What is the temperature of a sample of gas when the average translational kinetic energy of a molecule in the sample is 8.37 × 10 − 21 J ?

Complete question:

A parallel-plate capacitor has plates with an area of 405 cm² and an air-filled gap between the plates that is 2.25 mm thick. The capacitor is charged by a battery to 575 V and then is disconnected from the battery.   (a) How much energy is stored in the capacitor? (b) The separation between the plates is now increased to 4.50 mm.   How much energy is stored in the capacitor now?

Answer:

The energy stored in the capacitor when the plates is increased to 4.50 mm is 1.32 X 10⁻⁵ J

Explanation:

Given:

Area of the plates = 405 cm² = 405 X (10⁻²)² m²= 405 X 10⁻⁴m² = 0.0405m²

Energy stored in a capacitor = CV²/2

Where;

V is the voltage across the plates = 575 V

C is the capacitor =?

C = Kε(A/d)

K is constant = 1.0

ε is permittivity of free space = 8.885 X 10⁻¹²

d is the diameter of the two plates = 2.25 mm = 0.00225m

C = 1.0 x 8.885 X 10⁻¹² x (0.0405/0.00225)

C = 1.5993 X 10⁻¹⁰ F

(a) Energy stored in a capacitor = 0.5 X 1.5993 X 10⁻⁹ X 575²

= 2.64 X 10⁻⁵ J

(b) The separation between the plates is now increased to 4.50 mm.

C = Kε(A/d)

New diameter, d = 4.5 mm = 0.0045 m

C = 1.0 x 8.885 X 10⁻¹² x (0.0405/0.0045)

C = 7.9965 X 10 ⁻¹¹ F

Energy stored in a capacitor = 0.5 X 7.9965 X 10 ⁻¹¹ X  575²

= 1.32 X 10⁻⁵ J

Therefore, the energy stored in the capacitor when the plates is increased to 4.50 mm is 1.32 X 10⁻⁵ J

To solve this problem we will apply the concepts of Boiling Point elevation and Freezing Point Depression. The mathematical expression that allows us to find the temperature range in which these phenomena occur is given by

Boiling Point Elevation

[tex]\Delta T_b = K_b m[/tex]

Here,

[tex]K_b[/tex] = Constant ( Different for each solvent)

m = Molality

Freezing Point Depression

[tex]\Delta T_f = K_f m[/tex]

Here,

[tex]K_f =[/tex] Constant ( Different for each solvent)

m = Molality

According to the statement we have that

[tex]\Delta T_f = T_0 -T_f = 0-(-2.05)[/tex]

[tex]\Delta T_f = 2.05\°C[/tex]

From the two previous relation we can find the ratio between them, therefore

[tex]\frac{\Delta T_b}{\Delta T_f} = \frac{K_b}{K_f}[/tex]

[tex]\frac{\Delta T_b}{\Delta T_f} = \frac{0.512}{1.86}[/tex]

[tex]\frac{\Delta T_b}{\Delta T_f} = 0.275[/tex]

We already know the change in the freezing point, then

[tex]\Delta T_b = 0.275 (\Delta T_f)[/tex]

[tex]\Delta T_b = 0.275 (2.05)[/tex]

[tex]\Delta T_b = 0.5643\°C[/tex]

The temperature difference in the boiling point is 100°C (Aqueous solution), therefore

[tex]T_b -100 = 0.5643[/tex]

[tex]T_b = 100.56\°C[/tex]

Therefore the boiling point of an aqueous solution is [tex]100.56\°C[/tex]

Final answer:

The boiling point of the aqueous solution is 99.44 °C.

Explanation:

The boiling point of an aqueous solution can be determined using the formula:

Tbp = Tbpsolvent + (Kbp * m)

where:

Tbp is the boiling point of the solution

Tbpsolvent is the boiling point of the pure solvent

Kbp is the ebullioscopic constant for the solvent

m is the molality of the solution

In this case, we are given that the freezing point of the solution is -2.05 degrees C. We can use the formula for freezing point depression to find the molality:

AT = Kf * m

Rearranging the formula, we can solve for m:

m = AT / Kf

Substituting the given values:

m = (-2.05 °C) / (1.86 °C kg.mol-¹)

Rounding to 2 decimal places:

m = -1.10 mol/kg

Now, we can use the formula for boiling point elevation to find the boiling point:

Tbp = Tbpsolvent + (Kbp * m)

Substituting the given values:

Tbp = 100.00 °C + (0.512 °C/m * -1.10 mol/kg)

Calculating:

Tbp = 100.00 °C - 0.56 °C = 99.44 °C

Therefore, the boiling point of the aqueous solution is 99.44 °C.

Answer:

Explanation:

Given

Cannon is fired with a velocity of [tex]u=72.50\ m/s[/tex]

Using Equation of motion

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

where

[tex]y=displacement[/tex]

[tex]u=initial\ velocity[/tex]

[tex]a=acceleration[/tex]

[tex]t=time[/tex]

after time [tex]t=3.3 s[/tex]

[tex]y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2[/tex]

[tex]y=239.25-53.36[/tex]

[tex]y=185.89\ m[/tex]

So after 3.3 s cannon ball is at a height of 185.89 m

Answer:

0 MN/C, 2.697 MV

4.1953 MN/C, 1.2586 MV

19.2642857143 MN/C, 2.697 MV

Explanation:

k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]

Electric field at r = 8 cm

E = 0 (inside)

Electric potential is given by

[tex]V=\dfrac{kq}{R}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV[/tex]

Electric potential is 2.697 MV

r = 30 cm

[tex]E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3^2}\\\Rightarrow E=4195333.33\ MN/C=4.1953\ MN/C[/tex]

Electric field is 4.1953 MN/C

[tex]V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.3}\\\Rightarrow V=1258600\ V=1.2586\ MV[/tex]

Electric potential is 1.2586 MV

r = R = 14 cm

[tex]E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14^2}\\\Rightarrow E=19264285.7143\ N/C=19.2642857143\ MN/C[/tex]

The electric field is 19.2642857143 MN/C

[tex]V=\dfrac{kq}{r}\\\Rightarrow V=\dfrac{8.99\times 10^{9}\times 42\times 10^{-6}}{0.14}\\\Rightarrow V=2697000\ V=2.697\ MV[/tex]

The potential is 2.697 MV

To calculate the electric field and electric potential at different distances from the center of a spherical conductor, use the equations for electric field and electric potential due to a point charge. At a distance of 8.0 cm, the electric field is 7.87 * 10^3 N/C and the electric potential is 5.25 kV. At a distance of 30.0 cm, the electric field is 1.26 * 10^3 N/C, but the electric potential cannot be calculated without a reference point. At 14.0 cm, the electric field and electric potential are zero.

To calculate the electric field and electric potential at different distances from the center of a spherical conductor with a radius of 14.0 cm and a charge of 42.0 µC, we can use the equations for electric field and electric potential due to a point charge.

(a) At a distance of 8.0 cm from the center, the magnitude of the electric field can be calculated using the equation:

E = k * (Q/r²)

Substituting the given values, we get:

E = (9 * 10^9 Nm²/C²) * (42 * 10^-6 C) / (0.08 m)² = 7.87 * 10^3 N/C

To calculate the electric potential at this distance, we can use the equation:

V = k * (Q/r)

Substituting the given values, we get:

V = (9 * 10^9 Nm²/C²) * (42 * 10^-6 C) / (0.08 m) = 5.25 kV

(b) At a distance of 30.0 cm from the center, the magnitude of the electric field can be calculated in the same way:

E = (9 * 10^9 Nm²/C²) * (42 * 10^-6 C) / (0.3 m)² = 1.26 * 10^3 N/C

But the electric potential at this distance cannot be calculated using the given information, as we need the reference point for potential measurement.

(c) At a distance of 14.0 cm (the radius of the conductor) from the center, the electric field and electric potential are zero, as the conductor is in electrostatic equilibrium.

https://brainly.com/question/34973156

#SPJ6

Answer:

[tex] MA_1 = \frac{50 m/s}{\sqtr{1.288*188.9 J/Kg K * 1200 K}}=0.093[/tex]

[tex] MA_2 =\frac{1163.074 m/s}{\sqrt{1.288 *188.9 J/Kg K * 400 K}}=3.73[/tex]

Explanation:

Assuming this problem: "Carbon dioxide enters an adiabatic nozzle at 1200 K with a velocity of 50 m/s and leaves at 400 K. Assuming constant specific heats at room temperature, determine the Mach number (a) at the inlet and (b) at the exit of the nozzle. Assess the accuracy of the constant specific heat assumption."

Part a

For this case we can assume at the inlet we have the following properties:

[tex] T_1 = 1200 K, v_1 = 50 m/s [/tex]

We can calculate the Mach number with the following formula:

[tex] MA_1 = \frac{v_1}{c_1} = \frac{v_1}{\sqrt{kRT}}[/tex]

Where k represent the specific ratio given k =1.288 and R would be the universal gas constant for the carbon diaxide given by: [tex] R= 188.9 J/ Kg K[/tex]

And if we replace we got:

[tex] MA_1 = \frac{50 m/s}{\sqtr{1.288*188.9 J/Kg K * 1200 K}}=0.093[/tex]

Part b

For this case we can use the same formula:

[tex] MA_2 = \frac{v_2}{c_2} [/tex]

And we can obtain the value of v2 from the total energy of adiabatic flow process, given by this equation:

[tex] c_p T_1 + \frac{v^2_1}{2}=c_p T_2 + \frac{v^2_2}{2}[/tex]

The value of [tex] C_p = 0.8439 K /Kg K = 843.9 /Kg K[/tex] and the value fo T2 = 400 K so we can solve for [tex] v_2[/tex] and we got:

[tex] v_2= \sqrt{2c_p (T_1 -T_2) +v^2_1}=1163.074 m/s[/tex]

And now we can replace on this equation:

[tex] MA_2 = \frac{v_2}{c_2} [/tex]

And we got:

[tex] MA_2 =\frac{1163.074 m/s}{\sqrt{1.288 *188.9 J/Kg K * 400 K}}=3.73[/tex]

Final answer:

To determine the Mach number at the exit of the nozzle, we can use the isentropic flow equations.

Explanation:

To determine the Mach number at the exit of the nozzle, we need to use the isentropic flow equations. The Mach number (M) at the exit of the nozzle can be calculated using the equation:

M = sqrt( 2/(k-1) * ( (P/Pref) ^ ((k-1)/k) - 1) )

Where:

Given the information provided in the question, we can substitute the values into the equation to calculate the Mach number.

Answer:

The voltage across the capacitor decreases by a factor of 2

Explanation:

When we put a dielectric material on a capacitor, we generate a polarization on it that changes the capacitance of the capacitor. The dielectric constant is the ratio between the initial capacitance (Co) without the dielectric and the new capacitance (C) with the dielectric:

[tex]k=\frac{C}{C_0} [/tex]

So, if k is equal to 2:

[tex] 2= \frac{C}{C_0}[/tex]

[tex]C=2*C_0 [/tex]

So, the capacitance is doubled with the dielectric

Because the battery is disconnected, the charge on the plates of capacitor must be constant because conservation of charge.

The effective electric field of a capacitor is the electric filed when we put a dielectric on it, and it is:

[tex]E_{eff}=\frac{E}{k}=\frac{E}{2} [/tex]

With E the initial electric field, so the electric field is halved

Finally, because we know electric field and voltage (V) on a parallel capacitor with distance between plates (d) are related by:

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

Because electric field and voltage are directly proportional and d remains constant if Electric field is halved, then voltage is halved too. The voltage across the capacitor decreases by a factor of 2.

Answer:

a = (m₂-m₁) / (m₂ + m₁ + ½ m)

a = (m₂-m₁) / (m₂ + m₁ + 1)

Explanation:

An Atwood machine consists of two masses of different m1 and m2 value that pass through a pulley, in this case with mass. Let's use Newton's second law for this problem.

Assume that m₂> m₁, so the direction of descent of m₂ is positive, this implies that the direction of ascent of m₁ is positive

Equation of the side of m₁

              T₁ - W₁ = m₁ a

Equation of the side of m₂

              W₂ - T₁ = m₂ a

Mass pulley equation m; by convention the counterclockwise rotation is positive

              τ = I α

              T₁ R - T₂ R = I (-α)

The moment of inertia of a disk is

              I = ½ m R²

Angular and linear acceleration are related

             a = α R

             α = a / R

The rotation is clockwise, so it is negative

We replace

             (T₁ –T₂) R = ½ m R² (-a / R)

             T₁ -T₂ = - ½ m a

Let's write our three equations together

              T₁ - m₁ g = m₁ a

              m₂ g - T₂ = m₂ a

              T₁ –T₂ = -½ m a

Let's multiply the last equation by (-1) and add

               m₂ g - m₁ g = m₂ a + m₁ a + ½ m a

                a = (m₂-m₁) / (m₂ + m₁ + ½ m)

calculate

               a = (m₂ - m₁)/ (m₁ +m₂ + 1)

Based on the calculations, the acceleration of this system is equal to 0.58 [tex]m/s^2[/tex].

Given the following data:

First of all, we would determine the moment of inertia of this disk by using this formula:

[tex]I=\frac{1}{2} mr^2\\\\I=\frac{1}{2} \times 2 \times (0.248)^2\\\\I=0.0615\;Kgm^2[/tex]

Next, we would use a free body diagram to determine the tensional forces acting on the disk by applying Newton's Second Law of Motion as follows:

For the first force:

[tex]F_1g-F_{T1}=m_1a\\\\m_1g-F_{T1}=m_1a\\\\1.61(9.8)-F_{T1}=1.61a\\\\15.8-F_{T1}=1.61a\\\\F_{T1}=15.8-1.61a[/tex]

For the second force:

[tex]F_{T2}-F_2g=m_2a\\\\F_{T2}-m_2g=m_2a\\\\F_{T2}-1.38(9.8)=1.38a\\\\F_{T2}-13.5=1.38a\\\\F_{T2}=13.5+1.38a[/tex]

For the torque, we have:

[tex]\sum T =I\alpha \\\\F_{T1}r-F_{T2}r=I\alpha\\\\(F_{T1}-F_{T2})r=I\alpha\\\\(15.8-1.61a-[13.5+1.38a])0.248=0.0615 \times \frac{a}{0.248} \\\\(15.8-1.61a-13.5-1.38a)0.248=0.0615 \times \frac{a}{0.248} \\\\(2.3-2.99a)0.248=0.0615 \times \frac{a}{0.248}\\\\0.5704-0.7415a=\frac{0.0615a}{0.248}\\\\0.1415-0.1839a=0.0615a\\\\0.1839a+0.0615a=0.1415\\\\0.2454a=0.1415\\\\a=\frac{0.1415}{0.2454}[/tex]

Acceleration, a = 0.58 [tex]m/s^2[/tex]

Read more on acceleration here: brainly.com/question/24728358

Answer:

0.212V

Explanation:

The induced emf in circular loop is [tex]\epsilon=-d\frac{\phi_B}{dt}=-NA\frac{dB}{dt}[/tex]

N = Number of loops = 15

A = Cross sectional Area = [tex]\pi[/tex][tex]0.03^{2}[/tex] = 0.0009[tex]\pi[/tex]

dB = change in magnetic Field = 0-0.5 = -0.5

dt = time taken = 0.1sec

[tex]\epsilon=-15*0.0009\pi *\frac{-0.5}{0.1}[/tex] = 0.212V

Answer:

10.89 seconds

229.895 m

Explanation:

Distance van travels

[tex]x_v=170+5.5t+\dfrac{1}{2}0t^2\\\Rightarrow x_v=170+5.5t[/tex]

Position of car

[tex]x_c=0+32t+\dfrac{1}{2}-2t^2\\\Rightarrow x_c=32t-t^2[/tex]

They are equal

[tex]170+5.5t=32t-t^2\\\Rightarrow 17+5.5t-32t+t^2\\\Rightarrow t^2-26.5t+170=0\\\Rightarrow 10t^2-265t+1700=0[/tex]

[tex]t=\frac{-\left(-265\right)+\sqrt{\left(-265\right)^2-4\cdot \:10\cdot \:1700}}{2\cdot \:10}, \frac{-\left(-265\right)-\sqrt{\left(-265\right)^2-4\cdot \:10\cdot \:1700}}{2\cdot \:10}\\\Rightarrow t=15.6, 10.89\ s[/tex]

The collision occurs at 10.89 seconds

[tex]x_v=170+5.5\times 10.89\\\Rightarrow x_v=229.895\ m[/tex]

Collision occurs at 229.895 m from the starting point

Answer:

A) The car should be traveling at 31.9 m/s.

B) The speed of the car just before it lands on the other side is 37.0 m/s.

Explanation:

Hi there!

A) Please see the attached figure for a better description of the problem. When the car reaches the other side of the river, its position vector will be r1 in the figure. The components of this vector are r1x and r1y.

If we place the origin of the frame of reference at the edge of the cliff, the components of the vector r1 will be:

r1x = 61.0 m

r1y = -20.0 m + 2.1 m = -17.9 m

The equations for the x and y-components of the position vector of the car are the following:

x = x0 + v0 · t

y = y0 + 1/2 · g · t²  

Where:

x = horizontal position at a time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

y = vertical position at a time t.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Using the equation of the y-component of r1, we can find the time it takes the car to reach the other side of the river. We have to find the time at which the vector r1y is -17.9 m:

y = y0 + 1/2 · g · t²   (y0 = 0 because the origin of the frame of reference is located at the edge of the cliff).

y = 1/2 · g · t²

-17.9 m = -1/2 · 9.8 m/s² · t²

-17.9 m / -4.9 m/s² = t²

t = 1.91 s

Now, using the equation of the x-component, we can find the initial velocity. We know that at t = 1.91 s, the horizontal component of the vector r1 is 61.0 m:

x = x0 + v0 · t (x0 = 0 because the origin of the frame of reference is located at the edge of the cliff).

x = v0 · t

61.0 m = v0 · 1.91 s

v0 = 61.0 m / 1.91 s = 31.9 m/s

The car should be traveling at 31.9 m/s.

B) The equation of the velocity vector of the car is the following:

v = (v0, g · t)

The horizontal component of the velocity vector is v0, 31.9 m/s.

Let's calculate the value of the vertical component:

vy = g · t

vy = -9.8 m/s² · 1.91 s

vy = -18.7 m/s

Then, the velocity vector of the car just before it lands on the other side is the following:

v = (31.9, -18.7) m/s

The magnitude of this vector is calculated as follows:

|v| = √[(31.9 m/s)² + (-18.7 m/s)²]  

|v| = 37.0 m/s

The speed of the car just before it lands on the other side is 37.0 m/s.

Final answer:

To safely clear the river, the car must travel at 23.4 m/s at the cliff's edge, and it will land with a speed of 22.6 m/s on the other side.

Explanation:

A) To clear the river and land safely on the other side, the car should be traveling at a speed of 23.4 m/s as it leaves the cliff. This is calculated using the principles of projectile motion and conservation of energy.

B) The speed of the car just before it lands safely on the opposite side would be 22.6 m/s. This speed is also determined by energy considerations, such as the conversion of potential energy to kinetic energy.

To accomplish this feat, the entertainer must throw the ball upward with a minimum initial speed and reach a certain height above its initial position.

(a) While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We can find the time for this by using Equation 4.22:
y = yo + voyt - (1/2)gt².

If we take the initial position yo to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:

https://brainly.com/question/14117987

#SPJ12

Final answer:

The minimum initial speed the entertainer must throw the ball upward is 17.96 m/s, and the height of the ball just as she reaches the table is approximately 15.56 m.

Explanation:

Calculating the Minimum Initial Speed and Height of the Juggling Ball

To answer the student's question regarding the juggling entertainer's act, we need to apply concepts from kinematics, a subfield of classical mechanics in physics. First, let's find out the total time the entertainer has to throw the ball and return to the starting position. We can find the time taken to reach the table and come back with the given average speed and distance: Time = (2 × Distance) / Speed. The entertainer runs to and from a table which is 5.50 m away at an average speed of 3.00 m/s. Hence, the total time for the round trip is (2 × 5.50 m) / 3.00 m/s = 3.67 s.

Now, we use the equation of motion to calculate the initial vertical speed needed for the ball to be in the air for this duration: s = ut + 0.5 × a × t², where s is the displacement (which is 0 because the ball returns to the same position), u is the initial vertical speed, and a is the acceleration due to gravity (-9.81 m/s²). After rearranging the formulas, the initial speed u turns out to be 17.96 m/s.

To find the height of the ball as she reaches the table at 5.50 m away, we consider half the total time, which is 1.835 s. Putting this in the kinematic equation s = ut + 0.5 × a × t², we find the height to be approximately 15.56 m at that instant.

Answer:

9.027 Ω

Explanation:

Using,

R = R₀(1+αΔt)

R = R₀(1+α[t₂-t₁]).......................... Equation 1

Where R = the value of the resistance at the final temperature, R₀ = the value of the resistance at the initial temperature, α = Temperature coefficient of resistance, t₂ = Final temperature, t₁ = Initial temperature.

Given: R₀ = 8.6 Ω, t₂ = 42 °C, t₁ = 29 °C

Constant : 0.003819/°C

Substitute into equation 1

R = 8.6(1+0.003819[42-29])

R = 9.027 Ω.

Hence the resistance = 9.027 Ω

9.02Ω

The resistivity of a conductor increases as temperature increases. In this case, silver, which is a great conductor, will increase its resistivity linearly over a range of increasing temperatures . The relationship between resistance(R) and temperature(T) is given as;

R = R₀ (1 + α(T - T₀))     --------------(i)

where;

R and R₀ are the final and initial resistances of the material (silver in this case)

α = temperature coefficient of resistivity of the material (silver) = 0.0038/°C

T and T₀ are the final and initial temperatures.

From the question;

R₀ = 8.60Ω

T₀ = 29.0°C

T = 42.0°C

Substitute these values into equation (i);

R = 8.60 (1 + 0.0038(42.0 - 29.0))

R = 8.60(1 + 0.0038(13))

R = 8.60(1 + 0.0494)

R = 8.60(1.0494)

R = 9.02Ω

Therefore, the resistance at 42.0°C is 9.02Ω

Answer:

Explanation:

Given

Dimension of Plastic sheet is [tex]2\times 4\ m^2[/tex]

Charge on sheet [tex]Q=-10\ \mu C[/tex]

Charge density [tex]\sigma =\frac{q}{A}[/tex]

[tex]\sigma =\frac{-10\times 10^{-6}}{8}=1.25\times 10^{-6}\ C/m^2[/tex]

Sphere has mass of  [tex]m=4\ \mug[/tex]

If sphere is suspended motionless then its weight is balanced by repulsion force

Repulsive force [tex]F_r=qE[/tex]

where E=Electric field due to sheet

[tex]E=\frac{q}{2\epsilon _0}[/tex]

[tex]E=\frac{-10\times 10^{-6}}{2\times 8.85\times 10^{-12}}[/tex]

[tex]E=5.647\times 10^{5}\ N/C[/tex]

[tex]F_r=q\times 5.647\times 10^{5}[/tex]

[tex]F_r=mg[/tex]

[tex]q=\frac{mg}{E}[/tex]

[tex]q=\frac{4\times 10^{-6}\times 9.8}{5.647\times 10^{5}}[/tex]

[tex]q=6.9417\times 10^{-11}\ C[/tex]

Answer:

21.6 ohm

Explanation:

We are given that

EMF=E=9.63 V

Current=I=118 mA=[tex]118\times 10^{-3} A[/tex]

[tex]1 mA=10^{-3} A[/tex]

Resistance=[tex]R=60\Omega[/tex]

We have to find the internal resistance of the battery.

We know that

[tex]V=E-Ir[/tex]

We know that V=IR

[tex]IR=E-Ir[/tex]

[tex]IR+Ir=E[/tex]

[tex]I(R+r)=E[/tex]

[tex]R+r=\frac{E}{I}[/tex]

Substitute the values

[tex]60+r=\frac{9.63}{118\times 10^{-3}}[/tex]

[tex]60+r=81.6[/tex]

[tex]r=81.6-60[/tex]

[tex]r=21.6\Omega[/tex]

Hence, the internal resistance of the battery=21.6 ohm

Answer:

21.6 ohm

Explanation:

EMF of the battery, E = 9.63 V

Current, i = 118 mA = 0.118 A

Resistance, R = 60 ohm

Let the internal resistance of the cell is r.

[tex]i = \frac{E}{R + r}[/tex]

R + r = 9.63 / 0.118

60 + r = 81.6

r = 21.6 ohm

Answer:

Explanation:

It is given that red is 2 m from the mirror and green is 1 m from the mirror so the image of green and red will be  formed  1 and 2 m behind the mirror respectively.

Green will be seen at the same distance from the mirror when seen from different position to anyone else.

The above can be explained by the given diagram

Explanation:

X-ray telescopes have a different design than that of optical telescopes, since x-rays can reflect off from mirror if they are struck at a particular of  angle of the grazing. The X-rays are concentrated to a point in two reflections. By nesting the mirrors inside a one another , the X-ray telescope area can be enhanced.

Final answer:

X-ray telescope mirrors are designed to reflect high-energy X-rays at small angles, using precision-coated and aligned mirrors, while optical telescope mirrors only require front surface polishing and are used for visible light reflection, allowing them to be larger and more cost-effective.

Explanation:

The mirrors in X-ray telescopes, such as those used in the Chandra X-ray Observatory, are designed to reflect high-energy X-rays that are absorbed when incident perpendicular to the medium. Unlike mirrors in optical instruments, which reflect visible light usually by incident at direct angles, X-ray telescope mirrors reflect X-rays at small grazing angles, similar to a rock skipping across a lake. The design of these mirrors involves a long barrelled pathway with multiple pairs of mirrors, often coated with metals like iridium, to focus the rays at a specific point. They're precision-engineered to be extremely smooth for the most effective reflection.

In contrast, optical telescopes like reflectors use mirrors that only need the front surface polished accurately, avoiding issues like flaws and bubbles within the glass. These mirrors can be larger and more cost-effective than lenses, making them suitable for studying dimmer or more distant objects. Today's largest optical telescopes are reflectors for this reason.

Answer:

A) r•s = -21.99 units

B) rXs = 22.77 units (In the direction of k)

Completed question

Two vectors, r and s, lie in the xy plane. Their magnitudes are 4.25 and 7.45 units, respectively, and their directions are 312° and 86.0°, respectively, as measured counterclockwise from the positive x axis. What are the values of the following products?

A) r•s =

B) (r)X(s)=

Explanation:

A) dot product of r and s can be expressed mathematically as;

r•s = |r||s|cos(A) .......1

Where A is the angle between the two vectors.

|r| and |s| are the magnitude of vector r and s.

r•s = 4.25×7.45 ×cos(86-312) = -21.99 units

B) cross product of r and s can be expressed mathematically as;

rXs = |r||s|sin(A) .......2

Where A is the smallest angle between the two vectors

|r| and |s| are the magnitude of vector r and s.

A = 312-86 = 226

A = 360-226 = 134 smallest

rXs = 4.25 × 7.45 × sin134.

rXs = 22.77 units

In the direction of k

(a) -22.00 units

(b) 22.77 units

Given vectors are r and s

Where;

r = |r| = 4.25   and ∠r = 312°    measured anticlockwise

s = |s| = 7.45   and ∠s = 86°    measured anticlockwise

First, let's calculate the angle between vectors r and s by representing them in the figure below;

                                y |        /s

                                   |      /

                                   |    /

                                   |  /  

                                   |/ )86°                                x

                                   |\) 48°

                                   |  \

                                   |    \

                                   |      \

                                   |        \ r

To get the acute angle between r and the +x axis, subtract the reflex angle of r (312°) from 360° as follows;

360 - 312 = 48°

As shown in the diagram, the angle between vectors r and s is 48° + 86° = 134°

Now,

(a) The (r)(s) represents the dot or scalar product of the two vectors and it is given as;

(r) (s) = r x s cos θ            ---------------------------(i)

Where;

r = magnitude of vector r = 4.25

s = magnitude of vector s = 7.45

θ is the angle between the two vectors r and s = 134°

Substitute these values into equation (i) as follows;

(r) (s) = 4.25 x 7.45 cos 134°

(r) (s) = 4.25 x 7.45 x -0.6947

(r) (s) = 31.66 x -0.6947

(r) (s) = -22.00 units

(b) The (r) X (s) represents the vector product of the two vectors and it is given as;

(r) (s) = r x s sin θ            ---------------------------(ii)

Where;

r = magnitude of vector r = 4.25

s = magnitude of vector s = 7.45

θ is the angle between the two vectors r and s = 134°

Substitute these values into equation (ii) as follows;

(r) (s) = 4.25 x 7.45 sin 134°

(r) (s) = 4.25 x 7.45 x 0.7193

(r) (s) = 31.66 x 0.7193

(r) (s) = 22.77 units

Answer:

a.V=8tj+k

b.a=8j

Explanation:

Given:

Position r= i+4t^2j +tk

Nb r is position in metre and time in seconds

a.velocity is change in position/ change in time

v= ∆r/∆t =dr/dt

V=d ( i+ 4t^2j+tk)/dr

Differenting with respect to (t)

V=8tj+K

b.acceleration = change in velocity/change in time

a= ∆v/∆r =dv/dt

a=d (8tj+k)/dt

a= 8j

Answer:

(a) velocity, v = 8t j + k

(b) acceleration, a = 8 j

Explanation:

The position of the particle as a function of time is given as;

r = i + 4t² j + t k         --------------------(i)

(a) To get the expression of its velocity, v, find the derivative of its position with respect to time by differentiating equation (i) with respect to t as follows;

v = dr / dt = 0 + 8t j + k

v = dr / dt = 8t j + k

v = 8t j + k          ----------------------(ii)

Therefore, the equation/expression for the particle's velocity (v) is

v = 8t j + k

(b) To get the expression of its acceleration, a, find the derivative of its velocity with respect to time by differentiating equation (ii) with respect to t as follows;

a = dv / dt = t j + 0

a = dv / dt = t j

a = 8 j

Therefore, the expression for the particle's acceleration, a, is a = 8 j

Answer:

  P₂= 141.15 kPa  

Explanation:

Given that

Volume ,V= 3 L

V= 0.003 m³

Initial temperature ,T₁ = 273 K

Initial pressure ,P₁ = 105 kPa

Final temperature ,T₂ = 367 K

Given that volume of the cylinder is constant .

Lets take final pressure = P₂

We know that for constant volume process

[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}\\P_2=P_1\times \dfrac{T_2}{T_1}\\P_2=105\times \dfrac{367}{273}\ kPa\\\\P_2=141.15\ kPa[/tex]

Therefore the final pressure = 141.15 kPa

P₂= 141.15 kPa

Answer:

Volume of 22 drop will be 0.68 ml

Explanation:

We have given volume of one drop = 0.031 ml

We know that 1 liter = 1000 ml

So [tex]1ml=10^{-3}L[/tex]

So 0.031 ml will be equal to [tex]0.031\times 10^{-3}L[/tex]

We have to find the volume of 22 drop

For finding volume of 22 drop we have to multiply volume of one drop by 22

So volume of 22 drop will be [tex]=22\times 0.031\times 10^{-3}=0.682\times 10^{-3}L=0.68mL[/tex]

So volume of 22 drop will be 0.68 ml

Answer:

Visible Light and Radio waves

Explanation:

The earth's atmosphere is transparent to a few windows in the electromagnetic spectrum. it is completely transparent to allow observation from the ground in visible light rang 380 to 740 nano meters. Also in the range of radio wave as communication are done from space to ground in the form of radio waves.

it is Partially transparent to Microwave and infrared range.

The atmosphere is transparent to certain regions of the electromagnetic spectrum, such as visible light, radio waves, and parts of the infrared and microwave spectra.

The Earth's atmosphere is transparent to certain regions of the electromagnetic spectrum, allowing observations to be made from the ground. These regions include the visible light spectrum, the radio waves spectrum, and parts of the infrared and microwave spectra.

In these regions, electromagnetic waves can pass through the Earth's atmosphere relatively easily, allowing ground-based observations to be conducted.

https://brainly.com/question/32222934

#SPJ3

Answer:

(a). differential relation becomes dρ/ρ = -τρV2 dV/V

(b). fraction change in density; dρ/ρ = -8.7 ˣ 10⁻⁶

(c).  dρ/ρ = -8.7 ˣ 10⁻²

Explanation:

Let us begin,

(a).  given from the question we have that dp = -ρVdV

where dρ = ρ τ dp, i.e.

dp = dρ/ρτ ...............(1)

replacing value of dp we have,

-ρVdV = dρ/ρτ

so that dρ = -τp2 VdV

finally, dρ/ρ = -τp V2 dV/V

(b). from the question here, we were given Velocity to be = 10 m/s

density (ρ) =  1.23 kg/m3

pressure (p) =  1.01 x 10⁵ N/m2

from formula,

dρ/ρ = τs ρ V2 dV/V .............(2)

but τs = 1/γp = 1/(1.4× 1.01×10⁵) = 7.07 ˣ 10⁻⁶ m²/N

substituting value of τs  into equation (2) we have

dρ/ρ = τs ρ V2 dV/V =  (7.07 ˣ 10⁻⁶) ˣ (1.23) ˣ (10ˣ2) (0.01) = -8.7 ˣ 10⁻⁶

dρ/ρ = -8.7 ˣ 10⁻⁶

(c). from we question we know that dρ/ρ has a large ratio of (1000/10)²

so dρ/ρ = -8.7 ˣ 10⁻⁶ × (1000/10)² = -8.7 ˣ 10⁻²

dρ/ρ = -8.7 ˣ 10⁻².

comparing this result with part (b). we can see that when we increase the velocity of a factor 100, there is an increased factorial change in the density by a factor 104.

Answer:

Explanation:

Given

mass of object [tex]m=5\ kg[/tex]

Object is attached to the ceiling of an elevator by a rope

Suppose T is the Tension in the rope so

acceleration of the elevator [tex]a=4\ m/s^2[/tex]  (upward)

From Free Body diagram we can write as

[tex]T-mg=ma[/tex]

[tex]T=m(g+a)[/tex]

[tex]T=5(9.8+4)[/tex]

[tex]T=69\ N[/tex]

Answer:

a)The highest point reached by the rocket is 1412 m

b)The rocket crashes after 54.7 s

Explanation:

Hi there!

The equations of height and velocity of the rocket are the following:

h = h0 + v0 · t + 1/2 · a · t² (while the engines work).

h = h0 + v0 · t + 1/2 · g · t² (when the rocket is in free fall).

v = v0 + a · t (while the engines work).

v = v0 + g · t (when the rocket is in free fall).

Where:

h = height of the rocket at a time t.

h0 = initial height of the rocket.

v0 = initial velocity.

t = time.

a = acceleration due to the engines.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity of the rocket at a time t.

First, let's find the velocity and height reached by the rocket until the engines fail:

h = h0 + v0 · t + 1/2 · a · t²

Let's set the origin of the frame of reference at the launching point so that h0 = 0. Since the rocket starts from rest, v0 = 0. So after 30.0 s the height of the rocket will be:

h = 1/2 · a · t²

h = 1/2 · 2.5 m/s² · (30.0 s)²

h = 1125 m

Now let's find the velocity of the rocket at t = 30.0 s:

v = v0 + a · t (v0 = 0)

v = 2.5 m/s² · 30.0 s

v = 75 m/s

After 30.0s the rocket will continue to ascend with a velocity of 75 m/s. This velocity will be gradually reduced due to the acceleration of gravity. When the velocity is zero, the rocket will start to fall. At that time, the rocket is at its maximum height. So, let's find the time at which the velocity of the rocket is zero:

v = v0 + g · t

0 = 75 m/s - 9.8 m/s² · t (v0 = 75 m/s because the rocket begins its free-fall motion with that velocity).

-75 m/s / -9.8 m/s² = t

t = 7.7 s

Now, let's find the height of the rocket 7.7 s after the engines fail:

h = h0 + v0 · t + 1/2 · g · t²

The rocket begins its free fall at a height of 1125 m and with a velocity 75 m/s, then, h0 = 1125 m and v0 = 75 m/s:

h = 1125 m + 75 m/s · 7.7 s - 1/2 · 9.8 m/s² · (7.7 s)²

h = 1412 m

The highest point reached by the rocket is 1412 m

b) Now, let's calculate how much time it takes the rocket to reach a height of zero (i.e. to crash) from a height of 1412 m.

h = h0 + v0 · t + 1/2 · g · t² (v0 = 0 because at the maximum height the velocity is zero)

0 = 1412 m - 1/2 · 9.8 m/s² · t²

-1412 m / -4.9 m/s² = t²

t = 17 s

The rocket goes up for 30.0 s with an acceleration of 2.5 m/s².

Then, it goes up for 7.7 s with an acceleration of -9.8 m/s².

Finally, the rocket falls for 17 s with an acceleration of -9.8 m/s²

The rocket crashes after (30.0 s + 7.7 s + 17 s) 54.7 s

(a)The highest point reached by the rocket is 1412 m

(b)The rocket crashes after 54.7 s

The height reached by the rocket until the engines fail is:

[tex]h = h_0+ v_0 t + \frac{1}{2} a t^2[/tex]

here, h = height of the rocket at a time t.

h₀ = initial height of the rocket.

v₀ = initial velocity.

t = time.

a = acceleration due to the engines.

At the time of the launch h₀= 0 and v₀ = 0

[tex]h = \frac{1}{2} a t^2\\\\h = \frac{1}{2} \times2.5\times(30)^2\\\\h = 1125 m[/tex]

The velocity of the rocket at t = 30.0 s:

[tex]v = a t\\\\v = 2.5 \times30.0\\\\v = 75 m/s[/tex]

The rocket will continue to ascend with a velocity of 75 m/s until it is finally zero due to gravitational force. Now the time for which the rocket continues to ascend is given by:

[tex]0 = v - g t\\\\0 = 75 - 9.8 t \\\\\frac{-75}{ -9.8} = t\\\\t = 7.7 s[/tex]

The height gained by the rocket 7.7 s after the engines fail:

[tex]h' = h + v t + \frac{1}{2} g t^2\\\\h' = 1125+ 75\times 7.7 - .5\times9.8 \times (7.7 s)^2\\\\h' = 1412 m[/tex]

So, the highest point reached by the rocket is 1412 m

(b) Now,the time talen to crash is the time takn to fall down from the height of 1412m

[tex]h' =0 \times t +\frac{1}{2} g t^2\\\\1412 = 0.5\times9.8\times t^2\\\\t^2=\frac{1412}{9.8}\\\\ t = 17 s[/tex]

Total time taken to crash is  30.0s + 7.7s + 17s = 54.7s

Learn more about gravitation force:

https://brainly.com/question/24783651?referrer=searchResults

Answer:

Mass of the silver will be equal to 46.70 gram

Explanation:

We have given heat required to raise the temperature of silver by 24°C is 269 J , so [tex]\Delta T=24^{\circ}C[/tex]

Specific heat of silver = 0.240 J/gram°C

We have to find the mass of silver

We know that heat required is given by

[tex]Q=mc\Delta T[/tex], here m is mass, c is specific heat of silver and [tex]\Delta T[/tex] is rise in temperature

So [tex]269=m\times 0.240\times 24[/tex]

m = 46.70 gram

So mass of the silver will be equal to 46.70 gram

Answer:

A) equal in amount to the force your friend exerts on you.

Explanation:

According to the Newton's third law of motion every action has equal and opposite reaction. Here both the bodies exert equal and opposite forces on each other but it is just that I am able to stop myself from getting pulled by a greater force of friction by applying more normal force on the ground.

As we know that the force of friction is given as:

[tex]f=\mu.N[/tex]

where:

[tex]\rm f=\ force\ of\ friction\\N=\ applied\ normal\ force\ to\ the\ ground\\ \mu=\ coefficient\ of\ friction[/tex]

Answer:

The mass of the asteroids is 0.000334896182184 times the mass of the Earth.

39929.4542466 m

Explanation:

Total mass of the asteroids

[tex]m_a20000\times 10^{17}=2\times 10^{21}\ kg[/tex]

[tex]m_e[/tex] = Mass of Earth = [tex]5.972\times 10^{24}\ kg[/tex]

The ratio is

[tex]\dfrac{m_a}{m_e}=\dfrac{2\times 10^{21}}{5.972\times 10^{24}}\\\Rightarrow \dfrac{m_a}{m_e}=0.000334896182184[/tex]

The mass of the asteroids is 0.000334896182184 times the mass of the Earth.

Volume is given by

[tex]V=\dfrac{m}{\rho}\\\Rightarrow \dfrac{4\pi}{3\times 8} d^3=\dfrac{m}{\rho}\\\Rightarrow d^3=\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{10^{17}}{3000})^{\dfrac{1}{3}}\\\Rightarrow d=39929.4542466\ m[/tex]

The diameter is 39929.4542466 m

To find the temperature at 2000 feet above sea level, we need to use the average lapse rate of 3.5°F per 1000 feet. By calculating the temperature decrease with increasing altitude, we can determine that the temperature at 2000 feet would be 93°F.

To find the temperature at 2000 feet, we need to use the average lapse rate. The average lapse rate tells us how much the temperature decreases with increasing altitude. In this case, the average lapse rate is 3.5°F per 1000 feet. So for every 1000 feet increase in altitude, the temperature decreases by 3.5°F.

Since we want to find the temperature at 2000 feet, which is 2000 feet above sea level, we need to divide 2000 by 1000 to find how many intervals of 1000 feet we have. There are 2 intervals. So, we multiply the lapse rate of 3.5°F by 2 to get a temperature decrease of 7°F.

Therefore, the temperature at 2000 feet would be 100°F - 7°F = 93°F.

https://brainly.com/question/32246379

#SPJ3

If the temperature at the surface of Earth (at sea level) is 100°F, at 2000 feet if the average lapse rate is 3.5°F/1000 the temperature is 93°F.

If the temperature at the surface of Earth at sea level is 100°F, we can calculate the temperature at 2000 feet using the average lapse rate, which is given as 3.5°F per 1000 feet.

The temperature change is 3.5°F/1000 feet × 2000 feet = 7°F.

Therefore, the temperature at 2000 feet above sea level would be the surface temperature minus the temperature change due to the lapse rate:

100°F - 7°F

= 93°F.

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

[tex]\Delta W = \Delta KE[/tex]

[tex]\Delta W = \frac{1}{2} mv^2[/tex]

Here,

m = mass

v = Velocity

Our values are given as,

[tex]m = 79.7kg[/tex]

[tex]v = 4.77m/s[/tex]

Replacing,

[tex]\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2[/tex]

[tex]\Delta W = 907J[/tex]

Therefore the mechanical energy lost due to friction acting on the runner is 907J

When we Using formula of relative speed

When we put the value into the formula

When we put the value into the formula

Find out more information about distance here:

https://brainly.com/question/11495758

Final answer:

The fly travels a distance of approximately 115 mm before the steamrollers collide when both steamrollers and the fly start at the same time and approach each other at given constant speeds.

Explanation:

The question involves calculating the distance a fly travels, given that it is flying back and forth between two objects moving towards each other. We know the steamrollers start 115 mm apart and each moves at 1.10 m/s towards the other, while the fly travels at a constant speed of 2.20 m/s. To find the distance the fly travels before the steamrollers meet, we first need to determine how long it takes for the steamrollers to collide.

The steamrollers are moving towards each other, so their relative speed is the sum of their individual speeds, which is 1.10 m/s + 1.10 m/s = 2.20 m/s. Since they start 115 mm apart, which is 0.115 meters, the time it takes for them to meet is the distance divided by their relative speed, 0.115 m / 2.20 m/s = 0.05227 seconds. In this time, since the fly is traveling at 2.20 m/s, the distance it covers is the fly's speed multiplied by the time, which gives us 2.20 m/s * 0.05227 seconds = 0.11499 meters, or approximately 115 mm.