Explain the expression “light travels at a speed”.

The latent heat of fusion of alcohol is 50 kcal/kg and its melting point is -54oC. It has a specific heat of 0.60 in its liquid

sashaice [31]

Answer:

C

Explanation:

To melt the alcohol

Heat needed = M . L = 2 . 25 = 50 kcal

To warm up the alcohol

Heat needed = M . sp. ht. . ∆t = 2 . 0.6 . 100 = 120 kcal

Total heat needed = 170 kcal

Assuming that 0.6 kcal/ kg / ˚C is the specific heat and that the answer is wanted in kcal ( a rather odd unit to be in use here.)

5 0

3 years ago

What happens to the volume of a loaf of bread that is squeezed? The mass? The density?

nalin [4]

Explanation:

The volume of the bread decreases, making the bread appear more compact, and smaller in size. The mass stays the same, it won't change unless part of the bread is removed. The density increases, the air bubbles inside of the bread get squished down, causing the bread to be smaller, and in turn, causing it to be more solid.

I hope this helped!

Thanks!

Your friend in answering,

~Steve

3 0

3 years ago

This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a

Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = ?

t = time taken for decomposition = 3 days

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = $\frac{58}{100}\times 100=58g$

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

$k=\frac{2.303}{3}\log\frac{100}{58}$

$k=0.18days^{-1}$

a) Half-life of radon-222:

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days$

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant = $0.18days^{-1}$

t = time taken for decomposition = ?

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = $\frac{15}{100}\times 100=15g$

$t=\frac{2.303}{0.18}\log\frac{100}{15}$

$t=10.54days$

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0

3 years ago

A chef sanitized a thermometer probe and then checked the temperature of minestrone soup being held in a hot-holding unit. the t

ss7ja [257]

Answer: Option (A) is the correct answer.

Explanation:

A corrective action is defined as the action with the help of which a person can avoid a difficulty or problem that he/she was facing earlier.

For example, when the chef checked the temperature of soup using thermometer then it was 120 but his operation's critical limit was 135.

So, to avoid this problem he heated the soup to 165 at 15 seconds following which he got the result as desired.

Therefore, reheating the soup was his corrective action.

Thus, we can conclude that reheating the soup was the corrective action.

4 0

3 years ago

The lawn sprinkler consists of four arms that rotate in the horizontal plane. The diameter of each nozzle is 8 mm, and the water

sashaice [31]

Answer: the constant angular velocity of the arms is 86.1883 rad/sec

Explanation:

First we calculate the linear velocity of the single sprinkler;

Area of the nozzle = π/4 × d²

given that d = 8mm = 8 × 10⁻³

Area of the nozzle = π/4 × (8 × 10⁻³)²

A = 5.024 × 10⁻⁵ m²

Now total discharge is dived into 4 jets so discharge for single jet will be;

Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec

So using continuity equation ;

Q_single = A × V_single

V_single = Q_single/A

we substitute

V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)

V_single = 29.8566 m/s

Now resolving the forces as shown in the second image,

Vt = Vcos30°

Vt = 29.8566 × cos30°

Vt = 25.8565 m/s

Finally we calculate the angular velocity;

Vt = rω

ω_single = Vt / r

from the given diagram, radius is 300mm = 0.3m

so we substitute

ω_single = 25.8565 / 0.3

ω_single = 86.1883 rad/sec

Therefore the constant angular velocity of the arms is 86.1883 rad/sec

7 0

3 years ago