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2 years ago

Why do chillies burn even through they’re cold?

2 answers:

8 0

Well, Chillies burning in your mouth doesn't really deal with the temperature. It is because of the spices in it. This makes it feels hot in your lips and sometimes you would feel like rubbing lipstick on. Just to reduce the hotness. ;D

tigry1 [53]2 years ago

3 0

When chilies burn in our mouths, the surrounding temperature does not affect how hot the taste of the chilli is. It is only the spiciness that we taste in our mouths. This spiciness is caused by the chemical called capsaicin. This chemical is contained in the cells of the chilli. Capsaicin irritates the cells in your mouth so that is why you feel the discomfort in your mouth.

Hope it helps :)

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A ball has a mass of 2 kg and is thrown with a force of 8 Newtons for .35 seconds. What is the ball's change in

Anna [14]

Answer:

1.4s

Explanation:

Given parameters:

Mass of ball = 2kg

Force = 8N

Time = 0.35s

Unknown:

Change in velocity = ?

Solution:

To solve this problem, we use the expression obtained from Newton's second law of motion which is shown below:

Ft = m(v - u)

So;

Ft = m Δv

F is the force

t is the time

m is the mass

Δv is the change in velocity

8 x 0.35 = 2 x Δv

Δv = 1.4s

6 0

2 years ago

Two charges, -20 C and +4.7 C, are fixed in place and separated by 3.0 m. a. At what spot along a line through the charges is th

Pani-rosa [81]

The zero net electric field point is at a point that is 0.98 m away from 4.7C charge.If a 14C charge is placed at this point then, force acted on the charge placed at this point is equal to zero.

Explanation:

Let at A both net electric field is zero then

At A ,E1=E2

E1=k*Iq1I / (d+x)^2

E2=k*Iq2I /x^2

Equating both

6 0

3 years ago

b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul

Nat2105 [25]

Answer:

b) Betelgeuse would be $\approx 1.43 \cdot 10^{6}$ times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is $\approx 1.37 \cdot 10^{-5} }$ the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

$B = \displaystyle{\frac{L}{4\pi d^2}}$ (1)

where $B$ is the brightness, $L$ is the star luminosity and $d$ , the distance from the star to the point where the brightness is calculated (measured). Thus:

b) $B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}}$ and $B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}}$ where $L_{Sun}$ is the Sun luminosity ( $3.9 x 10^{26} W$ ) but we don't need to know this value for solving the problem. $ly$ is light years.

Finding the ratio between the two brightness we get:

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }$

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is $1.581 \cdot 10^{-5}\ ly$ . Then

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }$

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0

3 years ago

Someone please help??

stira [4]

The efficancy is .375

5 0

3 years ago

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves

MArishka [77]

Answer:

Explanation:

Given,

Work done by the rope 900 m/s.
Angle of inclination of the slope = $\theta\ =\ 12^o$
Initial speed of the skier = v = 1.0 m/s
Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

$\therefore W_r\ =\ W_g\ =\ 900\ J$

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt$

Part (c)

Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

4 0

3 years ago