All categories

3 years ago

What to do when having a cold

Physics

2 answers:

Novay_Z [31]3 years ago

6 0

Answer:

You can ease cold symptoms by resting, drinking plenty of water and using things like decongestants and painkillers.

To help you get better more quickly:

rest and sleep

keep warm

drink plenty of water (fruit juice or squash mixed with water is OK) to avoid dehydration

gargle salt water to soothe a sore throat

BigorU [14]3 years ago

4 0

Answer:I would prefer ginger with lemon and honey tea...Stay hydrated,Rest,

Explanation:

You might be interested in

What is the layer of rocks and moon dust called

lisov135 [29]

I believe it is called the "Lunar Crust" but not 100% sure.

5 0

3 years ago

Read 2 more answers

Two pans of a balance are 46.3 cm apart. The fulcrum of the balance has been shifted 0.633 away from the center by a dishonest s

soldier1979 [14.2K]

distance of each pan from the center or fulcrum is given as

$r = 23.15 cm$

now if dishonest shopkeeper shifted it by 0.633 cm from center

so distance on each side is given as

$d_1 = 23.15 - 0.633 = 22.52 cm$

$d_2 = 23.15 + 0.633 = 23.78 cm$

now the weight is balance as

$W_1d_1 = W_2d_2$

$W(22.52) = W_2(23.78)$

now we will have

$W_2 = 0.95W$

now we can find the percentage change as

$percentage = \frac{W - W_2}{W} \times 100$

$percentage = 5%$

5 0

3 years ago

A solid sphere has a temperature of 614 K. The sphere is melted down and recast into a cube that has the same emissivity and emi

krok68 [10]

Answer:581.87 K

Explanation:

Given

Sphere is melted to form a square

Let the radius of sphere be r and square has a side a

Therefore

$\frac{4\pi}{3}r^3=a^3$

Surface area of sphere $A_s=4\pi r^2$

Surface area of cube $A_c=6a^2$

Total emmisive remains same

Thus $P=A\epsilon \sigma T^4$

$A_sT_s^4=A_cT_c^4$

$\frac{T_c^4}{T_s^4}=\frac{A_s}{A_c}$

$\frac{T_c^4}{T_s^4}=\frac{1}{2}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{3}}$

$\frac{T_c}{T_s}=\frac{1}{2^{0.25}}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{12}}$

$T_c=T_s\times \frac{1}{2^{0.25}}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{12}}$

$T_c=614\times \frac{1.12679}{1.189}$

$T_c=581.87 K$

3 0

3 years ago

PLEASE HELP ILL MARK BRAINLIEST!!! A ball is initially thrown downwards with an initial speed of 20 m/s from the top of a 300 m

Sholpan [36]

Using the 3rd equation of motion:

= v² - u² = 2gs ------ [g = Acceleration due to gravity]

= v² - 20² = 2 × 10 × 300

= v² - 400 = 6000

= v² = 6000 - 400

= v = √5600

= v = 74.83 m/s

And yeah it's done :)

8 0

3 years ago

Can someone help me, it will get 5 stars

Tcecarenko [31]

vfdssmnfffffffffffffffffffffffffffffff

6 0

3 years ago

Read 2 more answers