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3 years ago

When cold milk

Physics

1 answer:

monitta3 years ago

7 0

Answer:

See explanation below.

Explanation:

The internal energy (kinetic energy) of the milk particles increases thus warming the milk up, while the internal energy of the coffee particles reduce their initial average speed as they are imparting energy to the slow milk particles.

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In order to be considered a semi-conductor the material must

wel

First one, for instance they become conductors or insulators depending on the temperature!

4 0

4 years ago

Read 2 more answers

A spherical snowball melts at a rate proportional to its surface area. (a) write a differential equation for its volume, v. (use

tresset_1 [31]

Answer:

Explanation:

The rate of change in volume is proportional to the surface area:

dV/dt = kA

Integrating:

V = kAt + C

At t=0, V = s, so:

s = kA(0) + C

C = s

Therefore:

V = kAt + s

6 0

3 years ago

Which of the following would decrease in size during the contraction of a sarcomere? The width of the I-bands The width of the A

ANEK [815]

Hi!

The correct answer would be: the width of I-bands

The sacromere is the smallest contractile unit of striated muscles. These units comprise of filaments (fibrous proteins) that, upon muscle contraction or relaxation, slide past each other. The sacromere consists of thick filaments (myosin) and thin filaments (actin).

Refer to the attached picture to clearly see the structure of a sacromere.

When a sacromere contracts, a series of changes take place which include:

- Shortening of I band, and consequently the H zone

- The A line remains unchanged

- Z lines come closer to each other (and this is due to the shortening of the I bands)

The only changes that take place occur in the zones/areas in the sacromere (as mentioned), not in the filaments (actin and myosin) that make the up the sacromere; hence all other options are wrong.

Hope this helps!

8 0

3 years ago

A wire of length 6cm makes an angle of 20° with a 3 mT

Crazy boy [7]

Answer:

Approximately $7.3 \times 10^{-3}\; \rm A$ (approximately $7.3\; \rm mA$ ) assuming that the magnetic field and the wire are both horizontal.

Explanation:

Let $\theta$ denote the angle between the wire and the magnetic field.

Let $B$ denote the magnitude of the magnetic field.

Let $l$ denote the length of the wire.

Let $I$ denote the current in this wire.

The magnetic force on the wire would be:

$F = B \cdot l \cdot I \cdot \sin(\theta)$ .

Because of the $\sin(\theta)$ term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is $90^\circ$ .)

In this question:

$\theta = 20^\circ$ (or, equivalently, $(\pi / 9)$ radians, if the calculator is in radian mode.)
$B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T$ .
$l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m$ .
$F = 1.5\times 10^{-4}\; \rm N$ .

Rearrange the equation $F = l \cdot I \cdot \sin(\theta)$ to find an expression for $I$ , the current in this wire.

$\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}$ .