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4 years ago

What is the best description of a wave

Physics

1 answer:

EleoNora [17]4 years ago

5 0

Answer:

In physics, mathematics, and related fields, a wave is a disturbance of a field in which a physical attribute oscillates repeatedly at each point or propagates from each point to neighboring points, or seems to move through space. The waves most commonly studied in physics are mechanical and electromagnetic.

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Ryan places 0.150 kg of boiling water in a thermos bottle. How many kgs of ice at –12.0 °C must Ryan add to the thermos so that

Greeley [361]

Answer:

The value is $m_i = 0.0234 \ kg$

Explanation:

Generally from the calorimetry principle we have that

$Heat \ loss\ by\ ice = heat\ gained\ by\ water\$

So here heat gained water is mathematically represented as i.e

$Q_w = m_w * c_w * (T_w - T )$

substituting 0.150 kg for $m_w$ , 4200 J/kg.°C for $c_w$ , 100°C for $T_w$ and 75°C for T

We have

$Q_w = 0.150 * 4200 * (100 - 75 )$

$Q_w =15750 \ J$

The Heat loss by the ice is mathematically represented as

$Q_i = Q_1 + Q_2 + Q_3$

Here $Q_1$ is the energy to move the ice to its melting point which is evaluated as

$Q_1 = m_i * c_i * ( T_o -T_i)$

Here $m_i$ is the mass of ice

$c_i$ is the specific heat of ice with value $2.05 * 10^3 J/kg.°C$

$T_o$ temperature of ice at melting point with value 0°C

$T_i$ is the temperature of ice with value -12°C

$Q_2$ is the energy to move the ice from its its melting point to liquid which is evaluated as

$Q_2 = m_i * L$

Here L is the Latent heat of melting of ice with value $334 * 10^3 J/kg$

$Q_3$ is the energy to move the ice from liquid to the equilibrium temperature which is evaluated as

$Q_1 = m_i * c_w * ( T -T_o)$

$Q_i = m_i [ c_i * ( T_o -T_i) + L + c_w * ( T -T_o) ]$

=> $Q_i = m_i [ 2.05 * 10^3 * ( 0 -(-12)) + 334 * 10^3 + 4200 * ( 75 - 0) ]$

From

$Heat \ loss\ by\ ice = heat\ gained\ by\ water\$

We have that

$m_i * 673600 =15750$

=> $m_i = \frac{15750}{673600}$

=> $m_i = 0.0234 \ kg$

8 0

4 years ago

A car going around a curve with a radius of 15m experiences a centripetal acceleration of 2.5 m/s2. How fast is the car going?

prisoha [69]

Answer:

Explanation:

The formula for centripetal acceleration is

$a_c=\frac{v^2}{r}$ . Filling in:

$2.5=\frac{v^2}{15}$ and

$v=\sqrt{2.5(15)}$ so

v = 6.1 m/s

8 0

4 years ago

Which type of graph would be best for showing the percentage of people in a family with different jobs? A. Circle graph B. Scatt

musickatia [10]

Answer:

Explanation:

A circle graph typically represents numbers in percentages, used to visualize a part to whole relationship or a composition

6 0

4 years ago

A race car is on a track has a velocity of 200 m/s. 15 seconds later the driver notices that he is going 350 m/s. What is his ac

shtirl [24]

Answer:

initial velocity (u) = 200m/s

final velocity (v) = 350 m/s

time (t) = 15s

acceleration (a) = ?

NOW,

a=v-u/t

a= 350-200/15

a= 50/15

a= 3.3333

Explanation:

it's too easy just u need to understand the question . and go according to it's content .

main thing to memorize is it's simple formula.

I HAVE SOLVE THIS QNA. IN VERY

SIMPLE AND UNDERSTANDABLE FORM.

I høpë u hađ uņdērstøöď ťhìs şølutîóñ

:verý ×wəłł.

8 0

4 years ago

In a laboratory on earth, all the air is pumped from a large tube. a feather and a steel ball are simultaneously released from r

Vlad [161]

By pumping out all air from a large tube, we basically formed an environment that is vacuum in nature inside the tube. So, air resistance is no longer a factor in the descent of the feather and the steel ball. These objects will fall at the same time to the end of the tube.

8 0

3 years ago