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zavuch27 [327]
3 years ago
13

An iron ball and an aluminum ball of mass 100 g each are heated to the same temperature and then cooled to a temperature of 20°

Physics
2 answers:
MrRa [10]3 years ago
7 0

Answer:

The specific heat of aluminum is greater.

Explanation:

It lost the most heat.

zepelin [54]3 years ago
3 0

Answer:

specific heat of aluminium is more than the specific heat of iron

Explanation:

mass of both the balls, m = 100 g

heat lost by iron is 3.6 kJ and the heat lost by aluminium is 7.2 kJ.

The amount of heat required to raise the temperature of substance by 1°C of mass 1 kg is defined as the specific heat of the substance.

Here, the amount of heat required to raise the temperature of aluminium by teh same amount as that of iron and mass remains same, so it means the opecific heat of aluminium is more than the specific heat of iron.

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First to answer will be the brainliest i need the answer ASAP
slavikrds [6]
The answer is chemical.
5 0
3 years ago
g Incandescent bulbs generate visible light by heating up a thin metal filament to a very high temperature so that the thermal r
Naddika [18.5K]

Answer:

2577 K

Explanation:

Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.

So, T = ⁴√(P/σεA)

Since P = 60 W, we substitute the vales of the variables into T. So,

T = ⁴√(P/σεA)

= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)

= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)

= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)

= ⁴√(0.00441 × 10¹⁶K⁴)

= 0.2577 × 10⁴ K

= 2577 K

6 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
spotlight on a boat is 2.5 m above the water, and the light strikes the water at a point that is 8.0 m horizontally displaced fr
ludmilkaskok [199]

Answer:

Explanation:

Let i be the angle of incidence and r be the angle of refraction .

From the figure

Tan ( 90 - i ) = 2.5 / 8

cot i = 2.5 / 8

Tan i = 8 / 2.5 = 3.2

i = 72.65°

From snell's law

sini / sin r = refractive index

sin 72.65 / sinr = 1.333

sin r = .9545  / 1.333

= .72

r = 46⁰

From the figure

Tan r  = d / 4

Tan 46 = d /4

d = 4 x Tan 46

= 4 x 1.0355

=4.14 m .

3 0
3 years ago
Can someone help me with this Physics question please?
strojnjashka [21]

Answer:

Explanation:

The formula for this, the easy one, is

N=N_0(\frac{1}{2})^{\frac{t}{H} where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:

N=48.0(\frac{1}{2})^{\frac{49.2}{12.3} and simplifying a bit:

N=48.0(.5)^4 and

N = 48.0(.0625) so

N = 3 mg left after 12.3 years

6 0
3 years ago
Read 2 more answers
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