What are the cues for a forehand stroke

which criteria are responsible for deciding whether a heterogeneous mixture is a colloid or a suspension

ad-work [718]

I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. The criteria responsible for deciding whether a heterogeneous mixture is a colloid or a suspension is whether the particles remain suspended for an extended period of time. I hope it helps you.

6 0

3 years ago

You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is

katen-ka-za [31]

The initial potential energy of the wagon containing gold boxes will enable

it roll down the hill when cut loose.

The Lone Ranger and Tonto have approximately 5.1 seconds.

Reasons:

Mass wagon and gold = 166 kg

Location of the wagon = 77 meters up the hill

Slope of the hill = 8°

Location of the rangers = 41 meters from the canyon

Mass of Lone Ranger, m₁ = 65 kg

Mass of Tonto m₂ = 66 kg

Solution;

Height of the wagon above the level ground, h = 77 m × sin(8°) ≈ 10.72 m

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Potential energy of wagon, P.E. ≈ 166 × 9.81 × 10.72 = 17457.0912

Potential energy of wagon, P.E. ≈ 17457.0912 J

By energy conservation, P.E. = K.E.

$K.E. = \mathbf{\dfrac{1}{2} \cdot m \cdot v^2}$

Where;

v = The velocity of the wagon a the bottom of the cliff

Therefore;

$\dfrac{1}{2} \times 166 \times v^2 = 17457.0912$

$v = \sqrt{\dfrac{17457.0912}{\dfrac{1}{2} \times 166} } \approx 14.5$

Velocity of the wagon, v ≈ 14.5 m/s

Momentum = Mass, m × Velocity, v

Initial momentum of wagon = m·v

Final momentum of wagon and ranger = (m + m₁ + m₂)·v'

By conservation of momentum, we have;

m·v = (m + m₁ + m₂)·v'

$\therefore v' = \mathbf{ \dfrac{m \cdot v}{(m + m_1 + m_2) }}$

Which gives;

$\therefore v' = \dfrac{166 \times 14.5}{(166 + 65 + 66) } \approx 8.1$

The velocity of the wagon after the Ranger and Tonto drop in, v' ≈ 8.1 m/s

$Time = \dfrac{Distance}{Velocity}$

$\mathrm{The \ time \ the\ Lone \ Ranger \ and \ Tonto \ have, \ t} = \dfrac{41 \, m}{8.1 \, m/s} \approx 5.1 \, s$

The Lone Range and Tonto have approximately 5.1 seconds to grab the

gold and jump out of the wagon before the wagon heads over the cliff.

Learn more here:

brainly.com/question/11888124

brainly.com/question/16492221

5 0

2 years ago

How loudness of a sound wave depends on the intensity of the sound wave?

Anestetic [448]

Loudness of a sound wave is directly proportional to the intensity of the sound wave. In other words, when one increases, other also increases and vice-versa

Hope this helps!

3 0

3 years ago

An automobile starter motor has an equivalent resistance of 0.0500Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal

alexandr1967 [171]

Answer

given,

resistance = 0.05 Ω

internal resistance of battery = 0.01 Ω

electromotive force = 12 V

a) ohm's law

V = IR

and volage

$V = \epsilon - Ir$

now,

$IR = \epsilon - Ir$

$I(R+r) = \epsilon$

$I= \dfrac{\epsilon}{R+r}$

inserting the values

$I= \dfrac{12}{0.05+0.01}$

I = 200 A

b) Voltage

V = I R

V = 200 x 0.05

V = 10 V

c) Power

P = I V

P = 200 x 10 = 2000 W

d) total resistance = 0.05 + 0.09 = 0.14 Ω

$I= \dfrac{\epsilon}{R+r}$

$I= \dfrac{12}{0.14+0.01}$

I = 80 A

V = 80 x 0.05 = 4 V

P = 4 x 80 = 320 W

6 0

3 years ago

Read 2 more answers

When the temperature of the air is 25 degress C, the velocity of a sound wave traveling through the air is approximately

dusya [7]

Assuming an ideal gas, the speed of sound depends on temperature
only. Air is almost an ideal gas.

Assuming the temperature of 25°C in a "standard atmosphere", the
density of air is 1.1644 kg/m3, and the speed of sound is 346.13 m/s.

The velocity can't be specified, since the question gives no information
regarding the direction of the sound.

6 0

3 years ago