Block A, with a mass of 4.0 kg, is moving with a speed of 3.0 m/s while block B, with a mass of 6.0 kg, is moving in the opposite direction with a speed of 5.0 m/s. What is the momentum of the two-block system

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:

Explained

Explanation:

The most dense planet in  our solar system is Earth. Earth is most dense planet because

The least dense planet of the solar system is Saturn because Saturn is mostly made of gases and its size is smaller than the Jupiter. Jupiter has more gravity hence its density is higher to Saturn. Moreover, Uranus and Neptune are ice giants. Although they are also made of gases but due their distance from sun most of these gases have solidified. Hence making them more dense than Saturn and Jupiter.

Saturn is the least dense planet, while Earth is the densest planet in our solar system.

The least dense planet in our solar system is Saturn, while the densest planet is Earth. Saturn is a gas giant composed mostly of hydrogen and helium, which have lower densities compared to the rocks and metals found on rocky planets like Earth. Earth, on the other hand, has a higher density due to its solid composition and a core made of iron and nickel.

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]

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:

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:

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

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:

(a) θ = 55.85 degree

(b) 7.89 km

Explanation:

Using vector notations

A = 1.88 km south = 1.88 (- j) km = - 1.88 j km

B = 9.05 km 47 degree north of east

B = 9.05 ( Cos 47 i + Sin 47 j) km

B = (6.17 i + 6.62 j) km

Net displacement is

D = A + B

D = - 1.88 i + 6.17 i + 6.62 j = 4.29 i + 6.62 j

(a) Angle made with positive X axis

tanθ = 6.62 / 4.29 = 1.474

θ = 55.85 degree

(b) distance = [tex]Distance = \sqrt{(4.29)^{2} + (6.62)^{2}}[/tex]

distance = 7.89 km

To find the car's final location, add the displacements in the north and east directions. The car's final location is 6.4 km north and 6.0 km east relative to the origin.

To find the car's final location, we need to add the displacements in both the north and east directions.

First, let's calculate the north displacement by using the distance formula:

North Displacement = 9.05 km * sin(47°) = 6.4 km

Next, let's calculate the east displacement by using the distance formula:

East Displacement = 9.05 km * cos(47°) = 6.0 km

Therefore, the car's final location relative to the origin is 6.4 km north and 6.0 km east.

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:

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:

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:

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:

(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:

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:

The voltage of the generator is 7.27 kV.

Explanation:

Given that,

Output of generator = 440 V

Current = 20 A

Resistance = 0.60 Ω

Power loss =0.010%

We need to calculate the total power of the generator

Using formula of power

[tex]P=VI[/tex]

Where, V = voltage

I = current

Put the value into the formula

[tex]P=440\times20[/tex]

[tex]P=8800\ W[/tex]

Th power lost on the transmission lines

[tex]P_{L}=0.010\% P[/tex]

[tex]P_{L} = 0.010\%\times8800[/tex]

[tex]P_{L}=0.88\ W[/tex]

The current passing through the transmission line

[tex]I'=\sqrt{\dfrac{P_{L}}{R}}[/tex]

[tex]I'=\sqrt{\dfrac{0.88}{0.60}}[/tex]

[tex]I'=1.211\ A[/tex]

We need to calculate the voltage of the generator

Using formula of voltage

[tex]V_{g}=\dfrac{P}{I'}[/tex]

Put the value into the formula

[tex]V_{g}=\dfrac{8800}{1.211}[/tex]

[tex]V_{g}=7.27\times10^{3}\ V[/tex]

[tex]V_{g}=7.27\ kV[/tex]

Hence, The voltage of the generator is 7.27 kV.

The voltage of the generator output needs to be stepped up with a transformer to limit power losses in transmission.

To limit power losses in transmission, the voltage of the generator output needs to be stepped up with a transformer. The formula to calculate power loss is P_loss = I^2 x R. Given that the power loss should not exceed 0.010% of the original power, we can calculate the maximum allowable power loss. By rearranging the formula, we can find the voltage required for the generator output with a transformer.

First, we calculate the original power using P = V x I. Then, we calculate the maximum allowable power loss as a percentage of the original power. Next, we substitute the given values into the formula and solve for the new voltage using the rearranged formula.

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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.

Learn more about the gravitational potential energy here;

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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:

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:

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:

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:

Part a)

x = 3.95 cm

Part b)

x = - 11.9 cm

Explanation:

Part a)

Since both charges are of same sign

so the position at which net force is zero between two charges is given as

[tex]\frac{kq_1}{r_1^2} = \frac{kq_2}{(15.8 - r)^2}[/tex]

here we know that

[tex]q_1 = Q[/tex]

[tex]q_2 = 9Q[/tex]

[tex]\frac{Q}{r^2} = \frac{9Q}{(15.8 - r)^2}[/tex]

square root both sides

[tex](15.8 - r) = 3r[/tex]

[tex]r = 3.95 cm[/tex]

Part b)

Since both charges are of opposite sign

so the position at which net force is zero will lie on the other side of smaller charges is given as

[tex]\frac{kq_1}{r_1^2} = \frac{kq_2}{(19.6 + r)^2}[/tex]

here we know that

[tex]q_1 = Q[/tex]

[tex]q_2 = -7Q[/tex]

[tex]\frac{Q}{r^2} = \frac{7Q}{(19.6 + r)^2}[/tex]

square root both sides

[tex](19.6 + r) = 2.64r[/tex]

[tex]r = 11.9 cm[/tex]

so on x axis it will be at x = - 11.9 cm

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:

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:

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:

[tex]\tau = 7.63 Nm[/tex]

Explanation:

As we know that moment of force is given as

[tex]\tau = \vec r \times \vec F[/tex]

now we have

[tex]\vec r = 1.2 m[/tex]

[tex]\vec F = 14 N[/tex]

now from above formula we have

[tex]\tau = r F sin\theta[/tex]

here we know that

[tex]\theta = 27 degree[/tex]

so we have

[tex]\tau = (1.2)(14) sin27[/tex]

[tex]\tau = 7.63 Nm[/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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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

A. C = 75μF and B. V = 10V.

We have to use the equation k = C/C₀ and k = V₀/V which both are the dielectric constant.

A. The capacitance after the slab is inserted.

With C₀ = 15μF and k = 5.0. Clear k for the equation k = C/C₀:

C = k*C₀

C = (5.0)(15x10⁻⁶F) = 0.000075F

C = 75μF

B. The voltage across the capacitor's plates after the slab is inserted.

With V₀ = 50V and k = 5.0. Clear V from the equation k = V₀/V:

V = V₀/k

V = 50V/5.0

V = 10V

Dielectric constant of the capacitor is the ratio between capacitance of capacitor before and after slab inserted.

The capacitance of capacitor is the ratio of the electric charge stored inside the capacitance to the potential difference.

The capacitance of the capacitor is 15-uF and it is connected to  50-V battery. The value of the dielectric constant is 5.0.

The dielectric constant of the slab is the ratio of capacitance of the capacitor after the slab and the capacitance of the before the slab inserted.

As the value of the dielectric constant is 5.0. Thus the capacitance of the capacitor after the slab is inserted given as,

[tex]0.5=\dfrac{C}{15\times10^{-6}}\\C=75\rm \mu F[/tex]

Thus, the capacitance of the capacitor after the slab is inserted is 75-uF.

The dielectric constant of the slab is the ratio of voltage across the capacitor's of the capacitor before the slab inserted and the voltage across the capacitor after the slab inserted.

As the value of the dielectric constant is 5.0. Thus the voltage across the capacitor's plates after the slab is inserted can be given as,

[tex]V=\dfrac{50}{5.0}\\C=10\rm V[/tex]

Thus, the voltage across the capacitor's plates after the slab is inserted is 10 volts.

Dielectric constant of the capacitor is the ratio between capacitance of capacitor before and after slab inserted.

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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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