A box of mass ????=21.0 kgm=21.0 kg is pulled up a ramp that is inclined at an angle ????=21.0∘θ=21.0∘ angle with respect to the horizontal. The coefficient of kinetic friction between the box and the ramp is ????k=0.285μk=0.285 , and the rope pulling the box is parallel to the ramp. If the box accelerates up the ramp at a rate of ????=2.09 m/s2a=2.09 m/s2 , calculate the tension ????TFT in the rope. Use ????=9.81 m/s2g=9.81 m/s2 for the acceleration due to gravity.

Answer:

[tex]P_{1}-P_{2}=5400lb/ft^{2}[/tex]

Explanation:

We shall use newtons second law to evaluate the pressure difference

For the system the forces that act on it as shown in the figure

Thus by Newton's second law

[tex]F_{1}-F_{2}=mass\times acceleration\\\\P_{1}\times Area-P_{2}\times Area=mass\times acceleration\\\\\because Force=Pressure\times Area\\\\\therefore P_{1}-P_{2}=\frac{mass\times acceleration}{Area}[/tex]

Mass of the gasoline can be calculated from it's density[tex]45lb/ft^{3}[/tex]

[tex]Mass=Density\times Volume\\\\Mass= 45lb/ft^{3}\times \pi \frac{d^{2}}{4}L\\\\Mass=45lb/ft^{3}\times\frac{\pi 8^{2}}{4}\times 24\\Mass=54286.72lbs[/tex]

Using the calculated values we get

[tex]P_{1}-P_{2}=\frac{54286.72\times 5}{\frac{\pi 8^{2}}{4}}[/tex]

[tex]P_{1}-P_{2}=5400lb/ft^{2}[/tex]

Answer:

Velocity of a proton, [tex]v=1.7\times 10^6\ m/s[/tex]    

Explanation:

It is given that,

Potential difference, [tex]V=15\ kV=15\times 10^3\ V[/tex]

Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.

Using the conservation of energy as :

[tex]\dfrac{1}{2}mv^2=qV[/tex]

q is the charge of proton

m is the mass of proton

[tex]v=\sqrt{\dfrac{2qV}{m}}[/tex]

[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}[/tex]

[tex]v=1695361.75\ m/s[/tex]

[tex]v=1.69\times 10^6\ m/s[/tex]

or

[tex]v=1.7\times 10^6\ m/s[/tex]

So, the velocity of a proton is [tex]1.7\times 10^6\ m/s[/tex]. Hence, this is the required solution.

Answer:

V=4v

Explanation:

To perform this operation, we will refer to Ohm's Law. This Law relates the terms current, voltage and resistance.

The current intensity that passes through a circuit is directly proportional to the voltage or voltage of the circuit and inversely proportional to the resistance it presents.

[tex]I=\frac{v}{r}[/tex]

Where

I=Current

V= Voltage

R=Resistance.

So, in this case:

R=200 ohm

I= 2mA  = 0.002 A

V= searched variable

[tex]V=r*I[/tex]

[tex]V=(200ohm)(0.002A)=4V[/tex]

in increasing order of kinetic energy, the joggers are: A, D, C, B

To rank the joggers in order of increasing kinetic energy, we'll use the formula for kinetic energy:

[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]

Where:

- ( m ) is the mass of the jogger

- ( v ) is the speed of the jogger

Let's calculate the kinetic energy for each jogger:

1. Jogger A:

[tex]\[ KE_A = \frac{1}{2} m v^2 \][/tex]

2.Jogger B:

[tex]\[ KE_B = \frac{1}{2} \left(\frac{m}{3}\right) (3v)^2 = \frac{1}{2} \left(\frac{m}{3}\right) 9v^2 = \frac{9}{2} \left(\frac{1}{3} m v^2\right) = \frac{9}{2} KE_A \][/tex]

3. Jogger C:

[tex]\[ KE_C = \frac{1}{2} (4m) \left(\frac{v}{3}\right)^2 = \frac{1}{2} (4m) \left(\frac{1}{9}\right) v^2 = \frac{2}{9} (4m v^2) = \frac{8}{9} KE_A \][/tex]

4. Jogger D:

[tex]\[ KE_D = \frac{1}{2} (3m) \left(\frac{v}{2}\right)^2 = \frac{1}{2} (3m) \left(\frac{1}{4}\right) v^2 = \frac{3}{8} (m v^2) = \frac{3}{4} KE_A \][/tex]

Now, let's rank the joggers based on their kinetic energies:

- Jogger A: [tex]\( KE_A \)[/tex]

- Jogger D: [tex]\( KE_D = \frac{3}{4} KE_A \)[/tex]

- Jogger C: [tex]\( KE_C = \frac{8}{9} KE_A \)[/tex]

- Jogger B: [tex]\( KE_B = \frac{9}{2} KE_A \)[/tex]

So, in increasing order of kinetic energy, the joggers are: A, D, C, B

Answer:

2.15 m

Explanation:

u = 2.2 m/s, v = 0, uk = 0.115

a = uk x g = 0.115 x 9.8 = 1.127 m/s^2

Let s be the distance traveled before stopping.

v^2 = u^2 - 2 a s

0 = 2.2^2 - 2 x 1.127 x s

s = 2.15 m

Answer:

Explanation:

(a) It is called conductors.

The conductors are the materials which can allow the current to pass through it.

(b) The example of conductor is copper, iron, etc.

The best conductor of electricity is silver.

(c) The example of semiconductor is silicon, germanium, etc.

The semiconductors are the materials which are insulators at normal temperature, but if the temperature increases, the conductivity of semiconductor increases.

(d) An example of insulator is wood.

Insulators are the materials which do not allow the current to pass through them.

Answer:

v/U=0.79

Explanation:

Given u=[tex]u=\frac{Uy}{\partial }=\frac{Uy}{cx^{1/2}}[/tex]

Now for the given flow to be possible it should satisfy continuity equation

[tex]\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0[/tex]

Applying values in this equation we have

[tex]\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0\\\\\frac{\partial u}{\partial x}=\frac{\partial (\frac{Uy}{cx^{1/2}})}{\partial x}\\\\\frac{\partial u}{\partial x}=\frac{-1}{2}\frac{Uy}{cx^{3/2}}\\\\[/tex]

Thus we have

[tex]\frac{\partial v}{\partial y}=\frac{1}{2}\frac{Uy}{cx^{3/2}}\\\\\therefore \int \partial v=\int \frac{1}{2}\frac{Uy}{cx^{3/2}}\partial x\\\\v=\frac{1}{4}\frac{Uy^{2}}{cx^{3/2}}\\\\v=\frac{1}{4}\frac{uy}{x}[/tex] Hence proved [tex]\because u=\frac{Uy}{cx^{1/2}}[/tex]

For maximum value of v/U put y =[tex]\partial[/tex]

[tex]v=\frac{1}{4}\frac{Uy^{2}}{cx^{3/2}}[/tex]

[tex]v=\frac{1}{4}\frac{Uy^{2}}{cx^{3/2}}\\\\\frac{v}{U}=\frac{\partial ^{2}}{4cx^{3/2}}\\\\[/tex]

Thus solving we get using the given values

v/U=0.79

The y-component of velocity in a boundary layer flow is derived using the principle of continuity for an incompressible fluid. Upon evaluation, the ratio v /U has a maximum of 0.25 at the specified location (x = 0.5 m and δ = 5 mm).

The y-component v of the velocity can be solved using the continuity equation, d/dx (u A) + d/dy (v A) = 0, where A is the cross-sectional area of the pipe. This equation arises from the principle of conservation of mass for an incompressible fluid. In this case, our A is the width of the plate (into the page) times the y-distance from the plate or y*b. When the equation derived for velocity, u = Uy/δ, and the equation for the boundary layer thickness, δ = cx1/2, are plugged into the continuity equation and simplified, we derive the expression v = uy/4x.  Substituting for δ, we get v = Uy/(4δ*sqrt(x)), and at x = 0.5 m and δ = 5 mm, the maximum value of v /U is 1/4 or 0.25.

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Answer:

a) Rotational Inertia = Decreases

Because the distance from the axis is decreasing

b) Angular momentum = Remains the same

because there is no external torque

c) Angular speed = increases

because here rotational inertia decreases due to which angular speed will increase

Explanation:

Here the Beetle is initially moving along the rim of the disc

So here during the motion of beetle there is no external force on the system of beetle and the disc.

So here we can also say that there is no torque acting on the system

so angular momentum of the disc + beetle system will remain conserved

so here we have

[tex]I_1\omega_1 = I_2 \omega_2[/tex]

here as the beetle crawls towards the centre of the disk then

a) Rotational Inertia = Decreases

Because the distance from the axis is decreasing

b) Angular momentum = Remains the same

because there is no external torque

c) Angular speed = increases

because here rotational inertia decreases due to which angular speed will increase

Answer:

a)  The elevator was  31.64 m high up when the bolt came loose.

b)  Speed of the bolt when it hits the bottom of the shaft = 25.41 m/s

Explanation:

a) Considering motion of bolt:-

Initial velocity, u =  5 m/s

Acceleration , a = -9.81 m/s²

Time = 3.1 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= 5 x 3.1 - 0.5 x 9.81 x 3.1²

    s = 0 x t + 0.5 x 9.81 x t²

    s = -31.64 m

The elevator was  31.64 m high up when the bolt came loose.

b) We have equation of motion v = u + at

  Initial velocity, u =  5 m/s

 Acceleration , a = -9.81 m/s²

 Time = 3.1 s  

Substituting

  v = u + at

  v  = 5 - 9.81 x 3.1 = -25.41 m/s

Speed of the bolt when it hits the bottom of the shaft = 25.41 m/s

Answer:

118.06 days

Explanation:

d = distance of the center of moon from surface of planet = 2.315 x 10⁵ km = 2.315 x 10⁸ m

R = radius of the planet = 4.15 x 10³ km = 4.15 x 10⁵ m

r = center to center distance between the planet and moon = R + d

M = mass of the planet = 7.15 x 10²² kg

T = Time period of revolution around the planet

Using Kepler's third law

[tex]T^{2}=\frac{4\pi ^{2}r^{3}}{GM}[/tex]

[tex]T^{2}=\frac{4\pi ^{2}(R + d)^{3}}{GM}[/tex]

[tex]T^{2}=\frac{4(3.14)^{2}((4.15\times 10^{5}) + (2.315\times 10^{8}))^{3}}{(6.67\times 10^{-11})(7.15\times 10^{22})}[/tex]

T = 1.02 x 10⁷ sec

we know that , 1 day = 24 h = 24 x 3600 sec = 86400 sec

T = [tex](1.02 \times 10^{7} sec)\frac{1 day}{86400 sec}[/tex]

T = 118.06 days

Answer:

Work done, W = 39.2 J

Explanation:

It is given that,

Mass of the block, m = 2 kg

The block is lifted vertically 2 m by the man i.e the distance covered by the block is, h = 2 m. The man is doing work against the gravity. It is given by :

[tex]W=mgh[/tex]

Where

g is acceleration due to gravity

[tex]W=2\ kg\times 9.8\ m/s^2\times 2\ m[/tex]

W = 39.2 J

So, the work done by the man is 39.2 J. Hence, this is the required solution.

Answer:

(iii) 6

Explanation:

Part a)

Since it is given that both ends are fixed and it is vibrating in 5th harmonic

So here it will have 5 number of loops in it

so we can draw it in following way

each loop will have 1 antinode and two nodes which means number of nodes is one more than number of anti-nodes

So there are 5 loops which means it will have 5 antinodes

and hence there will be 6 nodes in it

so correct answer will be

(iii) 6

Answer:

COP = 4.25

Explanation:

It is given that,

Cooling temperature, [tex]T_c=-26^{\circ}C=273-26=247\ K[/tex]

Heating temperature, [tex]T_h=50^{\circ}C=323\ K[/tex]

We need to find the coefficient of performance. It is given by :

[tex]COP=\dfrac{T_h}{T_h-T_c}[/tex]

[tex]COP=\dfrac{323}{323-247}[/tex]

COP = 4.25

So, the best coefficient of performance of a refrigerator is 4.45 Hence, this is the required solution.

Answer:  [tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]

Explanation:

Reduction is a process where electrons are gained and acidic solution means presence of [tex]H^+[/tex] ions.

Reduction of [tex]MnO_2[/tex] to [tex]Mn^{2+}[/tex]

Mn is in +4 oxidation state in [tex]MnO_2[/tex] which goes to +2 state in [tex]Mn^{2+}[/tex] by gain of 2 electrons.

[tex]MnO_2(s)\rightarrow Mn^{2+}(aq)[/tex]

In order to balance oxygen atoms:

[tex]MnO_2(s)\rightarrow Mn^{2+}(aq)+2H_2O[/tex]

In order to balance hydrogen atoms:

[tex]MnO_2(s)+4H^+(aq)\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]

In order to balance charges:

[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l[/tex]

Thus the net balanced half reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution is:

[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^{2+}(aq)+2H_2O(l)[/tex]

Final answer:

To balance the reduction of solid manganese dioxide (MnO2) to manganese ion (Mn2+) in acidic aqueous solution, the balanced half-reaction is MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l).

Explanation:

To balance the reduction of solid manganese dioxide (MnO2) to manganese ion (Mn2+) in acidic aqueous solution, we can follow these steps:

1. Write the half-reaction for reduction, adjusting the physical states of the reactants and products:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

2. Balance the number of Mn and O atoms on each side of the equation:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

3. Balance the H atoms by adding H+ ions:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

4. Balance the charges by adding electrons:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

Therefore, the balanced half-reaction for the reduction of solid manganese dioxide to manganese ion in acidic aqueous solution is:

MnO2(s) + 4H+(aq) + 2e- → Mn2+(aq) + 2H2O(l)

Answer:

670400 J

Explanation:

m = mass of water = 2 kg

T₀ = initial temperature of water = 20 ºC

T = initial temperature of water = 100 ºC

c = heat capacity of water = 4190 J /kg ºC

Q = energy needed to raise the temperature of water

Energy needed to raise the temperature of water is given as

Q = m c (T - T₀)

Inserting the values

Q = (2) (4190) (100 - 20)

Q = 670400 J

Answer:

The International Space Station move at 7.22 km/s.

Explanation:

Orbital speed of satellite is given by  [tex]v=\sqrt{\frac{GM}{r}}[/tex], where G is gravitational constant, M is mass of Earth and r is the distance to satellite from centre of Earth.

r = R + h = 6350 + 1400 = 7750 km = 7.75 x 10⁶ m

G = 6.673 x 10⁻¹¹ Nm²/kg²

M = 5.98 x 10²⁴ kg

Substituting

              [tex]v=\sqrt{\frac{6.673\times 10^{-11}\times 5.98\times 10^{24}}{7.75\times 10^6}}=7223.86m/s=7.22km/s[/tex]

  The International Space Station move at 7.22 km/s.      

Answer:

2.14×10⁸ m/s

Explanation:

L=Required Length=0.7 m

L₀=Initial length=1 m

c=speed of light=3×10⁸ m/s

[tex]From\ Lorentz\ Contraction\ relation\\L=L_0\sqrt{1-\frac {v^2}{c^2}}\\\Rightarrow 0.7=1\sqrt {1-\frac {v^2}{c^2}}\\\Rightarrow 0.49=1-\frac {v^2}{c^2}\\\Rightarrow 0.49-1=-\frac {v^2}{c^2}\\\Rightarrow -0.51=-\frac {v^2}{c^2}\\\Rightarrow 0.51=\frac {v^2}{c^2}\\\Rightarrow v^2=0.51\times c^2\\\Rightarrow v=\sqrt{0.51} \times c\\\Rightarrow v=0.71\times 3\times 10^8\\\Rightarrow v=2.14\times 10^8\ m/s\\\therefore velocity\ of\ stick\ should\ be\ 2.14\times 10^8\ m/s[/tex]

Answer:

(a) 15.4 second

(b) 769.23 m

Explanation:

u = 0, a = 6.5 m/s^2, v = 100 m/s

(a) Let t be the time taken

Use First equation of motion

v = u + a t

100 = 0 + 6.5 x t

t = 15.4 second

(b) Let height covered is h.

Use third equation of motion

v^2 = u^2 + 2 a h

100^2 = 0 + 2 x 6.5 x h

10000 = 13 x h

h = 769.23 m

Answer:

(a). The magnitude of the gravitational force is [tex]7\times10^{-7}\ N[/tex]

(b). The magnitude of the gravitational force is [tex]1.35\times10^{-6}\ N[/tex]

Explanation:

Given that,

Mass of baby = 4.20 kg

Mass of father = 100 kg

Distance = 0.200 m

We need to calculate the gravitational force

Using gravitational formula

[tex]F = \dfrac{G\times m_{b}m_{f}}{r^2}[/tex]

Put the value in to the formula

[tex]F=\dfrac{6.67\times10^{-11}\times4.20\times100}{0.200^2}[/tex]

[tex]F=7\times10^{-7}\ N[/tex]

(b). We need to calculate the gravitational force

Using gravitational formula

[tex]F = \dfrac{G\times m_{b}m_{f}}{r^2}[/tex]

Put the value in to the formula

[tex]F=\dfrac{6.67\times10^{-11}\times4.20\times1.9\times10^{27}}{(6.29\times10^{11})^2}[/tex]

[tex]F=1.35\times10^{-6}\ N[/tex]

Hence, (a). The magnitude of the gravitational force is [tex]7\times10^{-7}\ N[/tex]

(b). The magnitude of the gravitational force is [tex]1.35\times10^{-6}\ N[/tex]

The magnitude of the gravitational force is 7×10⁻⁷ N and the magnitude of the gravitational force due to Jupiter is 1.35×10⁻6 N

Gravitational potential energy is the energy which a body posses because of its position.

The gravitational potential energy of a body is given as,

[tex]G=\dfrac{Fr^2}{Mm}[/tex]

Here, (m) is the mass of the body, (F) is the gravitational force and (r) is the height of the body.

The mass of baby is 4.20 kg and mass of father is 100 kg. The distance between them is 0.200 m. Put the values in the above formula as,

[tex]6.67\times10^{-11}=\dfrac{F(0.200)^2}{100(4.20)}\\F=7\times10^{-7}\rm \; N[/tex]

Put the values in the above formula again as,

[tex]6.67\times10^{-11}=\dfrac{F(6.29\times10^{11})^2}{(1.9\times10^{27})(4.20)}\\F=1.35\times10^{-6}\rm \; N[/tex]

Thus, the magnitude of the gravitational force is 7×10⁻⁷ N and the magnitude of the gravitational force due to Jupiter is 1.35×10⁻6 N.

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Answer:

10.95 minute

Explanation:

V = 120 V, I = 2 A, m = 0.489 kg, c = 4186 J/kgC, T1 = 23 C , T2 = 100 C

Let time be the t.

Heat energy is equal to electrical energy

m x c (T2 - T1) = V x I x t

0.489 x 4186 x (100 - 23) = 120 x 2 x t

t = 656.73 second

t = 10.95 minute

Answer:

[tex]P = 2.94 \times 10^5 Pa[/tex]

Explanation:

Normal force due to four tires is counter balancing the weight of the car

So here we will have

[tex]4F_n = mg[/tex]

[tex]F_n = \frac{mg}{4}[/tex]

[tex]F_n = \frac{1.20 \times 10^3 \times 9.81}{4}[/tex]

[tex]F_n = 2943 N[/tex]

now we know that pressure in each tire is given by

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

Here we know that

[tex]A = 1.00 \times 10^2 cm^2 = 1.00 \times 10^{-2} m^2[/tex]

[tex]P = \frac{2943}{1.00 \times 10^{-2}}[/tex]

[tex]P = 2.94 \times 10^5 Pa[/tex]

Answer:

P = 294300Pa or 42.67psi by conversion.

Explanation:

Since Four tyres were inflated, we have that area of the four tyres are

4×1×10²cm²

Pressure is given as:

P = f/a but f = mg

P = m×g/a

Therefore,

P = 1.20x10³kg × 9.81m/s² / (4 ×1x10² cm²)

P = 1.20x10³kg×9.81m/s² / (0.04m²)

P = 294300Pa or 42.67psi by conversion.

Answer:

[tex]K.E =0.081eV[/tex]

[tex]\vec{\Omega}=(0.54\hat{i}+0.83\hat{j}+0.039\hat{k}[/tex]

Explanation:

Given:

Velocity vector [tex]\vec{v}=(2132\hat{i}+3300\hat{j}+154\hat{k})m/s[/tex]

the mass of neutron, m = 1.67 × 10⁻²⁷ kg

Now,

the kinetic energy (K.E) is given as:

[tex]K.E =\frac{1}{2}mv^2[/tex]

[tex]\vec{v}^2 = \vec{v}.\vec{v}[/tex]

or

[tex]\vec{v}^2 = (2132\hat{i}+3300\hat{j}+154\hat{k}).(2132\hat{i}+3300\hat{j}+154\hat{k})[/tex]

or

[tex]\vec{v}^2 =15.45\times 10^6 m^2/s^2[/tex]

substituting the values in the K.E equation

[tex]K.E =\frac{1}{2}1.67\times 10^{-27}kg\times 15.45\times 10^6 m^2/s^2[/tex]

or

[tex]K.E =1.2989\times 10^{-20}J[/tex]

also

1J = 6.242 × 10¹⁸ eV

thus,

[tex]K.E =1.2989\times 10^{-20}\times 6.242\times 10^{18}[/tex]

[tex]K.E =0.081eV[/tex]

Now, the direction vector [tex]\vec{\Omega}[/tex]

[tex]\vec{\Omega}=\frac{\vec{v}}{\left | \vec{v} \right |}[/tex]

or

[tex]\vec{\Omega}=\frac{(2132\hat{i}+3300\hat{j}+154\hat{k})}{\left |\sqrt{((2132\hat{i})^2+(3300\hat{j})^2+(154\hat{k}))^2} \right |}[/tex]

or

[tex]\vec{\Omega}=\frac{(2132\hat{i}+3300\hat{j}+154\hat{k})}{3930.65}[/tex]

or

[tex]\vec{\Omega}=0.54\hat{i}+0.83\hat{j}+0.039\hat{k}[/tex]

Answer:

Explanation:

When a body starts sliding on an inclined plane, the acceleration of body is due to its impotent 9f weight. The component of weight is mg Sin theta along the plane.

Thus, the acceleration is g Sin theta

= 9.8 × Sin 30 = 4.9 m/s^2

Answer:

a)

51764.7 ohm

b)

28.5 A

Explanation:

a)

t = time taken to charge to 90.0% of its final value = 16.2 s

T = time constant

C = capacitance = 136 x 10⁻⁶ F

R = resistance of the resistor

Q₀ = Maximum charge stored

Q = Charge after time "t" = 0.90 Q₀

Using the equation

[tex]Q = Q_{o}(1 - e^{\frac{-t}{T}})[/tex]

[tex]0.90 Q_{o} = Q_{o}(1 - e^{\frac{-16.2}{T}})[/tex]

[tex]0.90 = (1 - e^{\frac{-16.2}{T}})[/tex]

T = 7.04 s

Time constant is given as

T = RC

7.04 = R (136 x 10⁻⁶)

R = 51764.7 ohm

b)

V = Potential difference of voltage source = 265 volts

q = Amount of charge discharged = 0.90 Q₀ = 0.90 (CV) = (0.90) (136 x 10⁻⁶) (265) = 0.032436 C

t = time taken to discharge = 1.14 x 10⁻³ s

Current is given as

[tex]i = \frac{q}{t}[/tex]

i = 28.5 A

For the RC charging circuit of a camera flash unit has the value of resistance and current as,

RC circuit is the circuit in which the voltage or current passes through the resistor and capacitor consist by the circuit.

The formula for the RC circuit can be given as,

[tex]Q=Q_o\left(1-e^{\dfrac{-1}{RC}}\right )[/tex]

Here, [tex](Q_o)[/tex] is the initial charge and (R) is the resistance.

The  capacitor charges to 90.0% of its final value. Thus, the charge for the first case can be given as,

[tex]Q=0.90 Q_o[/tex]

The capacitance of 136 μF. Put the values in the above formula as,

[tex]0.90Q_o=Q_o\left(1-e^{\dfrac{-1}{R\times136\times10^{-6}}}\right )\\R=51765\rm ohm[/tex]

Thus, the value of the resistance R is 51765 ohms.

The voltage source of the RC circuit is 265 V  and the capacitor discharges 90.0% of its full charge in 1.14 ms. Thus, the amount of charge discharged is,

[tex]Q=0.9\times136\times10^{-6}\times265\\Q=3.2436\times10^{-2}\rm C[/tex]

The value of current is the ratio of discharge per second time (1.14 ms.). Therefore,

[tex]I=\dfrac{3.2436\times10^{-2}}{1.14\times10^{-3}}\\I=28.5\rm A[/tex]

Thus, for the RC charging circuit of a camera flash unit has the value of resistance and current as,

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Answer:

1.1 sec

Explanation:

m = mass of the box = 8 kg

k = spring constant of the spring = 69 N/m

v = initial speed of the box = 1.5 m/s

t = time period of oscillation of box in contact with the spring

Time period is given as

[tex]t = \pi \sqrt{\frac{m}{k}}[/tex]

Inserting the values

[tex]t = (3.14) \sqrt{\frac{8}{69}}[/tex]

t = 1.1 sec

D = 497.4x10⁻⁶m. The diameter of a mile of 24-gauge copper wire with resistance of 0.14 kΩ and resistivity of copper 1.7×10−8Ω⋅m is 497.4x10⁻⁶m.

In order to solve this problem we have to use the equation that relates resistance and resistivity:

R = ρL/A

Where ρ is the resistivity of the matter, the length of the wire, and A the area of ​​the cross section of the wire.

If a mile of 24-gauge copper wire has a resistance of 0.14 kΩ and the resistivity of copper is 1.7×10⁻⁸ Ω⋅m. Determine the diameter of the wire.

First, we have to clear A from the equation R = ρL/A:

A = ρL/R

Substituting the values

A = [(1.7×10⁻⁸Ω⋅m)(1.6x10³m)]/(0.14x10³Ω)

A = 1.9x10⁻⁷m²

The area of a circle is given by A = πr² = π(D/2)² = πD²/4, to calculate the diameter D we have to clear D from the equation:

D = √4A/π

Substituting the value of A:

D = √4(1.9x10⁻⁷m²)/π

D = 497.4x10⁻⁶m

The diameter of the wire is approximately 4.5 × 10^-6 meters.

To find the diameter of the wire, we can use the formula for resistance:

R = (ρ * L) / (A),

where R is the resistance, ρ is the resistivity, L is the length of the wire, and A is the cross-sectional area of the wire. Rearranging the formula to solve for the area:

A = (ρ * L) / R.

Given that the mile of 24-gauge copper wire has a resistance of 0.14 kΩ (or 0.14 * 10^3 Ω), the resistivity of copper is 1.7 × 10^-8 Ω ⋅ m, and 1 mile is equal to 1.6 km, we can substitute the values in the formula:

A = (1.7 × 10^-8 Ω ⋅ m * (1.6 * 10^3 m)) / (0.14 * 10^3 Ω).

Calculating the area, we find:

A ≈ 1.94 × 10^-11 m^2.

Next, we use the formula for the area of a circle (A = π * r^2) to find the radius (r) of the wire:

r = √(A / π).

Substituting the values, we have:

r ≈ √(1.94 × 10^-11 m^2 / π).

Calculating the radius, we find:

r ≈ 2.25 × 10^-6 m.

Finally, we can double the radius to find the diameter of the wire:

D = 2r ≈ 2 * 2.25 × 10^-6 m = 4.5 × 10^-6 m.

Therefore, the diameter of the wire is approximately 4.5 × 10^-6 meters.

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Answer:

- 0.23

0.56

Explanation:

[tex]\underset{v_{o}}{\rightarrow}[/tex] = initial velocity of the particle = [tex](5.24 Cos35.2) \hat{i} + (5.24 Sin35.2) \hat{j}[/tex] = [tex](4.28) \hat{i} + (3.02) \hat{j}[/tex]

[tex]\underset{v_{f}}{\rightarrow}[/tex] = final velocity of the particle = [tex](6.25 Cos57.5) \hat{i} + (6.25 Sin57.5) \hat{j}[/tex] = [tex](3.36) \hat{i} + (5.27) \hat{j}[/tex]

t = time interval = 4.00 sec

[tex]\underset{a}{\rightarrow}[/tex] = average acceleration = ?

Using the kinematics equation

[tex]\underset{v_{f}}{\rightarrow}[/tex] = [tex]\underset{v_{o}}{\rightarrow}[/tex] + [tex]\underset{a}{\rightarrow}[/tex] t

[tex](3.36) \hat{i} + (5.27) \hat{j}[/tex] = [tex](4.28) \hat{i} + (3.02) \hat{j}[/tex] + [tex]\underset{a}{\rightarrow}[/tex] (4)

[tex](- 0.92) \hat{i} + (2.25) \hat{j}[/tex] = 4 [tex]\underset{a}{\rightarrow}[/tex]

[tex]\underset{a}{\rightarrow}[/tex] = [tex](- 0.23) \hat{i} + (0.56) \hat{j}[/tex]

hence

Average acceleration along x-direction = - 0.23

Average acceleration along y-direction = 0.56

Answer:

Number of electrons, [tex]n=5.62\times 10^{21}[/tex]

Explanation:

It is given that,

Resistance, R = 4 ohms

Current, I = 3 A

Time, t = 5 min = 300 s

We need to find the number of electrons pass through the resistor during this time interval. Let the number of electron is n.

i.e. q = n e ...............(1)

And current, [tex]I=\dfrac{q}{t}[/tex]

[tex]I\times t=n\times e[/tex]

[tex]n=\dfrac{It}{e}[/tex]

e is the charge of an electron

[tex]n=\dfrac{3\ A\times 300\ s}{1.6\times 10^{-19}}[/tex]

[tex]n=5.62\times 10^{21}[/tex]

So, the number of electrons pass through the resistor is [tex]5.62\times 10^{21}[/tex]. Hence, this is the required solution.

Answer:

Q = 116.8 J

Explanation:

Here given that the temperature of 1 L hydrogen is increased by 90 degree C at constant pressure condition.

So here we will have

[tex]Q = n C_p \Delta T[/tex]

here we know that

n = number of moles

[tex]n = \frac{1}{22.4}[/tex]

[tex]n = 0.0446[/tex]

for ideal diatomic gas molar specific heat capacity at constant pressure is given as

[tex]C_p = \frac{7}{2}R[/tex]

now we have

[tex]Q = (0.0446)(\frac{7}{2}R)(90)[/tex]

[tex]Q = 116.8 J[/tex]

Answer:

a)

14 rad/s

b)

11.2 m/s

Explanation:

a)

d = diameter of tire = 0.80 m

r = radius of tire = (0.5) d = (0.5) (0.80) = 0.40 m

v = speed of bicycle = 5.6 m/s

w = angular speed of the tire

Speed of cycle is given as

v = r w

5.6 = (0.40) w

w = 14 rad/s

b)

v' = speed of blue dot

Speed blue of dot is given as

v' = v + rw

v' = 5.6 + (0.40) (14)

v' = 11.2 m/s

The angular speed of the bicycle tire is 14 rad/s. The speed of the blue dot when it is 0.80 m above the road is the same as the bicycle's speed which is 5.6 m/s.

This is a problem which involves the calculation  of the angular speed and linear speed of an object in circular motion, here the circular motion being the rotation of a bicycle tire.

Given the linear speed of the bicycle and the radius of the tire, we would use the equation that links linear speed v and angular speed a, expressed as v = ra. We can find angular speed by rearranging this formula as a = v/r. Hence, we can calculate the angular speed of the tire as 5.6 m/s divided by the radius of the tire, which is half of the diameter, so 0.4m, which equals 14 rad/s.

Next, when a point on the tire (the blue dot) is 0.8 m above the ground it is at the top of tire's circular path, so its speed is equivalent to the linear speed of the bicycle, 5.6 m/s. This is because at this point the dot is not in contact with the ground hence it isn't stationary relative to the ground unlike the point at the bottom of the tire.

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