Kevin decides to soup up his car by replacing the car's wheels with ones that have 1.4 times the diameter of the original wheels. Note that the speedometer in a car is calibrated based on the tire's diameter and on the distance the tire covers in each revolution. (a) Will the reading of the speedometer change

The image of this hollow sphere and uniform rod is missing, so i have attached it.

Answer:

A) J = 0.7443 kg•m²

B) T = 1.9169 N•m CCW

C) α = 2.5754 rad/s²

D) a = 3.966 m/s²

Explanation:

A) The moment of inertia J of the contraption around the fulcrum is given by the formula;

J = Jℓ + Jr

Let's calculate Jℓ

Jℓ = [((0.34²/3) × 0.50 × 0.34)/1.78] + (0.48 × (0.34 + 0.64)²)

Jℓ = 0.4647 kg•m²

Now, let's Calculate Jr

Jr = ((1.78 - 0.34)²/3) × ((1.78 - 0.34)/1.78) × 0.50

Jr = 0.2796 kg•m²

Thus;

J = 0.4647 + 0.2796

J = 0.7443 kg•m²

(b) Using CCW as positive, Torque in Nm is calculated as;

T = Tℓ - Tr

Let's calculate Tℓ

Tℓ = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81

Tℓ = 4.7739 N•m CCW

Now, let's Calculate Tr;

Tr = [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81

Tr = 2.857 N•m CW

Thus;

T = 4.7739 - 2.857

T = 1.9169 N•m CCW

(c) The angular acceleration α of the contraption, using CCW is gotten from the formula;

α = T/J

α = 1.9169/0.7443

α = 2.5754 rad/s²

(d) The linear acceleration a of the right end of the rod, using up as positive is given by;

a = α*(1.78 - 0.34)

a = 2.5754 × 1.54

a = 3.966 m/s²

A) the moment of inertia of the contraption is 0.7443 kgm²

B) The torque about the fulcrum is 1.9169 Nm

C) Angular acceleration of the contraption is 2.5754 rad/s²

D) The linear acceleration of the contraption is 3.966 m/s²

(A) The moment of inertia I of the contraption around the fulcrum is given by :

[tex]I = [(0.34^2/3) \times 0.50 \times 0.34)/1.78 + (0.48 \times (0.34 + 0.64)^2)] + [(1.78 - 0.34)^2/3) \times (1.78 - 0.34)/1.78) \times 0.50][/tex]

I = 0.4647 + 0.2796

I = 0.7443 kgm²

(B) Using CCW as positive, Torque in Nm is given by;

T = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81 - [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81

T = 4.7739 - 2.857

T = 1.9169 Nm

(C) The angular acceleration (α) of the contraption is given by:

α = T/I

since, torque is defined as T = Iα

α = 1.9169/0.7443

α = 2.5754 rad/s²

(D) The linear acceleration (a) of the right end of the rod

a = αr

where r is the distance from the pivot

a = α × (1.78 - 0.34)

a = 2.5754 × 1.54

a = 3.966 m/s²

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Answer:

5.0/s

Explanation:

Answer:

b and a it is this that abewsr

Answer:

they were fast ⛷⛷

Answer:

Heat loss per unit length = 642.358 W/m

The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m

Explanation:

From the information given:

Diameter D [tex]= 100 mm = 0.1 m[/tex]

Surface emissivity ε = 0.8

Temperature of steam [tex]T_s[/tex] = 150° C = 423K

Atmospheric air temperature [tex]T_{\infty} = 20^0 \ C = 293 \ K[/tex]

Velocity of wind V = 8 m/s

To calculate average film temperature:

[tex]T_f = \dfrac{T_s+T_{\infty}}{2}[/tex]

[tex]T_f = \dfrac{423+293}{2}[/tex]

[tex]T_f = \dfrac{716}{2}[/tex]

[tex]T_f = 358 \ K[/tex]

To calculate volume expansion coefficient

[tex]\beta= \dfrac{1}{T_f} \\ \\ \beta= \dfrac{1}{358} \\ \\ \beta= 2.79 \times 10^{-3} \ K^{-1}[/tex]

From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;

Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s

Thermal conductivity k = 30.608 × 10⁻³ W/m.K

Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s

Prandtl no. Pr = 0.698

Rayleigh No. for the steam line is determined as follows:

[tex]Ra_{D} = \dfrac{g \times \beta (T_s-T_{\infty}) \times D_b^3}{\alpha\times v}[/tex]

[tex]Ra_{D} = \dfrac{9.8 \times (2.79 *10^{-3})(150-20) \times (0.1)^3}{(31.244\times 10^{-6}) \times (21.7984\times 10^{-6})}[/tex]

[tex]Ra_{D} = 5.224 \times 10^6[/tex]

The average Nusselt number is:

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(Ra_D)^{1/6}}{[ 1+ (0.559/Pr)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(5.224\times 10^6)^{1/6}}{[ 1+ (0.559/0.698)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{5.0977}{[ 1.8826]^{8/27}}\Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2[/tex]

[tex]Nu_D = 23.29[/tex]

However, for the heat transfer coefficient; we have:

[tex]h_D = \dfrac{Nu_D\times k}{D_b} \\ \\ h_D = \dfrac{(23.29) \times (30.608 \times 10^{-3} )}{0.1}[/tex]

[tex]h_D = 7.129 \ Wm^2 .K[/tex]

Hence, Stefan-Boltzmann constant [tex]\sigma = 5.67 \times 10^{-8} \ W/m^2.K^4[/tex]

Now;

To determine the heat loss using the formula:

[tex]q'_b = q'_{ev} + q'_{rad} \\ \\ q'_b = h_D (\pi D_o) (T_t-T_{\infty})+\varepsilon(\pi D_b)\sigma (T_t^4-T_{\infty }^4)[/tex]

[tex]q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^{-8}) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}[/tex]

Now; here we need to determine the Reynold no and the average Nusselt number:

[tex]Re_D = \dfrac{VD_b}{v } \\ \\ Re_D = \dfrac{8 *0.1}{21.7984 \times 10^{-6}} \\ \\ Re_D = 3.6699 \times 10^4[/tex]

However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;

[tex]Nu_D = 0.3 + \dfrac{0.62 \times Re_D^{1/2}* Pr^{1/3}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\dfrac{Re_D}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = 0.3 + \dfrac{0.62 \times (3.6699*10^4)^{1/2}* (0.698)^{1/3}}{[1+(0.4/0.698)^{2/3}]^{1/4}} [1+ (\dfrac{3.669*10^4}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = (0.3 +\dfrac{105.359}{1.140}\times 1.218) \\ \\ Nu_D = 112.86[/tex]

SO, the heat transfer coefficient for forced convection is determined as follows afterward:

[tex]h_D = \dfrac{Nu_{D}* k}{D_b} \\ \ h_D = \dfrac{112.86*30.608 *10^{-3}}{0.1} \\ \\ h_D = 34.5 \ W/m^2 .K[/tex]

Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:

[tex]q'b = h_D (\pi D_b) (T_s-T_{\infty}) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^{-8}) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}[/tex]

Answer:

[tex]3.52\times 10^{25}\ \text{J}[/tex]

Explanation:

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]

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

[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]

[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]

Energy required is given by

[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]

The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].

Answer:

momentum=mass x velocity= 10 x 2 = 20kgm/s

Answer: 2000 J

Explanation: work W = F s

Answer:

If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.

Explanation:

Explanation:

रात में घूमने वाला arthaarat निशाचर

Answer:

(a) The volume of water is 100 cm³

(b) The volume of the rock is 20 cm³

(c) The density of the rock is 30 g/cm³

Explanation:

The given parameters of the perspex box are;

The area of the base of the box, A = 10 cm²

The initial level of water in the box, h₁ = 10 cm

The mass of the rock placed in the box, m = 600 g

The final level of water in the box, h₂ = 12 cm

(a) The volume of water in the box, 'V', is given as follows;

V = A × h₁

∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³

The volume of water in the box, V = 100 cm³

(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the  water, V, is given as follows;

[tex]V_T[/tex] = [tex]V_r[/tex] + V

[tex]V_T[/tex] = A × h₂

∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³

The total volume, [tex]V_T[/tex] = 120 cm³

The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V

∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³

The volume of the rock, [tex]V_r[/tex] = 20 cm³

(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)

∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³

Answer:

A. The bullet with 0.006kg has more energy

B. When the mass is doubled the kinetic energy increases

Explanation:

Kinetic energy increases when mass increases

kinetic energy increases when velocity increases

Answer:

everyone on the nighttime side of Earth is seeing it.

Explanation:

A lunar eclipse is a phenomenon that occurs when the Earth comes between the Moon and the Sun thereby causing it to cover the Moon with its shadow.

Simply stated, lunar eclipse takes place when the Moon passes or moves through the Earth's shadow thereby blocking any ray of sunlight from reaching the Moon. Thus, the full moon appears deep red (blood moon).

Also, a lunar eclipse would occur only when the Sun, Earth, and Moon are closely aligned to form a straight line known as the syzygy.

There are three (3) types of lunar eclipse and these are;

1. Total lunar eclipse.

2. Partial lunar eclipse.

3. Penumbra lunar eclipse.

Generally, if an observer on Earth sees a total lunar eclipse, everyone on the nighttime side of Earth is seeing it because it's quite easy to see a total lunar eclipse while the full moon passes through the innermost part of the shadow of the earth.

Answer:

the knee extensors must exert 15.87 N

Explanation:

Given the data in the question;

mass m = 4.5 kg

radius of gyration k = 23 cm = 0.23 m

angle ∅ = 30°

∝ = 1 rad/s²

distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m

using the expression;

ζ = I∝

ζ = mk²∝

we substitute

ζ = 4.5 × (0.23)² × 1

ζ  = 0.23805 N-m

so

from; ζ = rFsin∅

F = ζ / rsin∅

we substitute

F = 0.23805 / (0.03 × sin( 30 ° )

F = 0.23805 / (0.03 × 0.5)

F F = 0.23805 / 0.015

F = 15.87 N

Therefore, the knee extensors must exert 15.87 N

Answer:

8.45 m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 90 Kg

Initial velocity (u) = 13 m/s

Final velocity (v) = 0 m/s

Height (h) =?

NOTE: Acceleration due to gravity (g) = 10 m/s²

The height of the hill can be obtained as follow:

v² = u² – 2gh (since the cart is going against gravity)

0² = 13² – (2 × 10 × h)

0 = 169 – 20h

Rearrange

20h = 169

Divide both side by 20

h = 169/20

h = 8.45 m

Therefore, the height of the hill is 8.45 m

The student does zero work on the backpack because the upward force applied by the student is acting perpendicular to the backpack's displacement parallel to the ground.

Answer:

the potential energy decreases as it is converted to kinetic energy.

Explanation:

As things move, their potential energy converts to kinetic energy to power them along. When a ball rolls down the top of a ramp, all the potential energy it accumulated at the top of the ramp converts to kinetic energy to help it roll down. In other words, its potential energy decreases as its kinetic energy increases.

Answer:

Steel

Explanation:

Steel is a metal that magnets stick to because iron can be found inside steel

Answer:Magnets stick to any metal that contains iron, cobalt or nickel.

Explanation:Iron is found in steel, so steel attracts a magnet and sticks to it. Stainless steel, however, does not attract a magnet.

Answer:

1115560000 J

Explanation:

1/2 * 80,000 * 167^2 m/s = 1115560000 J

Answer:

decomposition

Explanation:

Answer:

1) 1000Nm

2)  95,625Nm

3) 1.05%

Explanation:

Mechanical Advantage is the ratio of the load  to the effort applied to an object.

MA = Load/Effort

1) Workdone on the load = Force(Load) * distance covered by the load

Workdone on the load = 500N * 2m

Workdone on the load = 1000Nm

2)  work done by the effort = Effort * distance moves d by effort

work done by the effort = 2125 * 45

work done by the effort = 95,625Nm

3) Efficiency = Workdone on the load/ work done by the effort * 100

Efficiency = 1000/95625 * 100

Efficiency = 1.05%

Hence the efficiency of the system is 1.05%

Answer:

8 feet

................

Answer:

The movement of Sun and moon

Explanation:

When the sun rise.it is am and when it sets .it is pm.

Answer:

ks= 133.2 N/m

Explanation:

       [tex]\Delta U + \Delta K = 0 (1)[/tex]

       [tex]k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)[/tex]

Answer:

Regular reflection

Explanation:

Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

i hope this helps a bit.

According to the context, a sharp image is formed when light reflects from a regular reflection.

It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.

This reflection of light happens when the angles that the two rays determine with the surface are equal.

Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.

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