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3 years ago

Write with all the steps and formulas and drawing if needed.

Physics

1 answer:

lianna [129]3 years ago

7 0

The answer is not contained detail explanation, just a solution and the required values.

All the details are in the pictures, the answers are marked with orange colour.

Note,

in the task no 20.:

$m_A - the \ mass \ of \ A; \ m_B-the \ mass \ of \ B \ balls.\\V_A \ and \ V_B-the \ velocities \ of \ the \ A&B \ balls \ before \ collision.\\V'_A \ and \ V'_B-the \ velocities \ of \ the \ A&B \ balls \ after \ collision.$

V - the velocity of the pair of the balls after collision.

in the task no 21:

m₁ - the mass of the copper ball; m₂ - the mass of the copper calorimeter; m₃ - the mass of the water; t₀ - the initial temperature of water in the copper calorimeter; θ - the final temperature in the calorimeter after the copper ball is transferred into a copper calorimeter; t₁ - the required initial temperature of the copper ball before it is transferred into the calorimeter.

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Which example best illustrates the transfer of energy between two waves?

Pachacha [2.7K]

Answer:

Vibrations of electric and magnetic fields.

Vibration of air particles

Vibration of the water particles.

Explanation:

We have here two groups of waves. Electromagnetic wave and mechanical waves.

For the first one, electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields.
For the second group, mechanical waves as a sound, for instance, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. Or In water waves, energy is transferred through the vibration of the water particles.

I hope it helps you!

4 0

4 years ago

Read 2 more answers

Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance

PSYCHO15rus [73]

Answer:

No she cannot.

Explanation:

Let $v_h$ be the horizontal component of the ball velocity when it's kicked, assume no air resistance, this is a constant. Also let $v_v$ be the vertical component of the ball velocity, which is affected by gravity after it's kicked.

The time it takes to travel 95m accross the field is

$t = 95 / v_h$ or $v_h = 95/t$

t is also the time it takes to travel up, and the fall down to the ground, which ultimately stops the motion. So the vertical displacement after time t is 0

$s = v_vt + gt^2/2= 0$

where g = -9.8m/s2 in the opposite direction with $v_v$

$v_vt - 4.9t^2 = 0$

$v_vt = 4.9t^2$

$v_v = 4.9t$

Since the total velocity that the goal keeper can give the ball is 30m/s

$v = v_v^2 + v_h^2 = 30^2 = 900$

$(4.9t)^2 + \left(\frac{95}{t})^2 = 900$

$24.01t^2 + \frac{9025}{t^2} = 900$

Let substitute x = $t^2$ > 0

$24.01 x + \frac{9025}{x} = 900$

We can multiply both sides by x

$24.01 x^2 + 9025 = 900x$

$24.01x^2 - 900x + 9025 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{900\pm \sqrt{(-900)^2 - 4*(24.01)*(9025)}}{2*(24.01)}$

As $(-900)^2 - 4*24.01*9025 = -56761 < 0$

The solution for this quadratic equation is indefinite

So it's not possible for the goal keeper to do this.

6 0

3 years ago

The ______________ scale has a single “anchor” point at the triple point of water.

Marysya12 [62]

The "Temperature" scale has a single "anchor" point at the triple point of water.

In short, Your Answer would be "Temperature"

Hope this helps!

8 0

4 years ago

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A) oil B) water C) vacuum D) diamond

Lesechka [4]

Answer:

its diamond ......................................

Explanation:

7 0

2 years ago

A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the stri

Dmitrij [34]

Answer:

(a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

Explanation:

Given that,

Radius = 8.00 cm

Mass = 0.180 kg

Height = 75.0 m

We need to calculate the angular speed of the rotating

Using conservation of energy

$\dfrac{1}{2}I\omega_{1}^2+\dfrac{1}{2}mv_{1}^{2}+mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

Here, initial velocity and angular velocity are equal to zero.

$mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

$mg(h_{1}-h_{2})=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}$

$mgH=\dfrac{1}{2}mr^2\omega_{2}^2+\dfrac{1}{2}m(r\omega_{2})^2$

Here, $H = h_{1}-h_{2}$

$gH=r^2\omega_{2}^2$

$\omega_{2}^2=\dfrac{gH}{r^2}$

$\omega_{2}=\sqrt{\dfrac{9.8\times75.0\times10^{-2}}{(8.00\times10^{-2})^2}}$

$\omega_{2}=33.8\ rad/s$

The angular speed of the rotating is 33.8 rad/s.

(b). We need to calculate the speed of its center

Using formula of speed

$v=r\omega$

Put the value into the formula

$v=8.00\times10^{-2}\times33.8$

$v=2.7\ m/s$

Hence, (a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

3 0

3 years ago

Read 2 more answers

Write with all the steps and formulas and drawing if needed.​

Write with all the steps and formulas and drawing if needed.