - 4\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) \][/tex]

  Simplify the negative signs:

  [tex]\[ \mathbf{E} = (-3\mathbf{i} - 3\mathbf{j}) + (\mathbf{i} + 4\mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) \][/tex]

  Combine like terms:

 [tex]\[ E_i = -3 + 1 - 2 = -4 \][/tex]

[tex]\[ E_j = -3 + 4 + 5 = 6 \][/tex]

So, the components of [tex]\( \mathbf{E} \)[/tex] are:

  [tex]\[ \mathbf{E} = -4\mathbf{i} + 6\mathbf{j} \][/tex]

2. Magnitude of [tex]\( \mathbf{E} \)[/tex]:

 [tex]\[ |\mathbf{E}| = \sqrt{E_i^2 + E_j^2} = \sqrt{(-4)^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.21 \, \text{meters} \][/tex]

3. Direction of [tex]\( \mathbf{E} \)[/tex]:

  The angle [tex]\( \theta_E \)[/tex] from the positive x-axis is:

  [tex]\[ \theta_E = \tan^{-1}\left(\frac{E_j}{E_i}\right) = \tan^{-1}\left(\frac{6}{-4}\right) = \tan^{-1}\left(-1.5\right) \][/tex]

Since [tex]\( E_i \)[/tex] is negative and [tex]\( E_j \)[/tex] is positive, [tex]\( \theta_E \)[/tex] is in the second quadrant:

 [tex]\[ \theta_E = 180^\circ - \tan^{-1}(1.5) \approx 180^\circ - 56.31^\circ = 123.69^\circ \][/tex]

Answer:

0.006 N

Explanation:

First find the mass of the horse which is the only unknown variable.

Then use it with the new data. Working out is in the attachment.

Answer : The cell emf for this cell is 0.077 V

Solution :

The balanced cell reaction will be,  

Oxidation half reaction (anode):  [tex]Zn(s)\rightarrow Zn^{2+}+2e^-[/tex]

Reduction half reaction (cathode):  [tex]Zn^{2+}+2e^-\rightarrow Zn(s)[/tex]

In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}{diluted}}{[Zn^{2+}{concentrated}]}[/tex]

where,

n = number of electrons in oxidation-reduction reaction = 2

[tex]E_{cell}[/tex] = ?

[tex][Zn^{2+}{diluted}][/tex] = 0.0111 M

[tex][Zn^{2+}{concentrated}][/tex] = 4.50 M

Now put all the given values in the above equation, we get:

[tex]E_{cell}=0-\frac{0.0592}{2}\log \frac{0.0111M}{4.50M}[/tex]

[tex]E_{cell}=0.077V[/tex]

Therefore, the cell emf for this cell is 0.077 V

The cell emf, or electromotive force, for a voltaic cell, like the Zn2+-Zn kind described, can be calculated with the Nernst equation. The emf is determined by the standard electrode potential as well as the concentrations of the redox species in each half-cell.

A voltaic cell runs on a spontaneous redox reaction occurring indirectly, with the oxidant and reductant redox couples contained in separate half-cells. In your example, both half-cells are Zn2+-Zn electrodes where the reaction is Zn2+ + 2e− → Zn (s) with a standard electrode potential, E°, of -0.763 V. The cell emf, or electromotive force, is calculated through the Nernst Equation:

Ecell = E° - (RT/nF) * ln(Q)

where Q is the reaction quotient, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the redox reaction, and F is Faraday's constant. Since we're dealing with a concentration cell, Q is given by the ratio of the concentrations of the reduced and oxidized species. Depending on the temperature and assuming that Zn2+ is reduced at the cathode and oxidized at the anode, the cell emf can be calculated accordingly.

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Answer:

314.92 m

Explanation:

Acceleration of the helicopter = 5.4 m/s² = a

Time taken by the helicopter to reach maximum height = 10.8 s = t

Initial velocity = 0 = u

Final velocity = v

Displacement = s

Equation of motion

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times 10.8+\frac{1}{2}5.4\times 10.8^2\\\Rightarrow s=314.92\ m[/tex]

Maximum height above ground reached by the helicopter is 314.92 m

Answer:

The distance traveled by the swan is 60.35 meters.

Explanation:

Given that,

A swan accelerate from rest (u = 0) to 6.5 m/s to take off.

Acceleration of the swan, [tex]a=0.35\ m/s^2[/tex]

We need to find the distance Δx it travel before becoming airborne. From the third equation of motion as :

[tex]\Delta x=\dfrac{v^2-u^2}{2a}[/tex]

[tex]\Delta x=\dfrac{(6.5)^2}{2\times 0.35}[/tex]

[tex]\Delta x=60.35\ m[/tex]

So, the distance traveled by the swan is 60.35 meters. Hence, this is the required solution.

Answer:

89.16pounds

Explanation:

The equation that defines this problem is as follows

W=k/X^2

where

W=Weight

K= proportionality constant

X=distance from the center of Earth

first we must find the constant of proportionality, with the first part of the problem

k=WX^2=131x3960^2=2054289600pounds x miles^2

then we use the equation to calculate the woman's weight with the new distance

W=2054289600/(4800)^2=89.16pounds

Answer:

F=<0,-29.4,0>N

D = <0,-10,0>m

Work=F·D=294 Joules

Explanation:

Force:

The gravity has a negative direction in the axis Y in our coordinate system:

F=<0,-mg,0>=<0,-3*9.8,0>=<0,-29.4,0>N

Displacement:

The initial position A is <0, 33, 0>m

The final position B is <0, 23,0>m

The displacement vector is D=B-A = <0,-10,0>m

Gravitational Work:

The work is the scalar product between the force and the displacement

Work=F·D=(-29.4)*(-10)=294 Joules

Answer:

airplane speed 135mph windspeed 45 mph

Explanation:

This information helps us to write down a system of linear equations

When going head wind, the speed of the wind is substracted from that of the airplane and on the return trip it is added, then:

A:=Airlplane speed

W:= Wind speed

(A+W)*1h=180mi (1)

(A-W)*2h=180mi (2)

then from (1) A=180-W (3), replacing this in (2) we get (180-W-W)*2h =180mi, then

360-4W=180, or 180=4W, then W=45 mph. Replacing this in (3) we have that A=180-45=135 mph.

The problem was solved by setting up equations based on the given distances and times. By solving these equations, it was determined that the plane's speed in still air is 105 miles per hour, and the wind's speed is 15 miles per hour.

To solve this problem, we can set up two equations using the relationship between distance, speed, and time. We will denote the plane's speed in still air as P and the wind's speed as W.

When flying against the wind, the plane's effective speed is P - W, and the time taken to cover the 180 miles is 2 hours. We can represent this with the equation:
180 = 2(P - W) ... (1)

On the return trip, with a tailwind, the plane's effective speed is P + W, and the time taken is 1.5 hours. This gives us a

second equation:
180 = 1.5(P + W) ... (2)

By solving these two equations simultaneously, we can find the values of P and W. Multiplying equation (2) by 2 to eliminate the fractions, we get:
360 = 3(P + W) ... (2')

Now, we subtract equation (1) from equation (2'):
360 - 180 = 3(P + W) - 2(P - W)
180 = 3P + 3W - 2P + 2W
180 = P + 5W

Using equation (1), we express P in terms of W:
180 = 2P - 2W
P = 90 + W ... (3)

Substituting (3) into the equation we got after subtracting:
180 = (90 + W) + 5W
180 = 90 + 6W
90 = 6W
W = 15 miles per hour

Now, we substitute the value of W back into equation (3) to find P:
P = 90 + 15
P = 105 miles per hour

The plane's speed in still air is 105 miles per hour and the wind's speed is 15 miles per hour.

Answer:

The lowest frequency is 95.6 Hz

Explanation:

The standing waves that can be formed in this system must meet some conditions, such as until this is fixed at the bottom here there must be a node (point without oscillation) and being free at its top at this point there should be maximum elongation (antinode)

For the lowest frequency we have a node at the bottom point and a maximum at the top point, this corresponds to ¼ of the wavelength, so the full wave has

      λ = 4L

As the speed any wave is equal to the product of its frequency by the wavelength

      v = f λ

      f = v / λ    

      f = v / 4L

      f = 2730 / (4 7.14)

      f=  95.6 1 / s = 95.6 Hz

The lowest frequency of the standing wave that can be formed on the flagpole is 95.59 Hz.

The velocity, frequency and wavelength of a wave are related according to the following equation:

Velocity (v) = wavelength (λ) × frequency (f)

v = λf

With the above formula, we can obtain the frequency as follow:

v = λf

2730 = 28.56 × f

Divide both side by 28..56

f = 2730 / 28.56

f = 95.59 Hz

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Answer:

20.7 s

Explanation:

The equation to calculate the velocity for a uniform acceleration a, time t and initial velocity v₀:

v = a*t + v₀

Solve for t:

t = (v - v₀)/a

Answer:

a) x component  = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr

A velocity wrt to velocity of B

[tex]\vec{V_{AB}} =\vec{V_A} - \vec{V_B}[/tex]

[tex]\vec{V_A} = 22 km/hr[/tex]

[tex]\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}[/tex]

                 [tex]= 31.52\hat{i} + 24.62 \hat{j}[/tex]

putting respective value to get velocity of  A with respect to B

[tex]\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})[/tex]

[tex]\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}[/tex]

a) x component  = -31.25 km/hr

b) y component = 46.64 km/hr

Answer:

a=8.1832m/s^2

Explanation:

Vo=initial speed=0m/s

Vf=final speed=26.8m/s

t=time=3.275s

the vehicle moves with a constant acceleration therefore we can use the following equation

A=aceleration=(Vf-Vi)/t

A=(26.8m/s-0m/s)/3.275s=8.1832m/s^2

the magnitude of the car´s aceleration is 8.1832m/s^2

Answer:

magnitude=34.45 m

direction=[tex]55.52\°[/tex]

Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance [tex]d[/tex] between two points:

[tex]d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}}[/tex] (1)

[tex]d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}}[/tex] (2)

[tex]d=\sqrt{1186.81 m^{2}}[/tex] (3)

[tex]d=34.45 m[/tex] (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

[tex]tan \theta=\frac{Y2-Y1}{X2-X1}[/tex]  (5)

[tex]tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}[/tex]  (6)

[tex]tan \theta=\frac{24.8}{19.5}[/tex]  (7)

Finding [tex]\theta[/tex]:

[tex]\theta= tan^{-1}(\frac{24.8}{19.5})[/tex]  (8)

[tex]\theta= 55.52\°[/tex]  (9) This is the direction of the vector

Answer:

A)t=1.375s

B)t=11s

Explanation:

for this problem we will assume that the east is positive while the west is negative, what we must do is find the relative speed between the wave and the powerboat, and then with the distance find the time for each case

ecuations

V=Vw-Vp  (1)

V= relative speed

Vw= speed of wave

Vp=Speesd

t=X/V(2)

t=time

x=distance=44m

A) the powerboat moves to west

V=18-(-14)=32m/s

t=44/32=1.375s

B)the powerboat moves to east

V=18-14=4

t=44/4=11s

Final answer:

The frequency at which the boat encounters a wave crest is 0.09 Hz regardless of the direction the boat is traveling.

Explanation:

To find the frequency at which the boat encounters a wave crest, we need to determine the time it takes for one wave crest to pass by the boat. The speed of the boat relative to the ocean floor doesn't affect the frequency, only the speed of the wave. The formula to calculate the frequency of a wave is:

Frequency = Wave Speed / Wavelength

Wave speed = Speed of the wave + Speed of the boat (since the boat is traveling in the opposite direction)
Frequency = (Speed of the wave + Speed of the boat) / Wavelength
Frequency = (18 m/s + (-14 m/s)) / 44 m
Frequency = 4 m/s / 44 m = 0.09 Hz

Wave speed = Speed of the wave - Speed of the boat (since the boat is traveling in the same direction)
Frequency = (Speed of the wave - Speed of the boat) / Wavelength
Frequency = (18 m/s - 14 m/s) / 44 m
Frequency = 4 m/s / 44 m = 0.09 Hz

Answer: [tex]p_{SO_2}=0.017atm[/tex]

Explanation:

We are given:

[tex]K_c=1.7\times 10^8[/tex]

Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:

[tex]K_p=K_c(RT)^{\Delta ng}[/tex]

Where,

[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = ?

[tex]K_c[/tex] = equilibrium constant in terms of concentration

R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature = [tex]600K[/tex]

[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]

[tex]\Delta ng[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}=2-3=-1[/tex]

Putting values in above equation, we get:

[tex]K_p=1.7\times 10^8\times (0.0821\times 600)^{-1}\\\\K_p=3.4\times 10^6[/tex]

The chemical reaction follows the equation:

                 [tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]

The expression for [tex]K_p[/tex] for the given reaction follows:

[tex]K_p=\frac{(p_{SO_3})^2}{ p_{O_2}\times {(p_{SO_2})^2}}[/tex]

We are given:

[tex]p_{SO_3}=300atm[/tex]   [tex]p_{O_2}=100atm[/tex]

Putting values in above equation, we get:

[tex]3.4\times 10^6=\frac{(300)^2}{100\times {(p_{SO_2})^2}}[/tex]

[tex]p_{SO_2}=0.017atm[/tex]

Hence, the partial pressure of the [tex]SO_2[/tex] at equilibrium is 0.017 atm.

Answer:

[tex]\Delta S = 2.11 J/K[/tex]

Explanation:

As we know that entropy change for phase conversion is given as

[tex]\Delta S = \frac{\Delta Q}{T}[/tex]

Here we know that heat required to change the phase of the water is given as

[tex]\Delta Q = mL[/tex]

here we have

[tex]m = 0.349 g = 3.49\times 10^{-4} kg[/tex]

L = 2250000 J/kg

now we have

[tex]\Delta Q = (3.49 \times 10^{-4})\times 2250000[/tex]

[tex]\Delta Q = 788.9 J[/tex]

also we know that temperature is approximately same as boiling temperature

so we have

[tex]T = 373 k[/tex]

so here we have

[tex]\Delta S = \frac{788.9}{373}[/tex]

[tex]\Delta S = 2.11 J/K[/tex]

Answer:

The height of the ball after 5 seconds is 162 ft.

Explanation:

First, replace the variable t with how many seconds the ball has dropped, which in this case is 5.

[tex]h = 512 - 14 {t}^{2} \\ h = 512 - 14 \times {5}^{2} [/tex]

Solve.

[tex]h = 512 - 14 \times {5}^{2} \\ h = 512 - 14 \times 25 \\ h = 512 - 350 \\ h = 162[/tex]

For a duration of 5 seconds, the ball had managed to drop 350 feet, with 162 feet left to go to touch ground level.

Recrystallization is a temperature-induced process that can vary among materials. Metal parts may recrystallize at different temperatures, and recrystallization can happen unevenly within a single part.

Recrystallization is a process where a material undergoes structural changes due to increased temperature. Metallic glasses can have different recrystallization temperatures based on the composition of the metals. Recrystallization can occur non-uniformly in a part, with some regions undergoing the process before others due to variations in temperature or cooling rates.

Answer:

75 N

Explanation:

t = Time taken = 2 seconds

u = Initial velocity

v = Final velocity

s = Displacement = 30 m

a = Acceleration

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 30=0\times 2+\frac{1}{2}\times a\times 2^2\\\Rightarrow a=\frac{30\times 2}{2^2}\\\Rightarrow a=15\ m/s^2[/tex]

The acceleration due to gravity on the planet is 15 m/s²

Force

F = ma

[tex]F=5\times 15\\\Rightarrow F=75\ N[/tex]

The gravitational force exerted on the object near the planet’s surface is 75 N

We have that for the Question it can be said that he gravitational force exerted on the 5kg object near the planet’s surface

F=75N

From the question we are told

A 5kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant. The mass of the planet is unknown. After 2s, the object has fallen 30m.

Air resistance is considered to be negligible. What is the gravitational force exerted on the 5kg object near the planet’s surface?

Generally the equation for the Motion   is mathematically given as

[tex]s=ut+1/2at^2\\\\Therefore\\\\30=0+1/2(a)(2)^2\\\\a=15m/s^2[/tex]

Therefore

F=ma

F=5*15

F=75N

Hence, the gravitational force exerted on the 5kg object near the planet’s surface

F=75N

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Answer:

The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chromatography chamber.

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures.

A strip of moist filter paper is put into the chromatography chamber so that its bottom touches the solvent and the paper lies on the chamber wall and reaches almost to the top of the container.

The container is closed and left for a few minutes to let the solvent vapors ascend the moist filter paper and saturate the air in the chamber.

The moist filter paper is in charge of preventing evaporation and ensuring the proper saturation of the air of the chamber.

Answer:

[tex]v_{f} =4.9\frac{m}{s}[/tex]  :  velocity of the crate after the person has pushed the crate a distance of 6 meters

Explanation:

Crate kinetics

Crate moves with uniformly accelerated movement

v f²=v₀²+2a*d (formula 1)

d:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Known data

v₀=0 The speed of the crate is equal to zero because part of the rest.

a= 2m/s²

d= 6m

Distance calculating

We replace data in the Formula (1)

v f²=0+2*2*d

v f²=2*2*6

v f²=24

[tex]v_{f} =\sqrt{24}[/tex]

[tex]v_{f} =4.9\frac{m}{s}[/tex]

To find the final velocity of a 50-kg crate after being pushed for 6 meters with an acceleration of 2.0 m/s², the kinematic equation v² = u² + 2as is used, resulting in a final velocity of 4.9 m/s.

To calculate the velocity of a 50-kg crate after being pushed a distance of 6 meters with an acceleration of 2.0 m/s2 from rest, we use the kinematic equation v2 = u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance moved. Since the crate starts from rest, the initial velocity, u, is 0 m/s. Substituting the given values we get:

v2 = 0 m/s + 2(2.0 m/s2)(6 m),

v2 = 24 m2/s2,

v = √(24 m2/s2),

v = 4.9 m/s.

Therefore, the velocity of the crate after being pushed 6 meters will be 4.9 m/s.

Answer:

The planetary system is 9 billion years old

Explanation:

In 4.5 billion years, the meteorite was half uranium and half lead. For this amount of uranium to be reduced by half again (reaching 1/4 uranium and 3/4 lead), another 4.5 billion years must have passed. Then, the planetary system should be 9 (2 x 4.5) billion years old.

Answer:

x = +7, x = -7

Explanation:

The equation to solve is:

[tex]\left|\frac{x}{7}\right|=1[/tex]

Which means that the absolute value of the fraction [tex]\frac{x}{7}[/tex] must be equal to 1. Since we have an absolute value, we have basically two equations to solve:

1) [tex]\frac{x}{7}=+1[/tex]

Solving this one, we find

[tex]\frac{x}{7}\cdot 7 = 1\cdot 7 \rightarrow x = +7[/tex]

2) [tex]\frac{x}{7}=-1[/tex]

Solving this one, we find

[tex]\frac{x}{7}\cdot 7 = -1\cdot 7 \rightarrow x = -7[/tex]

So the two solutions are +7 and -7.

Answer:

option (a)

Explanation:

the angular velocity of the carousel is same througout the motion, so the angular velocity of all the horses is same, but the linear velocity is different for different horses.

As the angular displacement of all the horses are same in the same time so the angular velocity is same.

The relation between the linear velocity and the angular velocity is given by

v = r ω

where, v is linear velocity and r be the distance between the horse and axis of rotation and ω be the angular velocity.

So, the angular velocity of Alice horse is same as the angular velocity of Bob horse.

ωA = ωB

Thus, option (a) is true.

The correct option is Option A ( ω A = ω B). Both Alice and Bob's horses on the carousel have the same angular velocity, regardless of their distance from the axis. Hence, ω A == ω B.

Thus, the correct answer is:
a. ω A = ω B

Answer:

0.000000113 or 1.13*[tex]10^{-7}[/tex] meters

Explanation:

One nanometer is [tex]10^{-9}[/tex] meters. So 113 nanometers would be 113*[tex]10^{-9}[/tex], or 1.13*[tex]10^{-7}[/tex] meters. That's expessed on "cientific notation." On the "usual" notation, it will be 0.000000113 meters.

Final answer:

The statement regarding receiving credit for correctly positioned endpoints using an endpoint mover is typically true. Credit is awarded for each endpoint positioned correctly in a digital math exercise or tool, which may involve placing endpoints according to specific criteria.

Explanation:

The statement "When using the endpoint mover, you will receive credit for each endpoint that is positioned correctly" is generally True. In the context of mathematical tools or interactive digital platforms, an endpoint mover likely refers to a feature that allows users to manipulate the endpoints of a line segment or other geometric object. When students are tasked with positioning these endpoints correctly according to given criteria, they would typically receive credit or points for each endpoint that is placed accurately.

For example, in a digital math exercise, if you're asked to position the endpoints of a line segment so that one end is at the point (3,2) and the other end is at the point (-1,5), and you succeed in placing them correctly, you would receive credit for both endpoints. However, details may vary based on the specific platform or educational program, so it's essential to follow the instructions provided.

Answer:

For cardboard = 29.4 g/cm²

For aluminium = 113.4 g/cm²

For lead = 193.8 g/cm²

Explanation:

Given:  

Density of the cardboard, d₁ = 0.6 g/cm³

Density of the aluminium, d₂ = 2.7 g/cm³

Density of the lead, d₃ = 11.4 g/cm³

Length of the cardboard,  L₁ = 49 cm

Length of the aluminium, L₂ = 42 cm

Length of the lead, L₃ = 17 cm

Now,

The absorber thickness is calculated as:

= Density × Length

therefore,

For cardboard = d₁ × L₁ = 0.6 × 49 = 29.4 g/cm²

For aluminium = d₂ × L₂ = 2.7 × 42 = 113.4 g/cm²

For lead = d₃ × L₃ = 11.4 × 17 = 193.8 g/cm²