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Vhich type of reaction is modeled by this chemical equation? PLi + CaCl2 - 2LICI + Ca

san4es73 [151]

Answer:

Double-replacement

Explanation:

2Li + CaCl2 → 2LiCl + Ca O

D. Double-replacement

7 0

2 years ago

You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the "oars") toward you, it moves a d

gulaghasi [49]

Answer: 94.6 N

Explanation:

P = F d / t

82 = F (1.3) / 1.5

F = (82 x 1.5) / 1.3

F = 94.6N

4 0

3 years ago

Hanging from a horizontal beam are nine simple pendulums of the following lengths: (a) 0.080, (b) 0.26, (c) 0.49, (d) 0.90, (e)

mote1985 [20]

Answer:

Explanation:

This is the case of forced oscillation . The pendulum having the same or matching time period or angular frequency with that of angular frequency of external periodic force , will be in resonance having largest amplitude.

Angular frequency of pendulum having length .9 m

= $\sqrt{\frac{g}{l } }$

l = .9

angular frequency

= $\sqrt{\frac{10}{0.9 } }$

= 3.33 rad / s

If we calculate angular frequencies of pendulum of all lengths given , we will find that other lengths do not give angular frequency falling between 2 and 4 radian . So only pendulum having length of .9 m will have vibration of maximum amplitude.

8 0

3 years ago

Which model below represents a pure substance?

san4es73 [151]

Figure 1, because it is a compound.

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Chuge%20%5Cfrak%7B%E0%BC%86%20%20%5C%3A%20%20%5C%3A%20question%20%5C%3A%20%20%5C%3A%20%E0%BC

My name is Ann [436]

The crate is moving at constant velocity when the forces acting on it are

balanced.

The value of the force F required to pull the crate with constant velocity is; $\underline{F = \sqrt{2} \cdot \mu \cdot m\cdot g}$

Reasons:

Mass of the crate = m

Cross section of through = Right angled

Orientation (inclination) of the through to horizontal = 45°

Coefficient of kinetic friction = μ

Required:

The value of the required to pull the crate along the through at constant

velocity.

Solution:

When the through is moving at constant velocity, we have;

Friction force acting on crate = Force pulling the crate

Friction force = Normal reaction × Coefficient of kinetic friction

Normal reaction on an inclined plane = $\mathbf{F_N}$

Each side of the through gives a normal reaction.

The vertical component of the normal reaction on each side of the through

is therefore;

$F_N$ ·j = $\mathbf{F_N}$ × sin(θ)

The sum of the vertical component = $F_N$ ·j + $F_N$ ·j = 2· $F_N$ ·j = 2· $F_N$ ×sin(θ)

The sum of the vertical component of the normal reactions = The weight of the crate

Therefore;

2· $F_N$ ×sin(θ) = m·g

θ = 45°

Therefore;

2· $F_N$ ×sin(45°) = m·g

$\displaystyle sin(45^{\circ}) = \mathbf{\frac{\sqrt{2} }{2}}$

Therefore;

$\displaystyle 2 \cdot F_N \cdot sin(45^{\circ}) = 2 \cdot F_N \times \frac{\sqrt{2} }{2} = \sqrt{2} \cdot F_N$

$\displaystyle F_N = \mathbf{ \frac{m \cdot g}{\sqrt{2} }}$

Which gives;

$\displaystyle Force \ required, \ F = Sum \of \ friction \ forces \ = 2 \times F_N \times \mu = \mathbf{ 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu}$

$\displaystyle Force \ required, \ F = 2 \times \frac{m \cdot g}{\sqrt{2} } \times \mu = \mathbf{ \sqrt{2} \cdot \mu \cdot m \cdot g}$

Force required to pull the crate at constant velocity, F = √2·μ·m·g

Learn more about force of friction here:

brainly.com/question/6561298

8 0

2 years ago

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