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denis-greek [22]
3 years ago
6

1. Which pair of physical quantities consists of two vectors?

Physics
2 answers:
Tems11 [23]3 years ago
8 0

Answer:

Velocity and force consists of two vectors as they both have magnitude and direction

Anni [7]3 years ago
6 0

Answer:

velocity and force

hope this helps

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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
3 years ago
Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat
nikklg [1K]

Answers:

a) T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds V are given by:

V=\omega L=\frac{2\pi}{T} L (1)

Where:

\omega is the angular frequency

L is the radius of the orbit of each satellite

T is the period of the orbit of each satellite

Isolating T:

T=\frac{2 \pi L}{V} (2)

Applying this equation to each satellite:

T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s (3)

T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s (4)

T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s (5)

T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s (6)

T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s (7)

T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s (8)

Ordering this periods from largest to smallest:

T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}

b) Acceleration a is defined as the variation of velocity in time:

a=\frac{V}{T} (9)

Applying this equation to each satellite:

a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2} (10)

a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2} (11)

a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2} (12)

a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2} (13)

a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2} (14)

a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2} (15)

Ordering this acceerations from largest to smallest:

a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}

4 0
3 years ago
What type of mixture does the foam demonstrate?
Ksenya-84 [330]
This question doesn't appear to be complete
4 0
3 years ago
Una patinadora de 50 kg parte del reposo y después de recorrer 3k alcanza una velocidad de 15 m/s. ¿Qué fuerza neta experimenta
e-lub [12.9K]

Answer:

F_{net} = 1.875\,N

Explanation:

Asúmase que la patinadora experimenta una aceleración constante. La fuerza neta experimentada por la patinadora:

F_{net} = (50\,kg)\cdot \left[\frac{\left(15\,\frac{m}{s}\right)^{2}-\left(0\,\frac{m}{s}\right)^{2} }{2\cdot (3000\,m)} \right]

F_{net} = 1.875\,N

6 0
3 years ago
17.
Zolol [24]

Answer:

I don't know this answer at all

Explanation:

I don't know about these problems

6 0
3 years ago
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