Why are common measurement systems important

Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergo

oksian1 [2.3K]

Answer:

e). $a = 0.066 m/s^2$

Explanation:

As we know that wheel is turned by 90 degree angle

so the angular speed of the wheel is given as

$\omega_f^2 - \omega_i^2 = 2\alpha \theta$

now we have

$\omega_f^2 - 0 = 2(0.01)(\frac{\pi}{2})$

$\omega = 0.177 rad/s$

now the centripetal acceleration of the point P is given as

$a_c = \omega^2 R$

$a_c = (0.177)^2(2)$

$a_c = 0.063 m/s^2$

tangential acceleration is given as

$a_t = R\alpha$

$a_t = 2(0.01)$

$a_t = 0.02 m/s^2$

now net acceleration is given as

$a = \sqrt{a_t^2 + a_c^2}$

$a = \sqrt{0.02^2 + 0.063^2}$

$a = 0.066 m/s^2$

8 0

3 years ago

A block of gelatin is 120mm by 120mm by 40mm whrn unstressed. A force of 49N is applied tangentially to the upper surface causin

Inessa05 [86]

Answer:

The shearing stress is 10208.3333 Pa

The shearing strain is 0.25

The shear modulus is 40833.3332 Pa

Explanation:

Given:

Block of gelatin of 120 mm x 120 mm by 40 mm

F = force = 49 N

Displacement = 10 mm

Questions: Find the shear modulus, Sm = ?, shearing stress, Ss = ?, shearing strain, SS = ?

The shearing stress is defined as the force applied to the block over the projected area, first, it is necessary to calculate the area:

A = 40*120 = 4800 mm² = 0.0048 m²

The shearing stress:

$Ss=\frac{F}{A} =\frac{49}{0.0048} =10208.3333Pa$

The shearing strain is defined as the tangent of the displacement that the block over its length:

$SS=tan\theta =\frac{Displacement}{L} =\frac{10}{40} =0.25$

Finally, the shear modulus is the division of the shearing stress over the shearing strain:

$Sm=\frac{10208.3333}{0.25} =40833.3332Pa$

6 0

3 years ago

In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wra

shtirl [24]

Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

Block falls 3m in 4.6 seconds

When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

The radius of the wheel is 600mm

R = 600mm = 0.6m

We want to find the moment of inertia of the flywheel

Taking moment about point G.

Then,

Clockwise moment = Anticlockwise moment

ΣM_G = Σ(M_G)_eff

M•g•R - Mf = I•α + M•a•R

Relationship between angular acceleration and linear acceleration

a = αR

α = a / R

M•g•R - Mf = I•a / R + M•a•R

Case 1, when y = 3 t = 4.6s

M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

a = 6 / 4.6²

a = 0.284 m/s²

M•g•R - Mf = I•a / R + M•a•R

12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6

70.632 - Mf = 0.4726•I + 2.0448

Re arrange

0.4726•I + Mf = 70.632-2.0448

0.4726•I + Mf = 68.5832 equation 1

Second case

Case 2, when y = 3 t = 3.1s

M= 24kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 3.1²

3 × 2 = 3.1²a

a = 6 / 3.1²

a = 0.6243 m/s²

M•g•R - Mf = I•a / R + M•a•R

24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6

141.264 - Mf = 1.0406•I + 8.99

Re arrange

1.0406•I + Mf = 141.264 - 8.99

1.0406•I + Mf = 132.274 equation 2

Solving equation 1 and 2 simultaneously

Subtract equation 1 from 2,

Then, we have

1.0406•I - 0.4726•I = 132.274 - 68.5832

0.568•I = 63.6908

I = 63.6908 / 0.568

I = 112.13 kgm²

8 0

4 years ago

A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and

irina [24]

The eccentricity of its orbit is $U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

Mass is a physical body's total amount of matter. It also serves as a gauge for the body's inertia or resistance to acceleration (change in velocity) in the presence of a net force. The strength of an object's gravitational pull to other bodies is also influenced by its mass.
The kilogram is the SI unit of mass (kg). In science and technology, a body's weight in a given reference frame is the force that causes it to accelerate at a rate equal to the local acceleration of free fall in that frame.
For instance, a kilogram mass weighs around 2.2 pounds at the surface of the planet. However, the same kilogram mass would weigh just about 0.8 pounds on Mars and about 5.5 pounds on Jupiter.
An object's mass is a crucial indicator of how much stuff it contains. Weight is a measurement of an object's gravitational pull. It is influenced by the object's location in addition to its mass. As a result, weight is a measurement of force.

The length of the semi-major axis is calculated as follows:

where $, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$$

$M=1.99 \times 10^{30} \mathrm{~kg}=$ mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$

To learn more about mass, refer to:

brainly.com/question/3187640

#SPJ4

4 0

2 years ago

When a potassium atom forms an ion, it loses one electron. What is the electrical charge of the potassium ion? *

Ganezh [65]

+1 An electron has a negative charge so losing a charge of -1 from an uncharged, or neutral, atom will leave an ion with a positive charge.

5 0

3 years ago

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