he work function of a certain metal is 1.90 eV. What is the longest wavelength of light that can cause photoelectron emission from this metal? (1 eV = 1.60 × 10-19 J, c = 3.00 × 108 m/s, h = 6.626 × 10-34 J ∙ s)

Answer:

Specific heat of the granite rock = 3387.05 Jkg⁻¹°C⁻¹

Explanation:

We have heat required, H = mcΔT

Mass of granite, m = 0.500 kg

Specific heat of granite, c = ?

Change in temperature, ΔT = 21°C

Heat energy, H = 8.5 kcal = 8500 x 4.184 = 35564 J

Substituting

         H = mcΔT

         35564 = 0.500 x c x 21

          c = 3387.05 Jkg⁻¹°C⁻¹

Specific heat of the granite rock = 3387.05 Jkg⁻¹°C⁻¹

Answer:

[tex]\frac{m_2}{m_1} = 0.020[/tex]

Explanation:

As we know that in this sling shot the kinetic energy given to the mass is equal to the elastic potential energy stored in it

now we shot two object in same sling shot so here the kinetic energy must be same in two objects

[tex]\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_2v_2^2[/tex]

now we have

[tex]m_1[/tex] = mass of pebble

[tex]m_2[/tex] = mass of bean

[tex]v_1 = 150 cm/s[/tex]

[tex]v_2 = 1050 cm/s[/tex]

now we have

[tex]\frac{m_2}{m_1} = \frac{v_1^2}{v_2^2}[/tex]

[tex]\frac{m_2}{m_1} = \frac{150^2}{1050^2}[/tex]

[tex]\frac{m_2}{m_1} = 0.020[/tex]

The velocities of the bean and pebble launched from the slingshot can be related to their masses under the assumption of equal elastic potential energy. However, a numerical ratio of the masses can't be provided without additional data.

Given that the procedure of launching both the pebble and the bean is the same, we assume that the same amount of elastic potential energy is converted into kinetic energy in both cases. By the formula for kinetic energy, K.E.= 1/2 mv^2, where m is the mass and v is the velocity, we can equate the kinetic energy of the two projectiles and use the known velocities to solve for the ratio of the masses. However, it's necessary to understand that without the masses or some other missing variables (such as the elasticity of the slingshot or air resistance), we cannot provide a numerical ratio of the bean to the pebble's mass. This question is mainly about the principles of energy conversion and the conservation of mechanical energy.

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Answer:

L = 4.32 m

Explanation:

Here we can use the energy conservation to find the distance that it will move

As per energy conservation we can say that the energy stored in the spring = gravitational potential energy

[tex]\frac{1}{2}kx^2 = mg(L + x)sin\theta[/tex]

[tex]\frac{1}{2}(1.35 \times 10^3)(0.10^2) = (0.180)(9.8)(L + 0.10)sin60[/tex]

now we need to solve above equation for length L

[tex]6.75 = 1.53(L + 0.10)[/tex]

[tex]L + 0.10 = 4.42[/tex]

[tex]L = 4.42 - 0.10[/tex]

[tex]L = 4.32 m[/tex]

Final answer:

The question involves calculating the distance a block moves up an incline after compressing a spring, using conservation of energy. It requires converting spring potential energy into gravitational potential energy and solving for the distance using trigonometry and principles of physics.

Explanation:

The question involves using energy considerations to determine how far up an incline a block moves before it stops. Initially, the block compresses a spring, converting mechanical energy into spring potential energy. This potential energy is then converted back into kinetic energy and finally into gravitational potential energy as the block moves up the incline. To calculate the distance, we use the conservation of energy principle, equating the spring potential energy at the beginning to the gravitational potential energy at the point where the block stops moving up the incline.

Given: mass of the block (m) = 180 g = 0.18 kg, spring constant (k) = 1.35 kN/m = 1350 N/m, compression distance (x) = 10.0 cm = 0.1 m, angle of incline (\(\theta\)) = 60.0\u00b0.

The spring potential energy (\(U_s\)) can be calculated using the formula \(U_s = \frac{1}{2}kx^2\). The gravitational potential energy (\(U_g\)) when the block has moved up the incline is given by \(U_g = mgh\), where h is the height above the initial position, which can be related to the distance along the incline (d) through trigonometry considering the angle of incline.

By setting \(U_s = U_g\) and solving for d, we find the distance d the block moves up the incline before stopping. This involves algebraic manipulation and application of trigonometric identities to relate height to distance on an incline.

Answer:

The largest voltage is 0.02 V.

(d) is correct option.

Explanation:

Given that,

Range = 20 dB

Smallest voltage = 200 mV

We need to calculate the largest voltage

Using formula of voltage gain

[tex]G_{dB}=10 log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})[/tex]

[tex]20 =10 log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})[/tex]

[tex]2=log_{10}(\dfrac{V_{out}^2}{V_{in}^{2}})[/tex]

[tex]10^2=\dfrac{V_{out}^2}{V_{in}^{2}}[/tex]

[tex]\dfrac{V_{out}}{V_{in}^}=10[/tex]

[tex]V_{in}=\dfrac{V_{out}}{10}[/tex]

[tex]V_{in}=\dfrac{200}{10}[/tex]

[tex]V_{in}=20\ mV[/tex]

[tex]V_{in}=0.02\ V[/tex]

Hence, The largest voltage is 0.02 V.

The largest voltage a display with a dynamic range of 20 dB can handle, when the smallest voltage is 200 mV, is 2.0 V.

If a display has a dynamic range of 20 dB and the smallest voltage it can handle is 200 mV, then the largest voltage it can handle can be calculated using the formula for decibels: 20 log(V1/V0), where V1 is the unknown voltage and V0 is the reference voltage, in this case, 200 mV. The dynamic range in decibels represents a ratio of the largest to smallest voltage it can handle.

To find the largest voltage (V1) the equation can be rewritten as V1 = V0 * 10^(dB/20). So, V1 = 200 * 10^(20/20) = 200 * 10 = 2000 mV or 2.0 V. Therefore, the correct answer is b. 2.0.

Answer:

6.30 L

Explanation:

P1 = P, V1 = 4.20 L, T1 = T

P2 = P/3, V2 = ?, T2 = T/2

Where, V2 be the final volume.

Use ideal gas equation

[tex]\frac{P_{1}\times V_{1}}{T_{1}} = \frac{P_{2}\times V_{2}}{T_{2}}[/tex]

[tex]V_{2} = \frac{P_{1}}{P_{2}}\times\frac{T_{2}}{T_{1}}\times V_{1}[/tex]

By substituting the values, we get

V2 = 6.30 L

answers:

m2 = 1.05 kg and m1 = 4.75 kg

Explanation:

the gravitational force is given by:

             Fg = Gm1×m2/(r^2)

9.20×10^-9 = [(6.67×10^-11)×(m1×m2)]/[(19×10-2)^2]

    (m1×m2) = 4.98 kg^2

             m1 = 4.98/m2 kg

but we given that:

m1 + m2 = 5.80 kg

4.98/m2 + m2 = 5.80

4.98 + (m2)^2 = 5.80×m2

(m2)^2 - 5.80×m2 + 4.98 = 0

by solving the quadratic equation above:

m2 = 4.75 kg or m2 = 1.05 kg

due to that from the information, m2 has a lighter mass, then m2 = 1.05 kg.

then m1 = 5.80 - 1.05 = 4.75 kg.

The average acceleration of the car is 2.45 m/s^2. If the car maintains this acceleration, it would be moving at a speed of 117.65 km/h 4.0 seconds later.

(a) To find the average acceleration, we can use the formula: average acceleration (a) = (final velocity - initial velocity) / time. Converting the velocities to m/s gives us 13.47 m/s and 22.28 m/s, respectively. Plugging in the values, we get: a = (22.28 - 13.47) m/s / 3.6 s = 2.45 m/s^2. Therefore, the average acceleration of the car is 2.45 m/s^2.

(b) Since the car is maintaining the same acceleration, we can use the kinematic equation: final velocity = initial velocity + acceleration * time. Converting the initial velocity to m/s, we have 22.28 m/s. Plugging in the values, we get: final velocity = 22.28 m/s + 2.45 m/s^2 * 4.0 s = 32.68 m/s. Converting back to km/h gives us 32.68 m/s * 3.6 km/h/m = 117.65 km/h.

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Final answer:

The average acceleration of the sports car is approximately 2.446 m/s^2, and if it maintained this acceleration for an additional 4.0 seconds, it would be moving at roughly 115.4 km/h.

Explanation:

Calculating the Average Acceleration and Future Velocity

To find the average acceleration (a) of the car, we use the formula a = (v_f - v_i) / t, where v_f is the final velocity, v_i is the initial velocity, and t is the time taken for the change in velocity. First, we need to convert the velocities from km/h to m/s by multiplying by (1000 m/km) / (3600 s/h).

v_i = 48.5 km/h = (48.5 * 1000) / 3600 m/s ≈ 13.472 m/s

v_f = 80.2 km/h = (80.2 * 1000) / 3600 m/s ≈ 22.278 m/s

Now, we can calculate the average acceleration:

a = (22.278 m/s - 13.472 m/s) / 3.6 s ≈ 2.446 m/s²

To find the velocity 4.0 seconds after the car reaches 80.2 km/h, we use the formula v = v_f + a * t:

v = 22.278 m/s + 2.446 m/s² * 4.0 s ≈ 32.062 m/s

To convert this back to km/h:

v = 32.062 m/s * (3600 s/h) / (1000 m/km) ≈ 115.4 km/h

Answer:

The value of D is [tex]7.753\times10^{-14}\ m^2/s[/tex]

Explanation:

Given that,

Temperature = 705°C

Maximum diffusion [tex]D_{0}=4.5\times10^{-5}\ m^2/s[/tex]

Activation energy [tex]Q_{d} = 164 kJ/mol[/tex]

We need to calculate the value of D

Using formula of diffusion coefficient

[tex]D=D_{0}\ exp\ (\dfrac{-Q_{d}}{RT})[/tex]

Where, D = diffusion coefficient

[tex]D_{0}[/tex] = Maximum diffusion coefficient

[tex]Q_{d}[/tex] = Activation energy

T = temperature

R = Gas constant

Put the value into the formula

[tex]D=4.5\times10^{-5}\ exp\ (\dfrac{-164\times10^{3}}{8.31\times705+273})[/tex]

[tex]D=7.753\times10^{-14}\ m^2/s[/tex]

Hence, The value of D is [tex]7.753\times10^{-14}\ m^2/s[/tex]

Final answer:

To find the diffusion coefficient D at 705°C, we use the Arrhenius equation, convert the activation energy to eV, and then calculate D using the given values of D0, Qd, Boltzmann's constant, and the temperature in Kelvin. After performing the calculations, we will obtain the required value for D at 705°C.

Explanation:

To calculate the value of D (diffusion coefficient) at 705°C for the diffusion of a species in a metal, we use the Arrhenius type equation for diffusion: D = D0 × exp(-Qd / (k × T)), where D0 is the pre-exponential factor, Qd is the activation energy for diffusion, k is Boltzmann's constant (8.617 x 10-5 eV/K), and T is the absolute temperature in Kelvin (K).

First convert the temperature from Celsius to Kelvin: T = 705°C + 273.15 = 978.15 K.

Then plug in the given values: D0 = 4.5 x 10-5 m2/s and Qd = 164 kJ/mol (which is 164000 J/mol).

Using these values, calculate D at 705°C:

D = 4.5 x 10-5 m2/s × exp(-164000 J/mol / (8.617 x 10-5 eV/K × 978.15 K))

Since 1 eV = 1.602 x 10-19 J, we can convert the activation energy to eV by dividing by this conversion factor:

Qd in eV = 164000 J/mol / (1.602 x 10-19 J/eV) = 1023629.84 eV/mol

Now insert the activation energy in eV into the equation:

D = 4.5 x 10-5 m2/s × exp(-1023629.84 eV/mol / (8.617 x 10-5 eV/K × 978.15 K))

After performing the calculations, we will obtain the required value for D at 705°C.

Answer:

Liquid to a gas

Explanation:

When the molecules in a body move with increased speed, it's possible that the body will change from liquid to gas. The speed is increasing means they have more kinetic energy. The molecules of gas are very far apart from each other. They have much space between them so that they can move freely.

So, when the molecules move with increased speed, the body will change from liquid to gas. Hence, the correct option is (d) " liquid to gas".

Answer:

[tex]0.0000584637\ m^{3}\\[/tex]

Explanation:

Hello

Density is a measure of mass per unit of volume

[tex]d=\frac{m}{v} \\\\\\[/tex]

and the weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity

[tex]W=mg[/tex]

let

[tex]d=13.6*10^{3} \frac{kg}{m^{3} }  \\ W=7.8 N\\W=mg\\m=\frac{W}{g} \\m=\frac{7.8 N}{9.81 \frac{m}{s^{2} } }\\m=0.8 kg\\\\d=\frac{m}{v} \\v=\frac{m}{d} \\v=\frac{0.8 kg}{13.6*10^{3} \frac{kg}{m^{3} }}\\Volume=0.0000584637\ m^{3}[/tex]

the volumen of the sample is 0.0000584637 m3

have a great day

Answer:

the motorcycle travels 42.4 meters until it stops.

Explanation:

Vi= 18 m/s

Vf= 0 m/s

t= 4.7 sec

Vf= Vi - a*t

deceleration:

a= Vi/t

a= 18m/s  / 4.7 sec =>  a=-3.82 m/s²

x= Vi*t - (a * t²)/2

x= 42.4m

The problem involves applying principles of soil mechanics to a direct shear test on a sand specimen. The calculations involve converting the specimen's physical dimensions to an area and applying formulas related to shear stress and friction angle.

The problem given is an application of soil mechanics principles in civil engineering. The situation is a direct shear test on a dry sand specimen. To answer this, we need to understand principles related to shearing force, shearing stress and their relationship with angle of internal friction (φ‘) and normal stress.

Shear stress can be calculated using the formula τ = F/A, where F is the force and A is the area over which the force is distributed. The area can be calculated based on the specimen's dimensions using A = πr² (where r is the radius of the specimen, which is half of the diameter). Given the normal stress and the shear stress, the angle of friction φ’ can be calculated using the formula tan(φ‘) = τ / σ, where σ is the normal stress.

To calculate the shear force required to cause failure under a different normal stress, we use the above formula in reverse, solving for τ (which represents the shear stress under the new normal stress), then multiply by the area to obtain the force. In other words, F = τA.

Please note that this is a simplified calculation ignoring potential complexities of real-world soil behavior.

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The angle of friction φ' is approximately 30.50°. For a normal stress of 200 kN/m², the required shear force to cause failure is approximately 367.97 N.

To determine the angle of friction (φ') and the required shear force at a different normal stress, we will conduct the following calculations:

a. Determining the angle of friction, φ’

Determine the cross-sectional area (A) of the specimen:

Diameter (d) = 63 mm = 0.063 m

Area (A) = π/4 × d² = (3.1416/4) × (0.063²) ≈ 0.003117 m²

Calculate the shear stress (τ) at failure:

τ = Shear Force (F) / Area (A)

F = 276 N

τ = 276 N / 0.003117 m² ≈ 88555.19 N/m² = 88.56 kN/m²

Calculate the angle of friction (φ'):

Normal stress (σ) = 150 kN/m²

The relationship between shear stress and normal stress in terms of the angle of friction (φ') is given by:

τ = σ × tan(φ')

88.56 kN/m² = 150 kN/m² × tan(φ')

tan(φ') = 88.56 / 150 ≈ 0.5904

φ' = atan(0.5904) ≈ 30.50°

b. For a normal stress of 200 kN/m², Shear Force required to cause failure:

Calculate the shear stress (τ) using the angle of friction:

σ = 200 kN/m²

τ = σ × tan(φ')

τ = 200 kN/m² × tan(30.50°) ≈ 118.08 kN/m²

Determine the required shear force (F):

τ = F / A

F = τ × A

F = 118.08 kN/m² × 0.003117 m² ≈ 367.97 N

Conclusion:

The angle of friction (φ') is approximately 30.50°.

The shear force required to cause failure at a normal stress of 200 kN/m² is approximately 367.97 N.

Answer:

8.6 kg

Explanation:

According to the question,

1 Kg = 2.21 pound

2.21 pound = 1 kg

1 pound = 1 / 2.21 kg

19 pound = 19 / 2.21 kg = 8.59 kg

By rounding to nearest tenth, it is equal to 8.6 kg.

Answer:

D). [tex]327 ^0 C[/tex]

Explanation:

As we know that temperature scale is linear so we will have

[tex]\frac{^0C - 0}{100 - 0} = \frac{K - 273}{373 - 273}[/tex]

now we have

[tex]\frac{^0 C - 0}{100} = \frac{K - 273}{100}[/tex]

so the relation between two scales is given as

[tex]^0 C = K - 273[/tex]

now we know that in kelvin scale the absolute temperature is 600 K

so now we have

[tex]T = 600 - 273 = 327 ^0 C[/tex]

so correct answer is

D). [tex]327 ^0 C[/tex]

Answer:

- 7.56 C

Explanation:

E = 105 N/C downwards

m = 81 kg

Let the charge on the man is q.

To lose te contact with the ground, the electrostatic foece should be balanced by the weight of the person.

The charge should be negative in nature so that the direction of electrostatic force is upwards and weight is downwards.

q E = m g

q = (81 x 9.8) / 105 = 7.56 C

Explanation:

a)

The circumference of the path is:

C = 2πr

C = 2π (0.235 m)

C = 1.48 m

Velocity = displacement / time

v = 1.48 m / 1.95 s

v = 0.757 m/s

b)

1 rev / 1.95 s = 0.513 Hz

c)

1 rev / 1.95 s × (2π rad / rev) = 3.22 rad/s

(a) The translational speed of the object is 0.76 m/s.

(b) The frequency of the object's motion is 0.51 Hz.

(c) The angular speed of the object is 3.22 rad/s.

The angular speed of the object is calculated as follows;

ω = 1 rev/ 1.95 s = 0.51 rev/s

ω =  0.51 rev/s x 2π rad

ω = 3.22 rad/s

The frequency of the object's motion is determined from the angular speed as shown below;

ω = 2πf

f = ω/2π

f = (3.22)/2π

f = 0.51 Hz

The translational speed of the object is calculated as follows;

v = ωr

v = 3.22 x 0.235

v = 0.76 m/s

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B = 17.5mT.

A solenoid is a coil formed by a wire (usually copper) wound into a cylindrical spiral shape capable of creating a magnetic field that is extremely uniform and intense inside, and very weak outside.

To calculate the magnetic field generated inside the solenoid through which a current flows is given by the equation:

B = μ₀nI

Where  μ₀ is the constant of magnetic proportionality of the vacuum (4π x 10⁻⁷T.m/A), n is the relation between the number of turns of wire and its length given by N/L and I is the current flowing through the solenoid.

Given a long solenoid of length 0.67m, 1700.00 turns of wire and a current flowing through the wire of 5.50A. Calculate the magnetic field inside the solenoid.

B = (4π x 10⁻⁷T.m/A)(1700turns/0.67m)(5.50A)

B = 0.0175T

B = 17.5mT

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

[tex]U=\dfrac{1}{2}kx^2[/tex]...(I)

Using gravitational potential energy

[tex]U' =mgh[/tex]....(II)

Using energy of conservation

[tex]E_{i}=E_{f}[/tex]

[tex]U_{i}+U'_{i}=U_{f}+U'_{f}[/tex]

[tex]\dfrac{1}{2}kx^2+0=0+mgh[/tex]

[tex]h=\dfrac{kx^2}{2mg}[/tex]

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

[tex]h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}[/tex]

[tex]h=11.653\ m[/tex]

Hence, The maximum height above the point of release is 11.653 m.

Answer:

Answer to the Question:

Explanation:

In this case, the total work done by the cable is zero, since in the aforementioned problem, the work depends only on the starting and ending point, these two being equal. Keeping its gravitational potential energy equal.

Answer:

The total work done is zero.

Explanation:

Given;

mass of the crate, m = 70.0 kg

height above ground through which the crate is lifted, h = 12.0 m

The only work associated in lifting and lowering this crate is gravitational potential energy.

Potential Energy during lifting = - mgh

                                                  = - 70 x 9.8 x 12

                                                  = - 8232 J

Potential Energy during lowering = mgh

                                                       =  70 x 9.8 x 12

                                                       =  8232 J  

Total total work done by the force = - 8232 J + 8232 J  = 0

Therefore, the total work done is zero.

Answer:

The speed of the skier after moving 100 m up the slope are of V= 25.23 m/s.

Explanation:

F= 280 N

m= 80 kg

α= 12º

μ= 0.15

d= 100m

g= 9,8 m/s²

N= m*g*sin(α)

N= 163 Newtons

Fr= μ * N

Fr= 24.45 Newtons

∑F= m*a

a= (280N - 24.5N) / 80kg

a= 3.19 m/s²

d= a * t² / 2

t=√(2*d/a)

t= 7.91 sec

V= a* t

V= 3.19 m/s² * 7.91 s

V= 25.23 m/s

Answer:

4.7 x 10⁵ J

Explanation:

m = mass of the car = 925 kg

d = distance traveled parallel to incline = 207 m

θ = angle of the slope = 14.5 deg

Force applied to move the car up is given as

F = mg Sinθ                                               eq-1

minimum work needed is given as

W = F d

using eq-1

W =  mgd Sinθ

inserting the values

W = (925)(9.8)(207) Sin14.5

W = 4.7 x 10⁵ J

Answer:

Heat energy required = 687.96 kJ

Explanation:

Heat energy required, H = mCΔT.

Mass of cooking oil, m = 2.34 kg = 2340 g

Specific heat of cooking oil, C = 1.75 J/(g⋅°C)

Initial temperature = 23 °C

Final temperature = 191 °C

Change in temperature, ΔT = 191 - 23 = 168 °C

Substituting values

            H = mCΔT

            H = 2340 x 1.75 x 168 = 687960 J = 687.96 kJ

Heat energy required = 687.96 kJ

Answer:

The electric field at x = 0.500 m is 0.02 N/C.

Explanation:

Given that,

Point charge at the origin[tex]q_{1} = 7.00\ nC[/tex]

Second point charge[tex]q_{2}=-250\ nC[/tex]  at x = +0.800 m

We calculate the electric field at x = 0.500 m

Using formula of electric field

[tex]E=\dfrac{kq}{r^2}[/tex]

The electric field at x = 0.500 m

[tex]E=\dfrac{k\times7\times10^{-9}}{(5)^2}+\dfrac{k\times(-2.50)\times10^{-9}}{(3)^2}[/tex]

[tex]E=9\times10^{9}(\dfrac{7\times10^{-9}}{25}-\dfrac{2.50\times10^{-9}}{9})[/tex]

[tex]E = 0.02\ N/C[/tex]

Hence, The electric field at x = 0.500 m is 0.02 N/C.

Answer: The electric field at x = 0.5 m is equal to 1.96 N/C, and the direction is in the postive x-axis (to the rigth)

Explanation:

I will use the notations (x, y, z)

the first particle is located at the point (0m, 0m, 0m) and has a charge q1 = 7.00 nC

the second particle is located at the point (0.8m, 0m, 0m) and has a charge q2 =  -2.50 nC

Now, we want to find the electric field at the point (0.5m, 0m, 0m)

First, we can see that we only work on the x-axis, so we can think about this problem as one-dimensional.

First, the electric field done by a charge located in the point x0 is equal to:

E(x) = Kc*q/(x - x0)^2

where Kc is a constant, and it is Kc = 8.9*10^9 N*m^2/C^2

then, the total magnetic field will be equal to the addition of the magnetic fields generated by the two charges:

E(0.5m) = Kc*q1/0.5m^2 + Kc*q2/(0.5m - 0.8m)

E(0.5m) = Kc*(7.0nC/(0.5m)^2 - 2.5nC/(0.3m)^2)

E(0.5m) = Kc*(0.22nC/m^2)

now, remember that Kc is in coulombs, so we must change the units from nC to C

where 1nC = 1*10^-9 C

E(0.5m) = (8.9*10^9 N*m^2/C^2)*(0.22x10^-9C/m^2) = 1.96 N/C

the fact that is positive means that it points in the positve side of the x-axis.

Answer:

340 kcal

Explanation:

Energy released by 2 g of Uranium is same as the energy released by 2 tons of coal.

Energy given by 2 tons of coal = 6.8 x 10^8 kcal

Energy given by 2 x 10^6 g of coal = 6.8 x 10^8 kcal

Energy released by 1 g of coal = (6.8 x 10^8) / (2 x 10^6) = 340 kcal

To find the energy released by burning 1.0 g of coal, we need to determine the proportion of energy released by burning coal compared to uranium-235. By converting two tons of coal to grams and using the information that 2.0 g of uranium-235 releases 6.8 × 108 kcal, we calculate that approximately 374.6 kcal is released when 1.0 g of coal is burned.

To calculate the amount of energy released when 1.0 g of coal is burned, we need to compare it with the energy released by uranium-235. The article states that the fission of 2.0 g of uranium-235 releases 6.8 × 108 kcal, which is the equivalent of burning two tons (4,000 lb) of coal. Therefore, the energy released by burning 1.0 g of coal can be found using a simple proportion.

First, let’s convert two tons of coal to grams:

Now, we have the total grams of coal that release the same amount of energy as 2.0 g of uranium-235:

4,000 lb × 453.592 g/lb = 1,814,368 g of coal

Next, we can set up the proportion to solve for the energy released by 1.0 g of coal:

(6.8 × 108 kcal) / (1,814,368 g) = x kcal / (1 g)

By cross-multiplying and solving for x, we find that:

x = (6.8 × 108 kcal) / (1,814,368 g)

x ≈ 374.6 kcal/g

Therefore, about 374.6 kcal of energy is released when 1.0 g of coal is burned.

Answer:

1.84 m from the initial point (3.16 m from the ceiling)

Explanation:

According to the law of conservation of energy, the initial kinetic energy of the ball will be converted into gravitational potential energy at the point of maximum height.

Therefore, we can write:

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

where

m = 2 kg is the mass of the ball

v = 6 m/s is the initial speed of the ball

g = 9.8 m/s^2 is the acceleration due to gravity

[tex]\Delta h[/tex] is the change in height of the ball

Solving for [tex]\Delta h[/tex],

[tex]\Delta h = \frac{v^2}{2g}=\frac{6^2}{2(9.8)}=1.84 m[/tex]

So, the ball raises 1.84 compared to its initial height.

Therefore:

- if we take the initial position of the ball as reference point, its maximum height is at 1.84 m

- if we take the ceiling as reference point, the maximum height of the ball will be

5 m - 1.84 m = 3.16 m from the ceiling

Answer:

Angle between the magnetic field and the wire's velocity is 15.66 degrees.

Explanation:

It is given that,

Potential difference or emf, V = 35 mV = 0.035 V

Length of wire, l = 12 cm= 0.12 m

Magnetic field, B = 0.27 T

Speed, v = 4 m/s

We need to find the angle between the magnetic field and the wire's velocity. We know that emf is given by :

[tex]\epsilon=Blv\ sin\theta[/tex]

[tex]sin\theta=\dfrac{\epsilon}{Blv}[/tex]

[tex]sin\theta=\dfrac{0.035\ V}{0.27\ T\times 0.12\ m\times 4\ m/s}[/tex]

[tex]sin\theta=0.25[/tex]

[tex]\theta=15.66^{\circ}[/tex]

So, the angle between the magnetic field and the wire's velocity is 15.66 degrees.

The angle between the wire's velocity and the magnetic field, when they are perpendicular to each other, is 90 degrees. This impacts the force exerted on the wire in the magnetic field.

The question pertains to the interaction of a moving conductor wire in a magnetic field - a fundamental concept in electromagnetism. From the question, the velocity of the wire is perpendicular to the magnetic field. This is the key because the magnetic force on a moving charge within a magnetic field depends on the angle between the charge's velocity and the direction of the magnetic field.

According to the fundamentals of physics, when velocity is perpendicular to the magnetic field, the angle between them is 90 degrees. The force exerted on the wire due to the magnetic field is then given by F = qvBsinθ, where q is the charge, v is the velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. With θ being 90 degrees, the sin(90°) equals 1, and this simplifies the calculation. So the angle developed between the magnetic field and the wire's velocity is 90 degrees in this case.

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Answer:

a) 45.87 m

b) 2.08 N

Explanation:

Mass of projectile=0.85 kg=m

Velocity of projectile=30 m/s=u = initial velocity

Final velocity =0

g= acceleration due to gravity=9.81 m/s²

h=maximum height of the projectile

a) In this case loss of Kinetic energy (K.E.) = loss in potential energy (P.E.)

ΔK.E.=ΔP.E.

[tex]\frac{1}{2}mu^2-\frac{1}{2}mv^2=mgh[/tex]

[tex]\frac{1}{2}m\times u^2=mgh\\\Rightarrow h=\frac {u^2}{2\times g}\\\Rightarrow h=\frac {30^2}{2\times 9.81}\\\therefore h=45.87\ m[/tex]

b) h'=height of the projectile=36.7 m

F=Average force due to air resistance

There will be a loss of P.E. due to air resistance

ΔP.E.=mg(h-h')

F×h'=mg(h-h')

F×36.7=0.85×9.81(45.87-36.7)

[tex]F=\frac{0.85\times 9.81(45.87-36.7)}{36.7}[/tex]

∴ F=2.08 Newton

Answer:

a. 76.92 MW

b. 26.92 MW

Explanation:

η = Efficiency of the power plant = 0.35

Q₂ = rate at which heat rejected to surrounding = 50 MW

Q₁ = Rate of Input heat = ?

Efficiency of the power plant is given as

[tex]\eta =1-\frac{Q_{2}}{Q_{1}}[/tex]

[tex]0.35 =1-\frac{50}{Q_{1}}[/tex]

Q₁ = 76.92 MW

Net Rate of work is given as

Q = Q₁ - Q₂

Q = 76.92 - 50

Q = 26.92 MW

Answer:

convection C

Explanation:

Answer:

C. convection

Explanation:

It is given that,

Let q₁ and q₂ are two small positively charged spheres such that,

[tex]q_1+q_2=5\times 10^{-5}\ C[/tex].............(1)

Force of repulsion between the spheres, F = 1 N

Distance between spheres, d = 2 m

We need to find the charge on the sphere with the smaller charge. The force is given by :

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

[tex]q_1q_2=\dfrac{F.d^2}{k}[/tex]

[tex]q_1q_2=\dfrac{1\ N\times (2\ m)^2}{9\times 10^9}[/tex]

[tex]q_1q_2=4.45\times 10^{-10}\ C[/tex]............(2)

On solving the system of equation (1) and (2) using graph we get,

[tex]q_1=0.0000384\ C=38.4\ \mu C[/tex]

[tex]q_2=0.0000116\ C=11.6\ \mu C[/tex]

So, the charge on the smaller sphere is 11.6 micro coulombs. Hence, this is the required solution.

Combined charge = 5.0 x 10-5 C
Distance between spheres = 2.0 m
Force = 1.0 N

Formula:
Coulomb's Law: F = k * (q1 * q2) / r2

Calculations:
q1 + q2 = 5.0 x 10-5 C (Given)
q1 = q2 - x (Assuming q2 is larger charge)
Substitute and solve for x

Answer:
The charge on the sphere with the smaller charge is 2.5 x 10-5 C or 25 micro-coulombs.