You are on your balcony and notice some bad squirrels digging in your garden directly below. You start tossing your pistachios at them to get them out of your yard. If the height of your guard rail is 43 inches above the deck, which is at a standard height of 10 feet above the ground, and you toss the pistachios straight down with a speed of 7.4 m/s, then how long does it take for the pistachio to reach the ground?

Final answer:

The diver takes approximately 2.18 seconds to reach the water after leaving the edge of the cliff. The diver travels approximately 3.21 meters horizontally from the cliff face before hitting the water.

Explanation:

To find the time it takes for the diver to reach the water after leaving the edge of the cliff, we can use the equation of motion:[tex]h = 1/2 * g * t^2,[/tex]where h is the height of the cliff and g is the acceleration due to gravity. Rearranging the equation to solve for t, we get t = sqrt(2h/g). Plugging in the values given, we have t = sqrt(2*30/9.8) ≈ 2.18 s.

To find the horizontal distance the diver travels, we can use the equation s = v * t, where s is the distance, v is the horizontal velocity, and t is the time. Rearranging the equation to solve for v, we get v = s/t. Plugging in the values given, we have v = 7/2.18 ≈ 3.21 m/s.

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is [tex]41.9^{\circ}[/tex]

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

[tex]s=u_y t + \frac{1}{2}at^2[/tex] (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

[tex]u_y = u sin \theta[/tex] is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

[tex]\theta=40.0^{\circ}[/tex] is the angle of projection

So

[tex]u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s[/tex]

[tex]a=g=-9.8 m/s^2[/tex] is the acceleration due to gravity (downward)

Substituting the numbers, we get

[tex]-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0[/tex]

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

[tex]d=u_x t[/tex]

where

[tex]u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s[/tex] is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

[tex]d=(9.19)(1.778)=16.34 m[/tex]

B)

In this second case,

[tex]\theta=42.5^{\circ}[/tex]

So the vertical velocity is

[tex]u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s[/tex]

So the equation for the vertical motion becomes

[tex]4.9t^2-8.1t-1.80=0[/tex]

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

[tex]u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s[/tex]

So, the range of the shot is

[tex]d=u_x t = (8.85)(1.851)=16.38 m[/tex]

C)

In this third case,

[tex]\theta=45^{\circ}[/tex]

So the vertical velocity is

[tex]u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s[/tex]

So the equation for the vertical motion becomes

[tex]4.9t^2-8.5t-1.80=0[/tex]

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

[tex]u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s[/tex]

So, the range of the shot is

[tex]d=u_x t = (8.49)(1.925)=16.34[/tex] m

D)

In this 4th case,

[tex]\theta=47.5^{\circ}[/tex]

So the vertical velocity is

[tex]u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s[/tex]

So the equation for the vertical motion becomes

[tex]4.9t^2-8.8t-1.80=0[/tex]

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

[tex]u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s[/tex]

So, the range of the shot is

[tex]d=u_x t = (8.11)(1.981)=16.07 m[/tex]

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is [tex]\theta=42.5^{\circ}[/tex].

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

[tex]s-u sin \theta t + \frac{1}{2}gt^2=0[/tex]

The solutions of this quadratic equation are

[tex]t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}[/tex]

This is the time of flight: so, the horizontal range is

[tex]d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta[/tex]

It can be found that the maximum of this function is obtained when the angle is

[tex]\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})[/tex]

Therefore in this problem, the angle which leads to the maximum range is

[tex]\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}[/tex]

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Answer:

V₂  =541.94 m L.

Explanation:

Given that

V₁ = 500 mL  

T₁ = 25°C  = 273 + 25 = 298 K

T₂ = 5°C = 273+50 =323  K

The final volume = V₂  

We know that ,the ideal gas equation  

If the pressure of the gas is constant ,then we can say that

[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]

[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}[/tex]

Now by putting the values in the above equation we get

[tex]V_2=500\times \dfrac{323}{298}\ mL\\V_2=541.94\ mL[/tex]

The final volume of the balloon will be 541.94 m L.

V₂  =541.94 m L.

The final volume of the balloon will be "541.94 mL".

Given:

Volume,

Temperature,

            [tex]= 273+25[/tex]

            [tex]= 298 \ K[/tex]

             [tex]=273+50[/tex]

             [tex]=323 \ K[/tex]

By using the Ideal gas equation, we get

→ [tex]\frac{V_2}{V_1} = \frac{T_2}{T_1}[/tex]

or,

→ [tex]V_2 = \frac{T_2\times V_1}{T_1}[/tex]

       [tex]= \frac{500\times 323}{298}[/tex]

       [tex]= 541.94 \ mL[/tex]

Thus the above approach is correct.  

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Answer:

(b) they are not detectable with current technology

Explanation:

The improvement in exoplanet detection methods in recent years, thanks to spaces telescopes such as the Kepler, has led to a revolution in the field of astronomy. The findings have gone from focusing on Jupiters hot to super-earth and, ultimately, to Earth-like planets.

Answer:

because of the raindrop velocity relative of the car has a vertical and horizontal component  

Explanation:

Answer:

Recall that the electric field outside  a uniformly charged solid sphere  is exactly the same as if the charge were all at a point in the centre of the  sphere:

[tex]E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}[/tex]

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

[tex]E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}[/tex]

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

[tex]V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r<R)=- \int\limits^r_\infty{E.dl\\\\= -\int\limits^R_\infty{\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } -\int\limits^r_R{\frac{1}{4\pi(e_{0)} }\frac{Q}{R^2}\frac{r}{R} dr\\[/tex]

=[tex]\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ][/tex]

∴NOTE: Graph is attached

The potential inside the uniformly charged solid sphere is given by a specific equation, while the potential outside the sphere is given by a different equation. The gradients of these potentials can also be calculated. A sketch of V(r) shows how the potential changes within and outside the sphere.

Inside the uniformly charged solid sphere:

The potential is given by the equation:

V(r) = k(q / R)((3R - r^2) / (2R^3))

The gradient of V inside the sphere is given by:

∇V(r) = -(kqr) / (R^3)

Outside the sphere:

The potential is given by the equation:

V(r) = (kq) / (r)

The gradient of V outside the sphere is given by:

∇V(r) = -(kq) / (r^2)

Sketch of V(r):

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A. To find the orientation angles of the dipole for which the torque is zero, we need to consider the torque equation for an electric dipole in an external electric field.

The torque ([tex]\(\tau\)[/tex]) acting on an electric dipole (p) in an electric field (E) is given by:

[tex]\[ \tau = p \cdot E \cdot \sin(\theta) \][/tex]

Where:

p is the magnitude of the electric dipole moment,

E is the magnitude of the electric field,

[tex]\(\theta\)[/tex] is the angle between the dipole moment (p) and the electric field (E).

For the torque to be zero, [tex]\(\sin(\theta)\)[/tex] must be zero. This happens when [tex]\(\theta = 0\)[/tex] or [tex]\(\theta = \pi\) (180 degrees)[/tex], as [tex]\(\sin(0) = 0\)[/tex] and [tex]\(\sin(\pi) = 0\)[/tex].

So, the two possible orientation angles for which the torque is zero are:

[tex]\(\theta = 0\)[/tex] degrees (aligned with the electric field)

[tex]\(\theta = 180\)[/tex] degrees (opposite to the electric field)

B. Now, let's analyze the stability of these orientations:

1. [tex]\(\theta = 0\)[/tex] degrees (aligned with the electric field):

In this orientation, the dipole moment is aligned with the electric field.

Any slight deviation from this orientation will result in a torque that tends to restore the dipole to its original position. Therefore, this orientation is stable.

2. [tex]\(\theta = 180\)[/tex] degrees (opposite to the electric field):

In this orientation, the dipole moment is opposite to the electric field. Similar to the previous case, any slight deviation from this orientation will result in a torque that tends to restore the dipole to its original position. Therefore, this orientation is also stable.

Both orientations are stable because they represent energy minima. Any deviation from these orientations results in a restoring torque that tries to bring the dipole back to its stable position.

Thus, both orientations ([tex]\(\theta = 0\) degrees and \(\theta = 180\)[/tex] degrees) are stable orientations for the electric dipole in a uniform electric field.

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Final answer:

The torque on an electric dipole in a uniform electric field is zero when the dipole is parallel to the field. This orientation is stable.

Explanation:

In general, the torque vector on an electric dipole, p, from an electric field, E, is given by the equation T = p x E, where the cross product represents the direction of the torque. To find all the orientation angles of the dipole for which the torque is zero, we set the cross product equal to zero:

p x E = 0

The torque is zero when the dipole and electric field vectors are parallel. Therefore, the dipole will experience a torque that tends to align it with the electric field vector. This means that the dipole is in a stable equilibrium when it is parallel to the electric field.

Therefore, the orientation angles that result in a zero torque are all angles that make the dipole parallel to the electric field.

The final velocity is [tex]v=35.75\ ft/s[/tex] and the final elevation is [tex]45ft[/tex].

The potential energy is the energy possessed by virtue of configuration and the kinetic energy is the energy possessed due to motion.

Given:

Initial velocity, [tex]u=40 ft/s[/tex]

Initial elevation, [tex]h_i=30 ft[/tex]

Weight of the object, [tex]w=100lbf[/tex]

Increase in potential energy, [tex]\Delta PE=1500 ft{-}lbf[/tex]

Decrease in kinetic energy, [tex]\Delta KE=-500 ft{-}lbf[/tex]

(a)

The final velocity of the object is computed as:

[tex]\frac{1}{2} \frac{w}{g}v^2-\frac{1}{2} \frac{w}{g}u^2= \Delta KE\\v^2=40^2 + 2 \times \frac{32.17}{100} \times (-500)\\v= \sqrt{1278.3}\\v=35.75 \ ft/s[/tex]

(b)

The final elevation of the object is computed as:

[tex]mgh_f-mgh_i= \Delta PE\\h_f= h_i+ \frac {\Delta PE}{mg}\\h_f= 30 + \frac{1500}{100}\\h_f=45 ft[/tex]

Therefore, the final velocity is [tex]v=35.75\ ft/s[/tex] and the final elevation is [tex]45ft[/tex].

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Complete question:

Electrons are ejected from a metallic surface with speeds ranging up to 4.60 x10⁵ m/s when light with a wavelength of 625nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?

Answer:

Part(a) The work function of the surface is 22.177 x 10⁻²⁰ J = 1.384 eV

Part(b) The cutoff frequency for this surface is 3.347 x 10¹⁴ Hz

Explanation:

The kinetic energy (KE) of the emitted photon:

KE = 0.5mv²

m is mass of electron = 9.1 X 10⁻³¹ kg

KE = 0.5 * 9.1 X 10⁻³¹  * (460000)² = 9.628 X 10⁻²⁰ J

in eV = 9.628 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 0.601 eV

The photon energy of the incoming radiation:

E = hf = hc/λ

c is speed of light (photon) = 3 x 10⁸

h is Planck's constant = 6.626 × 10⁻³⁴ J.s

E = (6.626 × 10⁻³⁴ *3 x 10⁸) /(625 X 10⁻⁹)

E = 31.805 X 10⁻²⁰ J

in eV = 31.805 X 10⁻²⁰ J x  6.242 X 10¹⁸ ev = 1.985 eV

Part (a) the work function of the surface

KE = hf - W

where;

W is work function

W = hf - KE

W =  31.805 X 10⁻²⁰ J - 9.628 X 10⁻²⁰ J = 22.177 x 10⁻²⁰ J

in eV = 22.177 x 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.384 eV

Part(b) the cutoff frequency for this surface

W =hf

f = W/h

f = (22.177 x 10⁻²⁰ J)/(6.626 × 10⁻³⁴ J.s)

f = 3.347 x 10¹⁴ Hz

The work function is the minimum energy required to eject an electron from a metallic surface, which can be calculated using the maximum kinetic energy of ejected electrons and the energy of the incident light. The cutoff frequency is the minimum frequency of light required to eject an electron.

The question is asking to calculate the work function of the surface and the cutoff frequency in a context related to the photoelectric effect. The photoelectric effect is a phenomenon in which electrons are ejected from a metallic surface when it is exposed to light of a certain frequency, in this case, light with a wavelength of 625 nm.

The energy of the incident light is used to expel electrons from the surface of the metal. Any remaining energy contributes toward the ejected electrons' kinetic energy. This can be modeled by the equation KE_maximum = hf - Φ, where KE_maximum is the electron's maximum kinetic energy, h is Planck's constant, f is the frequency of the light, and Φ is the work function of the material. The work function Φ is the minimum amount of energy required to eject an electron from the material's surface.

The cutoff frequency or threshold frequency is the minimum frequency of the incident light required to eject electrons. If the frequency of the light is less than this value, no electrons are ejected

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Answer:

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Explanation:

Given that

g₁ = 9.96 m/s²

g₂ = 9.72 m/s²

The actual value of  g = 9.8 m/s²

a)

The difference Δ g =  9.96 -9.72 =0.24  m/s²

[tex]The\ percentage\ difference=\dfrac{0.24}{9.72}\times 100=2.46\ percentage\\[/tex]

b)

For first one :

[tex]Error\ in\ the\ percentage =\dfrac{9.96}{9.81}\times 100 =101.52\ perncetage[/tex]

For second  :

[tex]Error\ in\ the\ percentage =\dfrac{9.72}{9.81}\times 100 =99.08\ perncetage[/tex]

c)

The mean g(mean )

[tex]g(mean )=\dfrac{9.96+9.72}{2}\ m/s^2\\g(mean)=9.84\ m/s^2[/tex]

[tex]The\ percentage=\dfrac{9.84}{9.8}\times 100=100.40\ percentage[/tex]

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

The percent difference between the two measurements is 2.44%. The percent error for the first measurement is 1.63%, and for the second measurement is 0.82%. The percent error of their mean is 0.41%.

In physics, the percent difference is calculated by subtracting the two values, taking the absolute value, dividing by the average of the two values, and then multiplying by 100. Therefore, the percent difference between the two measurements 9.96m/s² and 9.72m/s² is:

|(9.96-9.72)|/((9.96+9.72)/2)*100 = 2.44%

The percent error involves taking the absolute difference between the experimental value and the accepted value, divided by the accepted value, then multiplied by 100. So, the percent error for the two measurements with accepted value of 9.80m/s² are:

For 9.96m/s²: |(9.96-9.80)|/9.80*100 = 1.63%

For 9.72m/s²: |(9.72-9.8)|/9.8*100 = 0.82%

The percent error of the mean involves doing the above but using the mean of the experimental measurements:

|(Mean of measurements - Accepted value)|/Accepted value * 100 |(9.96+9.72)/2-9.8|/9.8*100 = 0.41%

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Complete question:

When lithium iodide is dissolved in water, the solution becomes hotter.

What can you conclude about the relative magnitudes of the lattice energy of lithium iodide and its heat of hydration?

Answer:

The heat of hydration is greater in magnitude than the lattice energy  and lattice energy is smaller in magnitude that the heat of hydration.

Explanation:

When the solution becomes hotter on addition of lithium iodide, it shows exothermic reaction and it means that the heat of hydration is greater than the lattice energy  and lattice energy is smaller in magnitude that the heat of hydration.

This can also be observed in the formula below;

[tex]H_{solution} = H_{hydration} + H_{lattice. energy}[/tex]

Heat of the hydration is thus greater in magnitude than that of the lattice energy and the lattice energy is smaller in the magnitude than the heat of hydration.

The lattice energy is defined as the energy needed to separate the mole of the icon into a slid gaseous ion. It can be measured empirically and is calculated by use of electrostatics.

The heat of hydration is the great energy generated when the water reacts with the contact of cement powder. This leas to high temperatures and cause thermal cracking and the reduction of mechanical properties.

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Answer:

66.02m/s

Explanation:

the equation describing the distance covered in the horizontal direction is

[tex]x=ucos\alpha t-(1/2)gt^{2}[/tex] but the acceleration in the horizontal path is zero, hence we have

[tex]x=ucos\alpha t[/tex]

Since the horizontal distance covered is 155m at 7.6secs, we have [tex]ucos\alpha =\frac{155}{7.6} \\ucos\alpha =20.38.............equation 1[/tex]

Also from the vertical path, the distance covered is expressed as

[tex]y=usin\alpha t-(1/2)gt^{2}[/tex]

since the horizontal distance covered in 7.6secs is 195m, then we have

[tex]y=usin\alpha t-(1/2)gt^{2}\\y=7.6usin\alpha -4.9(7.6)^{2}\\478.02=7.6usin\alpha \\usin\alpha =62.9...........equation 2[/tex]

Hence if we divide both equation 1 and 2 we arrive at

[tex]\frac{usin\alpha }{ucos\alpha } =\frac{62.9}{20.38} \\tan\alpha =3.08\\\alpha =tan^{-1}(3.08)\\\alpha =72.02^{0}\\[/tex]

Hence if we substitute the angle into the equation 1 we have

[tex]ucos72.02=20.38\\u=66.02m/s[/tex]

Hence the initial velocity is 66.02m/s

Answer:

b) Vectors A and B are in the same direction.

Explanation:

To understand this problem we will say that vector A has a magnitude of 5 units and vector B a magnitude of 3 units. In the subtraction of vectors the initial parts of vectors always bind together. And the vector resulting from the subtraction is traced from the end of the second vector (B) to the end of the first vector (A).

The length of the resultant vector will be 5 - 3 = 2

In the attached image, we analyze case a), b), and d)

For a)

As we can see in the attached image the resultant vector has a length of 8 units.

For d)

As we can see in the attached image the resultant vector has a length of 5.83 units.

For b)

The resultant vector has a length of 2 units.

Therefore the case given in b) is true

Final answer:

The condition required for |A-B| to equal |A| - |B| is when vectors A and B are in the same direction.

Explanation:

The correct answer is b. Vectors A and B are in the same direction. For two vectors A and B, the operation A - B is equivalent to adding -B to A, where -B is a vector of the same magnitude as B but in the opposite direction. When A and B are aligned and pointing in the same direction, the magnitude of their difference is equal to the difference of their magnitudes because the subtraction does not involve any trigonometric component due to the angle between them, since there is none. In mathematical symbols, |A - B| = |A| - |B|.

Answer:

5 m/s2

Explanation:

The total acceleration of the circular motion is made of 2 components: centripetal acceleration and linear acceleration of 4 m/s2. They are perpendicular to each other.

The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle

[tex]a_c = \frac{v^2}{r} = \frac{30^2}{300} = \frac{900}{300} = 3 m/s^2[/tex]

So the magnitude of the total acceleration is

[tex]a = \sqrt{a_c^2 + a_l^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 m/s^2[/tex]

Answer:

truu dat

 Explanation:

df

Answer:

0.528m

Explanation:

a)58.7 cm = 0.587 m

Let g = 9.8m/s2. When the frog jumps from ground to the highest point its kinetic energy is converted to potential energy:

[tex]E_p = E_k[/tex]

[tex]mgh = mv^2/2[/tex]

where m is the frog mass and h is the vertical distance traveled, v is the frog velocity at take-off

[tex]v^2 = 2gh = 2*9.8*0.587 = 11.5[/tex]

[tex]v = \sqrt{11.5} = 3.4 m/s[/tex]

b) Vertical and horizontal components of the velocity are

[tex]v_v = vsin(\alpha) = 3.4sin(58^0) = 2.877 m/s[/tex]

[tex]v_h = vcos(\alpha) = 3.4cos(58^0) = 1.8 m/s[/tex]

The time it takes for the vertical speed to reach 0 (highest point) under gravitational acceleration g = -9.8m/s2 is

[tex]\Delta t = \Delta v / g = \frac{0 - 2.877}{-9.8} = 0.293s[/tex]

This is also the time it takes to travel horizontally, we can multiply this with the horizontal speed to get the horizontal distance it travels

[tex]s_h = v_ht = 1.8*0.293 = 0.528 m[/tex]

Answer:

Q = 5267J

Explanation:

Specific heat capacity of copper (S) = 0.377 J/g·°C.

Q = MSΔT

ΔT = T2 - T1

ΔT=49.8 - 22.3 = 27.5C

Q = change in energy = ?

M = mass of substance =508g

Q = (508g) * (0.377 J/g·°C) * (27.5C)

Q= 5266.69J

Approximately, Q = 5267J

Final answer:

To raise the temperature of 508 g of copper from 22.3°C to 49.8°C, 5096.925 J of heat needs to be added, using the specific heat capacity of copper.

Explanation:

To calculate the amount of heat (Q) needed to raise the temperature of a substance, we use the formula Q = mcΔT, where m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. For 508 g of copper with a specific heat of 0.377 J/g°C, needing to increase from 22.3°C to 49.8°C, the change in temperature (ΔT) is 49.8°C - 22.3°C = 27.5°C.
Therefore, Q = (508 g)(0.377 J/g°C)(27.5°C) = 5096.925 J. Hence, 5096.925 J of heat needs to be added to the copper to achieve the desired temperature increase.

Answer:

1402.73 m

Explanation:

Mass of Azurite=3.25 lb

Percent of copper in AZurite mineral=55.1%

Diameter of  copper wire,d=0.0113  in

Radius of copper wire=[tex]r=\frac{d}{2}=\frac{0.0113}{2}=0.00565 in=\frac{565}{100000}=\frac{565}{100}\times \frac{1}{1000}=5.65\times 10^{-3}in[/tex]

[tex]\frac{1}{1000}=10^{-3}[/tex]

Density  of copper=[tex]\rho=8.96g/cm^3[/tex]

1 lb=454 g

3.25 lb=[tex]3.25\times 454=1475.5 g[/tex]

Mass of Azurite=[tex]1475.5 g[/tex]

Mass of copper=[tex]\frac{55.1}{100}\times 1475.5=813 g[/tex]

Density=[tex]\frac{Mass}{volume}[/tex]

Using the formula

[tex]8.96=\frac{813}{volume\;of\;copper}[/tex]

Volume of copper wire=[tex]\frac{813}{8.96}=90.7cm^3[/tex]

Radius of copper wire=[tex]5.65\times 10^{-3}\times 2.54=14.35\times 10^{-3} cm[/tex]

1 in=2.54 cm

Volume of copper wire=[tex]\pi r^2 h[/tex]

[tex]\pi=3.14[/tex]

Using the formula

[tex]90.7=3.14\times (14.35\times 10^{-3})^2\times h[/tex]

[tex]h=\frac{90.7}{3.14\times (14.35\times 10^{-3})^2}[/tex]

[tex]h=140273 cm[/tex]

1 m=100 cm

[tex]h=\frac{140273}{100}=1402.73 m[/tex]

Hence, the length of copper wire required=1402.73 m

Answer:

E = -3.6859 x 10∧6 N/C

Explanation:

Let q1 = 3.5 μF = 3.5 x 10∧-6 F and q2 =-7.6 μF = -7.6 x 10∧-6 F

are the charges placed at two corner of equilateral triangle. Electric Field Magnitude "E" on the third corner will be equal to the sum of Electric Filed Magnitude generated by q1 and q2.

E = E1 + E2

E = Ke × q1 / ([tex]d^{2}[/tex]) + Ke × q2 / ([tex]d^{2}[/tex])

E = Ke/[tex]d^{2}[/tex] (q1 + q2)       (taking common)

(Ke 8.99 × 10∧9 N[tex]m^{2}[/tex]/[tex]C^{2}[/tex] and distance d= 0.1 m)  

E =   8.99 × 10∧9 N[tex]m^{2}[/tex]/[tex]C^{2}[/tex] /[tex](0.1 m)^{2}[/tex] (3.5 x 10∧-6 F  -  7.6 x 10∧-6 F)

E = -3685.9 x 10∧3 N/C

E = -3.6859 x 10∧6 N/C

Answer:

[tex]t=5057.9167\ s[/tex]

Explanation:

Given:

Now the amount of heat required to heat the water by 50 K:

[tex]Q_w=m_w.c_w.\Delta T[/tex]

[tex]Q_w=0.051\times 4180\times 50[/tex]

[tex]Q_w=10659\ J[/tex]

Now the amount of heat required to heat the resistor by 50 K:

[tex]Q_r=m_r.c_r.\Delta T[/tex]

[tex]Q_r=0.008\times 3700\times 50[/tex]

[tex]Q_r=1480\ J[/tex]

Now the total heat to converted from the electrical energy:

[tex]Q=Q_w+Q_r[/tex]

[tex]Q=12139\ J[/tex]

Now Using Joule's law of heating:

[tex]Q=V.I.t[/tex]

[tex]12139=24\times 0.1\times t[/tex]

[tex]t=5057.9167\ s[/tex]

Final answer:

To determine the time needed to heat water by 5°K, calculate the total energy needed using the specific heat capacity of water, then use the power equation P = V × I to find the time. It will take approximately 444.125 seconds for the water to reach the desired temperature.

Explanation:

To calculate the time it will take for the water to heat up by 5°K using the provided values, first determine the total energy required to raise the temperature of the water by using the equation E = mw × cw × ΔT. Here, E is the energy in joules (J), mw is the mass of water, cw is the specific heat capacity of water, and ΔT is the change in temperature.

Using the given figures:
E = 51 g × 4.18 J/g°K × 5°K = 1065.9 J

Since power P is the rate at which energy is used and P = V × I where V is the voltage and I is the current, we can rearrange the equation to find time: t = E / P.

Substitute the power using the given voltage and current:
P = 24 V × 0.1 A = 2.4 W

The time in seconds to heat the water 5°K is thus:
t = 1065.9 J / 2.4 W = 444.125 seconds

Therefore, it will take approximately 444.125 seconds to heat up the water by 5°K.

Answer:

solution:

when the engine are fired the rocket has a linear constant acceleration motion:

[tex]V_{1} ^{2} =v_{0} ^{2} +2a_{1} (y_{1} -y_{0} )t=(0)^2+2(16)(y_{1}-0)\\V_{1} ^{2}=32y_{1}................. eq(1)[/tex]

[tex]V_{1}[/tex] is the final velocity of the rocket

when the engines are fired it become equal to the initial velocity of the rocket,

when the engines are shut off

[tex]V_{2} ^{2} =v_{1} ^{2} +2a_{2} (y_{2} -y_{1} )t=>v_{1} ^{2}-2(9.8)(960-y_{1})\\V_{2} ^{2} =v_{1} ^{2}-18816+19.6y_{1}...................eq(2)[/tex]

solve eq(1) and eq(2) we find

[tex](0)^2=(32y_{1} )-18816+19.6y_{1}[/tex]

solving for [tex]y_{1}[/tex]=364.65 m

Where [tex]y_{1}[/tex] is the distance travelled by the rockets for shutting off the engine

when the engines are fired:

[tex]y_{1} =y_{o} + v_{0}t_{1} +\frac{1}{2}at^{2} =>0+(0)T+\frac{1}{2}(16)t^{2\\\\\\364.65=8T^{2} -->T=6.75s[/tex]

NOTE:

DIAGRAM IS ATTACHED

Answer:

The heat capacity of the calorimeter is 104.65 J/K

Explanation:

Heat lost by the hot water = Heat gained by cold water + Heat gained by the calorimeter

Heat lost by hot water = mCΔT

m = 75 g, C = 4.186 J/g.K, ΔT = (80 - 42.5) = 37.5 K

Heat lost by hot water = 75 × 4.186 × 37.5 = 11773.125 J

Heat gained by cold water = mCΔT

m = 100 g, C = 4.186 J/g.K, ΔT = (42.5 - 20) = 22.5 K

Heat gained by cold water = 100 × 4.186 × 22.5 = 9418.5 J

Heat gained by the calorimeter = Heat lost by hot water - Heat gained by cold water

Heat gained by the calorimeter = 11773.125 - 9418.5 = 2354.625 J

Heat gained by the calorimeter = Heat capacity of the calorimeter × ΔT

ΔT = (42.5 - 20) = 22.5 K

Heat capacity of the calorimeter = (Heat gained by the calorimeter)/(ΔT) = 2351.25/22.5 = 104.65 J/K

Answer:

Charge enter a 0.100 mm length of the axon is [tex]8.98\times 10^{-12} C[/tex]

Explanation:

Electric field E at a point due to a point charge is given by

[tex]E=k \frac{q}{r^2}[/tex]

where [tex]k[/tex] is the constant =[tex]9.0 \times 10^9 Nm^2 / C^2[/tex]

[tex]q[/tex] is the magnitude of point charge and [tex]r[/tex] is the distance from the point charge

Charges entering one meter of axon is [tex]5.\times 10^{11} \times (+e)[/tex]

Charges entering 0.100 mm of axon is [tex]5.\times 10^{11} \times (+e) \times (0.1 \times 10^{-3}[/tex]

substituting the value of [tex]+e=1.6\times 10^{-19} C[/tex] in above equation, we get charge enter a 0.100 mm length of the axon is

[tex]q=5.\times 10^{11} \times1.6\times 10^{-19} \times (0.1 \times 10^{-3}\\q=8.98\times 10^{-12} C[/tex]

Answer:

Q = 3.363 x 10⁻² C

Explanation:

given,

Voltage, V= 169 V

Capacitance of the capacitor, C = 199 μF

Charge in the capacitor = ?

We know,

Q = CV

Q = 169 x 199 x 10⁻⁶

Q = 3.363 x 10⁻² C

Hence, the Charge stored in the capacitor is equal to Q = 3.363 x 10⁻² C

a) Time at which velocity is +20.0 m/s: 2.04 s

b) Time at which velocity is -20.0 m/s: 6.12 s

c) Time at which the displacement is zero: t = 0 and t = 8.16 s

d) Time at which the velocity is zero: t = 4.08 s

e) i) ii) iii) The acceleration of the boulder is always [tex]9.8 m/s^2[/tex] downward

f) See graphs in attachment

Explanation:

a)

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

[tex]a=g=-9.8 m/s^2[/tex]

downward (acceleration due to gravity). So, we can use the following suvat equation:

[tex]v=u+at[/tex]

where:

v is the velocity at time t

u = 40.0 m/s is the initial velocity

a=g=-9.8 m/s^2 is the acceleration

We want to find the time t at which the velocity is

v = 20.0 m/s

Therefore,

[tex]t=\frac{v-u}{a}=\frac{20-40}{-9.8}=2.04 s[/tex]

b)

In this case, we want to find the time t at which the boulder is moving at 20.0 m/s downward, so when

v = -20.0 m/s

(the negative sign means downward)

We use again the suvat equation

[tex]v=u+at[/tex]

And substituting

u = +40.0 m/s

a=g=-9.8 m/s^2

We find the corresponding time t:

[tex]t=\frac{v-u}{a}=\frac{-20-(+40)}{-9.8}=6.12 s[/tex]

c)

To solve this part, we can use the following suvat equation:

[tex]s=ut+\frac{1}{2}at^2[/tex]

where

s is the displacement

u = +40.0 m/s is the initial velocity

[tex]a=g=-9.8 m/s^2[/tex] is the acceleration

t is the time

We want to find the time t at which the displacement is zero, so when

s = 0

SUbstituting into the equation and solving for t,

[tex]0=ut+\frac{1}{2}at^2\\t(u+\frac{1}{2}a)=0[/tex]

which gives two solutions:

t = 0 (initial instant)

[tex]u+\frac{1}{2}at=0\\t=-\frac{2u}{a}=-\frac{2(40)}{-9.8}=8.16 s[/tex]

which is the instant at which the boulder passes again through the initial position, but moving downward.

d)

To solve this part, we can use again the suvat equation

[tex]v=u+at[/tex]

where

u = +40.0 m/s is the initial velocity

[tex]a=g=-9.8 m/s^2[/tex] is the acceleration

We want to find the time t at which the velocity is zero, so when

v = 0

Substituting and solving for t, we find:

[tex]t=\frac{v-u}{a}=\frac{0-(40)}{-9.8}=4.08 s[/tex]

e)

In order to evaluate the acceleration of the boulder, let's consider the forces acting on it.

If we neglect air resistance, there is only one force acting on the boulder: the force of gravity, acting downward, with magnitude

[tex]F=mg[/tex]

where m is the mass of the boulder and [tex]g[/tex] the acceleration of gravity.

According to Newton's second law, the net force on the boulder is equal to the product between its mass and its acceleration:

[tex]F=ma[/tex]

Combining the two equations, we get

[tex]ma=mg\\a=g[/tex]

So, the acceleration of the boulder is [tex]g=9.8 m/s^2[/tex] downward at any point of the motion, no matter where the boulder is (because the force of gravity is constant during the motion).

f)

Find the three graphs in attachment:

- Position-time graph: the position of the boulder initially increases as the boulder goes upward; however, the slope of the curve decreases as the boulder goes higher (because the velocity decreases). The boulder reaches its maximum height at t = 4.08 s (when velocity is zero), then it starts going downward, until reaching its initial position at t = 8.16 s

- Velocity-time graph: the initial velocity is +40 m/s; then it decreases linearly (because the acceleration is constant), and becomes zero when t = 4.08 s. Then the velocity becomes negative (because the boulder is now moving downward) and its magnitude increases.

- Acceleration-time graph: the acceleration is constant and it is [tex]-9.8 m/s^2[/tex], so this graph is a straight horizontal line.

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Answer:

[tex] E_f = \frac{V}{\frac{d}{2}}= 2 \frac{V}{d}= 2* 6x10^4 \frac{N}{C} = 1.2x10^5 \frac{N}{C}[/tex]

Explanation:

For this case we know that the electric field is given by:

[tex] E= 6 x 10^4 \frac{N}{C}[/tex]

And we want to find the final electric field assuming that the separation is halved and becomes d/2.

For this case we can use the following two equations:

[tex] C = \epsilon_o \frac{A}{d} = \frac{Q}{V}[/tex]   (1)

[tex] E = \frac{\sigma}{\epsilon_o}[/tex]   (2)

Where Q represent the charge, V the voltage, d the distance, A the area.

We can rewrite the equation (2) like this:

[tex] E = \frac{\sigma}{\epsilon_o} = \frac{Q}{A \epsilon_o}[/tex]   (3)

And we can solve for Q from equation (1)like this:

[tex] Q = \frac{\epsilon_o A V}{d}[/tex]

And if we replace into equation (3) the previous result we got:

[tex] E = \frac{\epsilon_o A V}{A d \epsilon_o} = \frac{V}{d}[/tex]

And since the the electric field not change and the distance would be the half we have that the final electric field would be given by:

[tex] E_f = \frac{V}{\frac{d}{2}}= 2 \frac{V}{d}= 2* 6x10^4 \frac{N}{C} = 1.2x10^5 \frac{N}{C}[/tex]

Explanation:

a) By conservation of energy we can write

mgh on earth = mgh on mars.

[tex]mg_Eh_E=mg_Mh_M[/tex]

M,E are earth and mars respectively.

[tex]h_m =\frac{g_E}{g_M}\times h_E[/tex]

[tex]h_m=\frac{9.81}{3.71}\times h_E[/tex]

h_m= 2.64 h_E

b) Consider the time taken for the rock to reach the top of its trajectory. By symmetry, this is T/2. Inserting this into the kinematics equation v = u+at, we get the following two sets of equations:

final velocities will be zero v= 0

[tex]0= v- g_E\frac{T}{2}[/tex]

[tex]0=v- g_M\frac{T_M}{2}[/tex]

This gives [tex]2v= g_ET_E=g_mT_m[/tex] and

therefore,

[tex]T_M= 2.64T_E[/tex]

The rocks ejected by a volcano on Mars with the same initial velocity as one on Earth would reach a maximum height about 2.69 times greater, and stay aloft approximately 2.69 times longer.

The maximum height a rock reaches is given by the formula H = (v^2) / (2g), where v is the initial velocity and g is the acceleration due to gravity. If we ignore potential differences in air resistance and other factors, and the initial velocity of the rocks is the same on both planets, then the change in maximum height is directly proportional to the change in gravity because the initial velocity is constant.

(a) Mars has 37.1% of Earth's gravitational force. Thus, if a volcano on Mars ejected rocks with the same initial velocity as a volcano on Earth, they would reach approximately 2.69 times the height H.

(b) The time a rock stays in the air is given by the formula T = 2v / g. So the rock will stay aloft on Mars approximately 2.69 times as long as T, the time that would stay on Earth given the same initial velocity.

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This question asks for calculations of energy separations between specific quantum states (n) of an electron in a one-dimensional box, using Quantum Physics principles. The energy difference is calculated using a standard formula, and the values are converted into different units. This demonstrates the important foundational concepts of Quantum Physics.

In Quantum Physics, the energy difference between specific states (n) can be calculated using the formula for the potential energy inside a one-dimensional box, which is defined as E = n^2h^2 / (8mL^2), where h is the Planck constant, m is the mass of the particle (electron in this case), and L is the length of the box.

The energy separation between n = 1 and n = 2 can be found by simply finding the difference in energy levels. This can also be done for n = 5 and n = 6. Thus, a straightforward calculation will yield the energy separations in joules (J). To convert to alternative units, use convenient relations: 1 J = 0.239006 kJ/mol, 1 J = 6.242×10^18 eV, and 1 cm^-1 = 1.98644582 x 10^-23 J.

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The distance fallen before the parachute opens is 485 m. The correct option is (a).

Velocity is a vector quantity that describes the rate of change of an object's position with respect to time. It specifies the object's speed and the direction of its motion. In other words, velocity is a measure of how fast an object is moving in a particular direction. The SI unit for velocity is meters per second (m/s). Velocity can be positive or negative depending on the direction of motion. For example, a car traveling northward has a positive velocity, while a car traveling southward has a negative velocity.

Here in the Question,

We can use the following equations of motion to set up the differential equations for the velocity and position of the skydiver:

F = ma

v = dx/dt

The net force on the skydiver is given by:

F_net = F_g - F_D

where F_g is the force due to gravity, F_D is the force due to air resistance, and the negative sign indicates that F_D is opposite to the direction of motion.

So we have:

F_g - F_D = ma

where a is the acceleration of the skydiver.

Substituting F_D = -bv into the above equation, we get:

mg - bv = ma

Dividing both sides by m, we get:

g - (b/m)v = a

Now we can set up the differential equations as follows:

dv/dt = g - (b/m)v

dx/dt = v

To solve these differential equations, we can use separation of variables:

dv/(g - (b/m)v) = dt

dx/v = dt

Integrating both sides with respect to t, we get:

-(m/b) ln(g - (b/m)v) = t + C1

ln(v) = t + C2

where C1 and C2 are constants of integration.

Solving for v and simplifying, we get:

v = (g*m/b) - Ce^(-bt/m)

where C = (g*m/b - v0) is another constant of integration, and v0 is the initial velocity (which is zero).

Using the initial condition v(0) = 0, we get:

C = g*m/b

So we have:

v = (gm/b) - (gm/b)e^(-bt/m)

To find the distance fallen before the parachute opens, we need to integrate v with respect to time from t = 0 to t = 10 seconds (when the parachute opens):

x = ∫[0 to 10] v dt

Substituting v, we get:

x = ∫[0 to 10] (gm/b) - (gm/b)e^(-bt/m) dt

x = (g*m/b) t + (m^2/b^2) (e^(-bt/m) - 1)

Substituting g = 9.81 m/s^2, m = 82.0 kg, and b = 0.75 kg/s, we get:

x = 485.1 m

Therefore, the distance fallen before the parachute opens is approximately 485 m. The answer is (a).

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Answer:

L > 0.08944 m or L > 8.9 cm

Explanation:

Given:

- Flux intercepted by antenna Ф = 0.04 N.m^2 / C

- The uniform electric field E = 5.0 N/C

Find:

- What is the minimum side length of the antenna L ?

Solution:

- We can apply Gauss Law on the antenna surface as follows:

                             Ф = [tex]\int\limits^S {E} \, dA[/tex]

- Since electric field is constant we can pull it out of integral. The surface at hand is a square. Hence,

                             Ф = E.(L)^2

                             L = sqrt (Ф / E)

                             L > sqrt (0.04 / 5.0)

                             L > 0.08944 m

The area of a square antenna needed to intercept a flux of 0.040 N⋅m2/C in a uniform electric field of magnitude 5.0 N/C is 0.008 m². Consequently, each side of the antenna must be about 0.089 meters (or 8.9 cm) long.

The question pertains to the relationship between electric field and flux. The electric flux through an area is defined as the electric field multiplied by the area through which it passes, oriented perpendicularly to the field.

We are given that the electric field (E) is 5.0 N/C and the flux Φ must be 0.040 N⋅m2/C.

Hence, to intercept this amount of flux, the antenna must have an area (A) such that A = Φ / E.

That is, A = 0.040 N⋅m2/C / 5.0 N/C = 0.008 m².

Since the antenna is square, each side will have a length of √(0.008) ≈ 0.089 meters (or 8.9 cm).

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