A rod of 2.0-m length and a square (2.0 mm × 2.0 mm) cross section is made of a material with a resistivity of 6.0 × 10−8 Ω ⋅ m. If a potential difference of 0.50 V is placed across the ends of the rod, at what rate is heat generated in the rod?

Answer:

Induced emf through a loop of wire is 3.5 V.

Explanation:

It is given that,

Initial magnetic flux, [tex]\phi_i=1.7\ Wb[/tex]

Final magnetic flux, [tex]\phi_f=0.3\ Wb[/tex]

The magnetic flux through a loop of wire decreases in a time of 0.4 s, t = 0.4 s

We need to find the average value of the induced emf. It is equivalent to the rate of change of magnetic flux i.e.

[tex]\epsilon=-\dfrac{\phi_f-\phi_i}{t}[/tex]

[tex]\epsilon=-\dfrac{0.3\ Wb-1.7\ Wb}{0.4\ s}[/tex]

[tex]\epsilon=3.5\ V[/tex]

So, the value of the induced emf through a loop of wire is 3.5 V.

Answer:

[tex]d_{o}[/tex] = 154 cm

[tex]d_{i}[/tex] = 216.6 cm

The image is real

Explanation:

[tex]h_{o}[/tex] = height of the object = 3.20 cm

[tex]h_{i}[/tex] = height of the image = 4.50 cm

f = focal length of the converging lens = 90 cm

[tex]d_{o}[/tex] = object distance from the lens = ?

[tex]d_{i}[/tex] = image distance from the lens = ?

using the equation for magnification

[tex]\frac{h_{i}}{h_{o}}= \frac{ d_{i}}{d_{o}}[/tex]

[tex]\frac{4.50}{3.20}= \frac{d_{i}}{d_{o}}[/tex]

[tex]d_{i}[/tex] = 1.40625 [tex]d_{o}[/tex]                        eq-1

using the lens equation

[tex]\frac{1}{d_{i}} + \frac{1}{d_{o}} = \frac{1}{f}[/tex]

using eq-1

[tex]\frac{1}{( 1.40625)d_{o}} + \frac{1}{d_{o}} = \frac{1}{90}[/tex]

[tex]d_{o}[/tex] = 154 cm

Using eq-1

[tex]d_{i}[/tex] = 1.40625 [tex]d_{o}[/tex]  

[tex]d_{i}[/tex] = 1.40625 (154)

[tex]d_{i}[/tex] = 216.6 cm

The image is real

Answer:

25.2 cm

Explanation:

K = 1050 N/m

m = 1.95 kg

h = 1.75 m

By the conservation of energy, the potential energy of the block is converted into the potential energy stored in the spring

m g h = 1/2 x k x y^2

Where, y be the distance by which the spring is compressed.

1.95 x 9.8 x 1.75 = 1/2 x 1050 x y^2

33.44 = 525 x y^2

y = 0.252 m

y = 25.2 cm

The spring will compress by approximately 25.3 cm when a 1.95 kg block is dropped from a height of 1.75 m due to the conversion of gravitational potential energy into elastic potential energy.

The student's question is regarding the compression of a spring due to the gravitational potential energy of a mass that was dropped onto it. This problem can be solved by using the conservation of energy principle, where the potential energy of the block (due to its height) is converted into the spring's elastic potential energy when the spring is compressed.

To find the compression distance (x), we use the following steps:

First we calculate the gravitational potential energy:

PE = mgh = (1.95 kg)(9.81 m/s2)(1.75 m) = 33.637875 J

Then we set this equal to the spring's potential energy and solve for x:

33.637875 J = 1/2 (1050 N/m) x2

x2 = (2 * 33.637875 J) / (1050 N/m)

x2 = 0.064067619

x = 0.253 J

Therefore, the spring compresses by approximately 0.253 meters or 25.3 centimeters.

Answer:

70 Hz

Explanation:

The Doppler equation describes how sound frequency depends on relative velocities:

fr = fs (c + vr)/(c + vs),

where fr is the frequency heard by the receiver,

fs is the frequency emitted at the source,

c is the speed of sound,

vr is the velocity of the receiver,

and vs is the velocity of the source.

Note: vr is positive if the receiver is moving towards the source, negative if away.  

Conversely, vs is positive if the receiver is moving away from the source, and negative if towards.

When the car is approaching you:

fr = 76 Hz

vr = 0 m/s

When the car is moving away from you:

fr = 65 Hz

vr = 0 m/s

c, vs, and fs are constant.

We can write two equations:

76 = fs c / (c − vs)

65 = fs c / (c + vs)

If we divide the two equations:

76/65 = [fs c / (c − vs)] / [fs c / (c + vs)]

76/65 = [fs c / (c − vs)] × [(c + vs) / (fs c)]

76/65 = (c + vs) / (c − vs)

76 (c − vs) = 65 (c + vs)

76c − 76vs = 65c + 65vs

11c = 141vs

vs = 11/141 c

Substitute into either equation to find fs.

65 = fs c / (c + 11/141 c)

65 = fs c / (152/141 c)

65 = 141/152 fs

fs = 70 Hz

The question involves the Doppler effect in sound waves. To find the original frequency of the car's horn, the mean of the frequencies heard when the car was approaching and receding is calculated. This gives a result of 70.5 Hz.

This question involves the Doppler effect, which is a change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. In this case, the source of the sound is the car's horn.

To calculate the actual frequency of the car's horn, you need to take the frequency you heard when the car was approaching (76 Hz) and when it was leaving (65 Hz) and find the mean of these two values. So, the frequency of the car horn is ((76+65)/2) = 70.5 Hz.

This calculation assumes that your movement is minimal compared to that of the car. As such, most of the perceived frequency change is due to the motion of the car, not the observer. Therefore, the actual frequency of the horn is somewhat between the heard frequencies when the car was approaching and receding. This happens because of the change in relative velocity between the source of sound (car) and the observer when the car goes by.

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Answers:  

a) How high does the rock get?=1.933m

b)How far downrange from the pedestal does the rock land?=21.25m

Explanation:

This situation is a good example of projectile motion or parabolic motion, in which the travel of the rock has two components: x-component and y-component. Being their main equations as follows:  

x-component:  

[tex]x=V_{o}cos\theta t[/tex]   (1)  

[tex]V_{x}=V_{o}cos\theta[/tex]   (2)  

Where:  

[tex]V_{o}=18m/s[/tex] is the rock's initial speed  

[tex]\theta=20\°[/tex] is the angle

[tex]t[/tex] is the time since the rock is propelled until it hits the ground  

y-component:  

[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex]   (3)  

[tex]V_{y}=V_{o}sin\theta-gt[/tex]   (4)  

Where:  

[tex]y_{o}=10m[/tex]  is the initial height of the rock

[tex]y=0[/tex]  is the final height of the rock (when it finally hits the ground)  

[tex]g=9.8m/s^{2}[/tex]  is the acceleration due gravity

Knowing this, let's begin with the anwers:

Here we are talking about the maximun height [tex]y_{max}[/tex] the rock has in its parabolic motion. This is fulfilled when [tex]V_{y}=0[/tex].

Rewritting (4) with this condition:

[tex]0=V_{o}sin\theta-gt[/tex]   (5)  

Isolating [tex]t[/tex]:

[tex]t=\frac{V_{o}sin\theta}{g}[/tex]  (6)  

Substituting (6) in (3):

[tex]y_{max}=y_{o}+V_{o}sin\theta(\frac{V_{o}sin\theta}{g})-\frac{1}{2}g(\frac{V_{o}sin\theta}{g})^{2}[/tex]   (7)  

[tex]y_{max}=\frac{V_{o}^{2}sin^{2}\theta}{2g}[/tex]   (8)  

Solving:

[tex]y_{max}=\frac{(18m/s)^{2}sin^{2}(20\°)}{2(9.8m/s^{2})}[/tex]   (9)  

Then:

[tex]y_{max}=1.933m[/tex]   (10) This is the maximum height the rock has.

Here we are talking about the maximun horizontal distance [tex]x_{max}[/tex] the rock has in its parabolic motion (this is fulfilled when [tex]y=0[/tex]):

[tex]0=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (11)  

Isolating [tex]t[/tex] from (11):

[tex]t=\frac{2V_{o}sin\theta}{g}[/tex] (12)  

Substituting (12) in (1):

[tex]x_{max}=V_{o}cos\theta (\frac{2V_{o}sin\theta}{g})[/tex]   (13)

[tex]x_{max}=\frac{V_{o}^{2}(2cos\theta sin\theta)}{g}[/tex]   (14)

Knowing [tex]sin(2\theta)=2cos\theta sin\theta[/tex]:

[tex]x_{max}=\frac{V_{o}^{2}sin2\theta}{g}[/tex]   (15)

Solving:

[tex]x_{max}=\frac{(18m/s)^{2}sin2(20)}{9.8m/s^{2}}[/tex]   (16)

Finally:

[tex]x_{max}=21.25m[/tex]   (17)

The mass of water that needs to evaporate from the skin to reduce a body temperature by 0.45°C is calculated using the body's specific heat capacity, body mass and the latent heat of vaporization of water. First, we find the energy required to cool the body by the temperature change and then find the mass of water that would embody that amount of energy.

The question is asking for the calculation of the mass of water that should evaporate from the skin to reduce the body temperature by 0.45°C. To find this, we need to first understand that evaporation is a main method for body cooling, and it involves a considerable amount of energy being taken from the skin as water changes into vapor.

The energy for evaporating water is explained by the equation Qv = m * Lv where Qv stands for heat energy, m represents mass and Lv is the latent heat of vaporization of water. Given that the specific heat capacity of the body is 4.0 J/g °C and the body mass is 95 kg, the amount of energy required to cool the body by 0.45°C is calculated by multiplying these values (body mass in grams * temperature change in °C * specific heat capacity).

After calculating this energy, we get how much heat needs to be removed from the body to achieve the desired temperature reduction. Lastly, to find the mass of water to be evaporated, we use the equation Qv = m * Lv again but rearrange it as m = Qv / Lv (as Lv = 2256 kJ/kg). This gives us the amount of water that needs to evaporate from the body to reduce the temperature by 0.45°C.

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Answer:

So police car will overtake the speeder after 5.64 s

Explanation:

Initially the distance between police car and the speeder when police car is about to accelerate

[tex]d = (v_1 - v_2)t[/tex]

[tex]v_1 = 110 km/h = 110 \times \frac{1000}{3600}m/s = 30.55 m/s[/tex]

[tex]v_2 = 95 km/h = 95 \times \frac{1000}{3600}m/s = 26.39 m/s[/tex]

[tex]d = (30.55-26.39)(2) = 8.32 m[/tex]

now we have

now velocity of police with respect to speeder is given as

[tex]v_r = v_2 - v_1 = 26.39 - 30.55 = -4.17 m/s[/tex]

relative acceleration of police car with respect to speeder

[tex]a_r = a = 2 m/s^2[/tex]

now the time taken to cover the distance between police car and speeder is given as

[tex]d = v_i t + \frac{1}{2}at^2[/tex]

[tex]8.32 = -4.17 t + \frac{1}{2}(2)(t^2)[/tex]

[tex]t^2 -4.17 t - 8.32 = 0[/tex]

[tex]t = 5.64 s[/tex]

Answer: t = 7.61s

Explanation: The initial speed of the police's car is Vp = 95 km/h.

The initial speed of the car is Vc = 110km/h

The acceleration of the police's car is 2m/s^2

Now, we should write this the quantities in the same units, so lets write the velocities in meters per second.

1 kilometers has 1000 meters, and one hour has 3600 seconds, so we have that:

Vp = 95*1000/3600 m/s = 26.39m/s

Vc = 110*1000/3600 m/s = 30.56m/s

now, after the police car starts to accelerate, the velocity equation will be now.

The positions of the cars are:

P(t) = (a/2)*t^2 + v0*t + p0

Where a is the acceleration, v0 is the initial velocity, and p0 is the initial position.

We know that the difference in the velocity of the cars is two seconds, so after those the speeding car is:

(30.56m/s - 26.39m/s)*2s = 8.34m/s

Now we can write the position equations as:

Pp = 1m/s*t^2 + 26.39m/s*t + 0

Here i assume that the initial position of the car is at the 0 units in one axis.

Pc = 30.56m/s*t + 8.34m/s

Now we want to find the time at wich both positions are the same, and after that time the police car will go ahead of the speeding car.

1m/s*t^2 + 26.39m/s*t  = 30.56m/s*t + 8.34m/s

t^2 + (26.39 - 30.56)*t - 8.34 = 0

t^2 - 4.15*t - 8.34 = 0

Now we need to solve this quadratic equation:

t = (4.15 +/- √(4.15^2 - 4*1*(-8.34))/2 = (4.15 +/- 7.11)/2

From here we have two solutions, one positive and one negative, and we need to take the positive one:

t = (4.15 + 7.11)/2s = 11.26/2 s= 5.61s

And remember that the police car started accelerating two secnods after that the speeding car passed it, so the actual time is:

t = 5.61s + 2s = 7.61s

Answer:

7.07 x 10⁸ N/m²

Explanation:

F = Force applied to the wire = 68 N

d = diameter of the wire = 0.7 mm = 0.7 x 10⁻³ m

r = radius of the wire = (0.5) d = (0.5) (0.7 x 10⁻³) = 0.35 x 10⁻³ m

Area of cross-section of wire is given as

A = (0.25) πr²

A = (0.25) (3.14) (0.35 x 10⁻³)²

A = 9.61625 x 10⁻⁸ m²

Tensile stress is given as

[tex]P = \frac{F}{A}[/tex]

[tex]P = \frac{68}{9.61625\times 10^{-8}}[/tex]

P = 7.07 x 10⁸ N/m²

Given:

Applied Force on wire = 68 N

Diameter of wire, d = 0.7 mm = [tex]0.7\times 10^{-3}[/tex] m

Radius of wire, r = [tex]\frac{d}{2}[/tex] = 0.35 mm = [tex]0.35\times 10^{-3}[/tex] m

Formula used:

Stress =  [tex]\frac{Applied Force}{cross-sectional area}[/tex]

Explanation:

Cross-sectional area, A = [tex]\pi r^{2}[/tex]  =  [tex]\pi (0.35\times 10^{-3})^{2}[/tex]

A = [tex]3.84\times 10^{-7} m^{2}[/tex]

Using the formula for stress:

Stress =  [tex]\frac{68}{3.84\times 10^{-7}}[/tex] =  [tex]1.76\times 10^{8}[/tex]

Answer:

232 J

Explanation:

Heat gained = mass × specific heat × increase in temperature

q = m C (T − T₀)

Given m = 75.1 g, C = 0.14 J/g/°C, T = 53.8°C, and T₀ = 31.7°C:

q = (75.1 g) (0.14 J/g/°C) (53.8°C − 31.7°C)

q = 232 J

Answer:

[tex]Q=232.36J[/tex]

Explanation:

The heat capacity (C) of a physical system depends on the amount of substance of that system. For a system formed by a single homogeneous substance, it is defined as:

[tex]C=mc(1)[/tex]

Here m is the mass of the system and c is the specific heat capacity.

The heat capacity is defined as the ratio between the heat absorbed by the system and the resulting temperature change:

[tex]C=\frac{Q}{\Delta T}(2)[/tex]

We equal (1) and (2) and solve for Q:

[tex]\frac{Q}{\Delta T}=mc\\Q=mc\Delta T\\Q=mc(T_f-T_i)\\Q=75.1g(0.14\frac{J}{g^\circ C})(53.8^\circ C-31.7^\circ C)\\Q=232.36J[/tex]

Answer:

i = 323 A

Explanation:

Initial flux due to magnetic field from the coil is given as

[tex]\phi = NB.A[/tex]

here we will have

[tex]N = 190 [/tex]

[tex]B = 1.60 T[/tex]

[tex]A = 0.340 m^2[/tex]

now the flux is given as

[tex]\phi_1 = (190)(1.60)(0.340) = 103.36 T m^2[/tex]

finally current in the electromagnet changed to zero

so final flux in the coil is zero

[tex]\phi_2 = 0[/tex]

now we know that rate of change in flux will induce EMF in the coil

so we will have

[tex]EMF = \frac{\phi_1 - \phi_2}{\Delta t}[/tex]

[tex]EMF = \frac{103.36 - 0}{20 \times 10^{-3}}[/tex]

[tex]EMF = 5168 Volts[/tex]

now induced current is given as

[tex]i = \frac{EMF}{R}[/tex]

[tex]i = \frac{5168}{16} = 323 A[/tex]

Answer:

acceleration in the new orbit is 8 time of acceleration of planet in old orbit

[tex]a_{new} = 8a.[/tex]

Explanation:

given data:

radius of orbit = R

Speed pf planet = v

new radius = 0.5R

new speed = 2v

we know that acce;ration is given as

[tex]a = \frac{v^{2}}{R},[/tex]

[tex]a_{new} =\frac{(2v)^{2}}{0.5R},[/tex]

           [tex]= \frac{4v^{2}}{0.5R}[/tex]

          [tex]= \frac{8 v^{2}}{R}[/tex]

[tex]a_{new} = 8a.[/tex]

acceleration in the new orbit is 8 time of acceleration of planet in old orbit

The magnitude of the probe’s acceleration in the new orbit is 8 times the magnitude of the probe’s acceleration in the old orbit.

The acceleration acted on the body moving in a closed circular path towards the center of the curve is known as centripetal acceleration. Due to centripetal acceleration, the body is able to move in a closed circular path.

[tex]a_C=\frac{V^{2} }{R}[/tex]

For old orbit

the radius of orbit = R

Speed pf planet = v

[tex]a_C=\frac{V^{2} }{R}[/tex]

For new orbits

new radius = 0.5R

new speed = 2v

[tex]a_C=\frac{(2V)^{2} }{0.5R}[/tex]³

[tex]a_C=\frac{8V^{2} }{R}[/tex]

Hence,

[tex]a_C_{new}=8a_C__{old}[/tex]

The magnitude of the probe’s acceleration in the new orbit is 8 times the magnitude of the probe’s acceleration in the old orbit.

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Answer:

Mass of the climber = 69.38 kg

Explanation:

Change in length

        [tex]\Delta L=\frac{PL}{AE}[/tex]

Load, P = m x 9.81 = 9.81m

Young's modulus, Y = 0.37 x 10¹⁰ N/m²

Area

       [tex]A=\frac{\pi (8.3\times 10^{-3})^2}{4}=5.41\times 10^{-5}m^2[/tex]

Length, L = 15 m

ΔL = 5.1 cm = 0.051 m

Substituting

       [tex]0.051=\frac{9.81m\times 15}{5.41\times 10^{-5}\times 0.37\times 10^{10}}\\\\m=69.38kg[/tex]  

Mass of the climber = 69.38 kg

To find the diffusion rate in the second channel, we can use a formula that takes into account the cross-sectional area and length of the channels. Substituting the given values and simplifying the expression will give the diffusion rate in the second channel.

The diffusion rate is determined by several factors, including the concentration difference, the diffusion constant, temperature, and the size of the molecules. In this case, the diffusion rate in the first channel is given as 4.0 x 10^(-11) kg/s. To find the diffusion rate in the second channel, we can use the formula:

diffusion rate = (cross-sectional area of second channel / cross-sectional area of first channel) x (length of second channel / length of first channel) x diffusion rate of first channel

Substituting the given values:

diffusion rate = (0.30 cm^2 / 0.50 cm^2) x (0.10 cm / 0.25 cm) x (4.0 x 10^(-11) kg/s)

Simplifying the expression will give the diffusion rate in the second channel.

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The new diffusion rate in a channel with a cross-sectional area of 0.30 cm² and a length of 0.10 cm would be 6.0 x 10⁻¹¹ kg/s.

The diffusion rate for a solute in a solvent-filled channel is directly proportional to the cross-sectional area and inversely proportional to the length of the channel. Given that the initial diffusion rate is 4.0 x 10-11 kg/s in a channel with an area of 0.50 cm² and length 0.25 cm, to calculate the diffusion rate in a new channel with a different area and length, we use the equation:

Diffusion rate (new) = (Area (new) / Area (old)) x (Length (old) / Length (new)) x Diffusion rate (old)

Substituting the given values:

Diffusion rate (new) = (0.30 cm² / 0.50 cm²) x (0.25 cm / 0.10 cm) x 4.0 x 10⁻¹¹ kg/s

Diffusion rate (new) = (0.60) x (2.50) x 4.0 x 10⁻¹¹ kg/s

Diffusion rate (new) = 6.0 x 10⁻¹¹ kg/s

This calculation shows how the diffusion rate changes when altering the channel's dimensions.

Answer:

Explanation: r₂ = 15 x 10⁻³ m ; r = 12 x 10⁻³m ; r₁ = 10 x 10⁻³ m

     E = Q /4π∈r²x ( r³ - r₁ ³) / ( r₂³ - r₁³ )

         = 15 x 10⁻⁶ x 8.99 x 10⁹x( 12³ - 10³ ) / ( 15³- 10³ ) X 12³

          = 23.92 N/C

The electric field is the force acting on a charge located in a region in space. The electric field at a distance r from the center of the shell is 2.8x10⁶N/C

The electric field (E) can be defined as a region in the space that interacts with electric charges or charged bodies through electric forces.

It is a vectorial field in which an electric charge (q) suffers the effect of the electric force (F). In other words, it represents the force that is acting on a charge located in a certain region.

E = F/q

It is expressed in newton/coulomb (N/C)

In the exposed example, we have the following data,

We need to get the value of the electric field at the given distance r.

E = Q/4πεr² = kQ/r²

Since the r value is located between r1 and r2, we need to get the electric field at that intermedia distance, r. This is a new ratio.

The total charge Q is distributed in the whole area of the shell thickness or density, delimited by r1 and r2.

But to get the electric field at the intermedia distance, r, we only need to get the charge q lower than Q and distributed between the r and r1, which represents a volume v lower than the total volume V of the shell.

So to get the q value, we need to consider the charge volumetric density, σ.

σ = Q/V

Where

But to get Q/V, we need to get the total shell volume value, V.

V = 4/3 π r³

V = 4/3 π (r2³ - r1³)

V = 4/3 π (0.15³ - 0.1³)

V = 4/3 π (0.00237)

V = 0.00993m³

Now that we have the total volume, and we know the value of the total charge, we need to get the density of the shell.

σ = Q/V

σ = 15 μC / 0.00993m³

σ = 15x10⁻⁶ C / 0.00993m³

σ = 0.00151 C/m³

Now, we will calculate the volume v between r and r1, which is what we are interested in.

v = 4/3 π r³

v = 4/3 π (r³ - r1³)

v = 4/3 π (0.12³ - 0.1³)

v = 4/3 π (0.0007)

v = 0.00305m³

Finally, using this new v value and the σ value, we can get the charge q distributed  throughout the volume v.

q = vσ

q = 0.00305m³ x 0.00151 C/m³

q = 4.6x10⁻⁶ C

Now we can get the electric field at a distance r from the center of the shell

E = kq/r²

E = ((8.99 × 10⁹ N ∙ m²/C²) x (4.6x10⁻⁶ C) ) / 0.12²m

E = 2.8x10⁶N/C

Hence, the electric field at a distance r is 2.8x10⁶N/C

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Answer:

Part a)

Q = 3198 J

Part b)

It is compression of gas so this is energy transferred to the gas

Explanation:

Part a)

Energy transfer during compression of gas is same as the work done on the gas

In isothermal process work done is given by the equation

[tex]W = nRT ln(\frac{V_2}{V_1})[/tex]

now we know that

n = 1.4 moles

T = 27 degree C = 300 K

[tex]V_2 = 2.5 m^3[/tex]

[tex]V_1 = 1 m^3[/tex]

now we have

[tex]W = (1.4)(8.31)(300)(ln\frac{2.5}{1})[/tex]

[tex]Q = 3198 J[/tex]

Part b)

It is compression of gas so this is energy transferred to the gas

Answer:

Explanation:

It is given that,

Mass of lump, m₁ = 0.05 kg

Initial speed of lump, u₁ = 12 m/s

Mass of the cart, m₂ = 0.15 kg

Initial speed of the cart, u₂ = 0

The lump of clay sticks to the cart as it is a case of inelastic collision. Let v is the speed of the cart and the clay after the collision. As the momentum is conserved in inelastic collision. So,

[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]

[tex]v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]

[tex]v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}[/tex]

v = 3 m/s

So, the speed of the cart and the clay after the collision is 3 m/s. Hence, this is the required solution.

The speed of the cart and clay after the collision is 3 m/s.

The speed of the cart and clay after the collision is determined by applying the principle of conservation of linear momentum as shown below;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

0.05(12) + 0.15(0) = v(0.05 + 0.15)

0.6 = 0.2v

v = 0.6/0.2

v = 3 m/s

Thus, the speed of the cart and clay after the collision is 3 m/s.

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Answer:

F = 0.018 N

Explanation:

Magnetic force between two parallel current carrying wires is given by

[tex]F = \frac{\mu_0 i_1 i_2 L}{2\pi d}[/tex]

here we know that

[tex]i_1 = 34 A[/tex]

[tex]i_2 = 69 A[/tex]

d = 15 cm

L = 5.9 m

now from above formula we can say

[tex]F = \frac{(4\pi \times 10^{-7})(34 A)(69 A)5.9}{2\pi (0.15)}[/tex]

now the force between two wires is given as

[tex]F = 0.018 N[/tex]

Answer:

The weight of the block on the moon is 15 kg.

Explanation:

It is given that,

The acceleration of the block, a = 7.5 m/s²

Force applied to the box, F = 70 N

The mass of the block will be, [tex]m=\dfrac{F}{a}[/tex]

[tex]m=\dfrac{70\ N}{7.5\ m/s^2}[/tex]

m = 9.34 kg

The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :

[tex]W=m\times g[/tex]

[tex]W=9.34\ kg\times 1.62\ m/s^2[/tex]

W = 15.13 N

or

W = 15 N

So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.

The mass of the block is approximately 9.33 kg, and when you multiply that by the acceleration due to gravity on the moon (1.62 m/s^2), you get a weight of approximately 15 N. Therefore, the closest answer is 15 N.

To solve this problem, we need to find the mass of the block first. We know on earth, Force (F) = mass (m) * acceleration (a). Given that the force is 70N, and the acceleration is 7.5 m/s, we can solve for m. So, m = F/a = 70N / 7.5 m/s = 9.33 kg (approximately).

Now, let's figure out the weight of the same block on the moon. Weight is calculated as mass times the acceleration due to gravity (Weight = m*g). On the moon, the acceleration due to gravity is 1.62 m/s^2, so Weight = 9.33 kg * 1.62 m/s^2 = 15.1 N (approximately).

So, the closest answer will be 15 N.

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The question requires determining angles for bright and dark fringes in a double-slit experiment using the formula d sin(\theta) = m\lambda. The student should apply this formula, adjusting it for dark fringes, and solve for the angles for each order m, considering the wavelength and slit separation given.

The question relates to Young's double-slit experiment, which demonstrates the wave nature of light through the interference pattern produced when light passes through two closely spaced slits. The angles locating the fringes can be calculated using the formula

d sin(\theta) = m\lambda, where d is the separation between the slits, \theta is the angle of the fringe from the central maximum, m is the order of the fringe (which can be an integer or half-integer value depending on whether it's a bright or dark fringe), and \lambda is the wavelength of the light.

(a) For the dark fringe with m=0, we expect no fringe to appear as m=0 corresponds to the central maximum which is bright, not dark.

(b) For the bright fringe with m=1, we rearrange the formula to \theta = arcsin(m\lambda / d). Substituting the given values, the angle can be calculated.

(c) For the dark fringe with m=1, the condition is changed to d sin(\theta) = (m + 0.5)\lambda, as dark fringes occur at half-wavelength shifts from the bright fringes.

(d) The calculation for the bright fringe with m=2 follows the same procedure as for m=1, using the appropriate value for m.

Answer:

[tex]B = \frac{\mu_0 \rho r^2 \omega}{2}[/tex]

Explanation:

Let the position of the point where magnetic field is to be determined is at distance "r" from the axis of cylinder

so here total charge lying in this region is

[tex]q = \rho(\pi r^2 L)[/tex]

now magnetic field inside the cylinder is given as

[tex]B = \frac{\mu_0 N i}{L}[/tex]

here current is given as

[tex]i = \frac{q\omega}{2\pi}[/tex]

[tex]i = \frac{\rho (\pi r^2 L) \omega}{2\pi}[/tex]

[tex]i = \frac{\rho r^2 L \omega}{2}[/tex]

now magnetic field is given as

[tex]B = \frac{\mu_0 \rho r^2 L \omega}{2L}[/tex]

[tex]B = \frac{\mu_0 \rho r^2 \omega}{2}[/tex]

Answer:

Gravitational force = 89.18 x 10^-31 N

Electric force = 2.4 x 10^-17 N

Magnetic force = 7.68 x 10^-17 N

Explanation:

B = 80 micro tesla = 80 x 10^-6 T  north

E = 150 N/C downward

v = 6 x 10^6 m/s east

Gravitational force = m g = 9.1 x 10^-31 x 9.8 = 89.18 x 10^-31 N

Electric force = q E = 1.6 x 10^-19 x 150 = 2.4 x 10^-17 N

Magnetic force = q v B Sin 90 = 1.6 x 10^-19 x 6 x 10^6 x 80 x 10^-6

                          = 7.68 x 10^-17 N

Answer:

The correct option is b) In galaxy clusters

Explanation:

A type of galaxy that appear elliptical in shape and have an almost featureless and smooth image is known as the elliptical galaxy.

An elliptical galaxy is three dimensional and consists of more than one hundred trillion stars which are present in random orbits around the centre.

Elliptical galaxy is generally found in the galaxy clusters.

Answer:

365°C

Explanation:

°C=1.10*23.2/0.0700

°C=365°C//

Answer:

The function has a maximum in [tex]x=3[/tex]

The maximum is:

[tex]f(3) = 39[/tex]

Explanation:

Find the first derivative of the function for the inflection point, then equal to zero and solve for x

[tex]f(x)' = -4*2x + 24=0[/tex]

[tex]-4*2x + 24=0[/tex]

[tex]8x=24[/tex]

[tex]x=3[/tex]

Now find the second derivative of the function and evaluate at x = 3.

If [tex]f (3) ''< 0[/tex] the function has a maximum

If [tex]f (3) '' >0[/tex] the function has a minimum

[tex]f(x)''= 8[/tex]

Note that:

[tex]f(3)''= -8<0[/tex]

the function has a maximum in [tex]x=3[/tex]

The maximum is:

[tex]f(3)=-4(3)^2+24(3) + 3\\\\f(3) = 39[/tex]

Answer:

[tex]E = 2 \times 10^{-3} V[/tex]

Explanation:

As we know that rate of change in flux will induce EMF

So here we can

[tex]EMF = \frac{d\phi}{dt}[/tex]

now we have

[tex]EMF = \pi r^2\frac{dB}{dt}[/tex]

now we also know that induced EMF is given by

[tex]\int E. dL = \pi r^2\frac{dB}[dt}[/tex]

[tex]E (2\pi r) = \pi r^2\frac{dB}{dt}[/tex]

[tex]E = \frac{r}{2}(\frac{dB}{dt})[/tex]

now plug in all values in it

[tex]E = \frac{0.0133}{2}(0.12 t)[/tex]

[tex]E = 8 \times 10^{-4} (2.50) = 2 \times 10^{-3} V/m[/tex]

Answer:

[tex]v_a = 2 v_b[/tex]

Explanation:

As we know that the speed of car A and car B is given by

[tex]v_a[/tex] & [tex]v_b[/tex]

now we know that both cars are decelerated by same deceleration and stopped finally

so the distance moved by the car is given by the equations

[tex]v_f^2 - v_i^2 = 2a d[/tex]

[tex]0 - v_a^2 = 2(-a) d_a[/tex]

[tex]d_a = \frac{v_a^2}{2a}[/tex]

similarly we have

[tex]d_b = \frac{v_b^2}{2a}[/tex]

now we know that

[tex]d_a = 4 d_b[/tex]

[tex]\frac{v_a^2}{2a} = 4 \frac{v_b^2}{2a}[/tex]

[tex]v_a = 2 v_b[/tex]

Answer: [tex]0.000001475s=1.475\mu s[/tex]

Explanation:

The index of refraction [tex]n[/tex] is a number that describes how fast light propagates through a medium or material.  

Being its equation as follows:  

[tex]n=\frac{c}{v}[/tex] (1)

Where [tex]c=3(10)^{8}m/s[/tex] is the speed of light in vacuum and [tex]v[/tex] its speed in the other medium.

So, from (1) we can find the velocity at which the light travels and then the time it requires to travel : [tex]v=\frac{c}{n}[/tex] (2)

For medium 1:

[tex]n_{1}=1.33[/tex]

[tex]v_{1}=\frac{c}{n_{1}}[/tex] (3)

[tex]v_{1}=\frac{3(10)^{8}m/s}{1.33}=225563909.8m/s[/tex] (4)

For medium 2:

[tex]n_{2}=1.51[/tex]

[tex]v_{2}=\frac{c}{n_{2}}[/tex] (5)

[tex]v_{2}=\frac{3(10)^{8}m/s}{1.51}=198675496.7m/s[/tex] (6)

On the other hand, the velocity [tex]v[/tex] is the distance [tex]d[/tex] traveled in a time [tex]t[/tex]:

[tex]v=\frac{d}{t}[/tex] (7)

We can isolate [tex]t[/tex] from (7) and find the value of the required time:

[tex]t=\frac{d}{v}[/tex] (8)

In this case the total time will be:

[tex]t=t_{1}+t_{2}=\frac{d_{1}}{v_{1}}+\frac{d_{2}}{v_{2}}[/tex] (9)

Where:

[tex]d_{1}=331cm=3.31m[/tex] is the distance the light travels in medium 1

[tex]d_{2}=151cm=1.51m[/tex] is the distance the light travels in medium 2

[tex]v_{1}=225563909.8m/s[/tex] is the velocity of light in medium 1

[tex]v_{2}=198675496.7m/s[/tex] is the velocity of light in medium 2

[tex]t=t_{1}+t_{2}=\frac{3.31m}{225563909.8m/s}+\frac{1.51m}{198675496.7m/s}[/tex] (10)

Finally:

[tex]t=0.000001475s=1.475(10)^{-6}s=1.475\mu s[/tex] (10)

Light takes different amounts of time to travel through different media due to refraction. The time can be calculated by dividing the distance traveled in each medium by the speed of light in that medium.

When light travels from one medium to another, it changes direction, a phenomenon called refraction. The time it takes for light to travel from point A to point B in this case can be calculated by dividing the distance traveled in each medium by the speed of light in that medium. In medium 1, the distance traveled is 331 cm and the index of refraction is 1.33. In medium 2, the distance traveled is 151 cm and the index of refraction is 1.51.

Using the equation time = distance / speed, we can calculate the time it takes for light to travel in each medium.

In medium 1: time1 = 331 cm / speed1

In medium 2: time2 = 151 cm / speed2

Ham como assim pq ta em ingles

Explanation:

Given that,

Height of object = 4.31 cm

Distance of the object = -12.6 cm

Distance of the image = -8.77 cm

For concave mirror,

Using mirror's formula

[tex]\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}[/tex]

[tex]\dfrac{1}{f}=\dfrac{1}{-12.6}-\dfrac{1}{8.77}[/tex]

[tex]\dfrac{1}{f}=-\dfrac{10685}{55251}[/tex]

[tex]f=-\dfrac{55251}{10685}[/tex]

[tex]f = -5.17\ cm[/tex]

Radius of the mirror is

[tex]f = |\dfrac{R}{2}|[/tex]

[tex]r=2f[/tex]

[tex]r=2\times5.17[/tex]

[tex]r=10.34\ cm[/tex]

The magnification of the mirror,

[tex]m=-\dfrac{v}{u}[/tex]

[tex]\dfrac{h_{i}}{h_{o}}=\dfrac{v}{u}[/tex]

[tex]h_{i}=-h_{o}\times\dfrac{v}{u}[/tex]

[tex]h_{i}=-4.31\times\dfrac{8.77}{12.6}[/tex]

[tex]h_{i}=-2.99\ cm[/tex]

Now, For convex mirror,

Using mirror's formula

[tex]\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}[/tex]

[tex]\dfrac{1}{f}=\dfrac{1}{-12.6}+\dfrac{1}{8.77}[/tex]

[tex]\dfrac{1}{f}=\dfrac{1915}{55251}[/tex]

[tex]f=\dfrac{55251}{1915}[/tex]

[tex]f = 28.85\ cm[/tex]

Radius of the mirror is

[tex]f = \dfrac{R}{2}[/tex]

[tex]r=2f[/tex]

[tex]r=2\times28.85[/tex]

[tex]r=57.7\ cm[/tex]

The magnification of the mirror,

[tex]m=-\dfrac{v}{u}[/tex]

[tex]\dfrac{h_{i}}{h_{o}}=\dfrac{v}{u}[/tex]

[tex]h_{i}=-h_{o}\times\dfrac{v}{u}[/tex]

[tex]h_{i}=4.31\times\dfrac{8.77}{12.6}[/tex]

[tex]h_{i}=2.99\ cm[/tex]

Hence, This is the required solution.

Answer:

3.7 x 10⁷ m/s

Explanation:

ΔV = Potential difference through which the proton moves = 7.15 MV = 7.15 x 10⁶ Volts

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

v = speed of the proton as it reach zero potential region

m = mass of the proton = 1.66 x 10⁻²⁷ kg

Using conservation of energy

Kinetic energy gained by proton = Electric potential energy lost

(0.5) m v² = q ΔV

(0.5) (1.66 x 10⁻²⁷) v² = (1.6 x 10⁻¹⁹) (7.15 x 10⁶)

v = 3.7 x 10⁷ m/s

Final answer:

The speed of protons in a Van de Graaff accelerator transitioning from a 7.15 MV potential to zero potential is about 3.70 x 10^7 m/s, option c

Explanation:

Protons being accelerated in a Van de Graaff accelerator from a 7.15 MV potential to zero potential can be analyzed using the principle of conservation of energy.

The kinetic energy gained by the protons equals the initial potential energy they had, leading to the formula: 1/2 mv^2 = qV.

Calculating this, the speed of the protons when they reach the zero potential region is approximately 3.70 x 10^7 m/s (C).