How can Ohm's Law relate to a house fire?

The weight y of a fiddler crab is directly proportional to the 1.25 power of the weight x of its claws. A crab with a body weigh

frez [133]

Answer:

The weight of body is 1.3040 gram.

Explanation:

Given that,

The weight y of a fiddler crab is directly proportional to the 1.25 power of the weight x of its claws.

Suppose a crab with a body weight of 1.8 gram has claws weighing 1.1 gram.

Estimate the weight of a fiddler crab with claws weighing 0.85 gram.

Determine the weight of crab body

We need to calculate the value of proportional constant

$y\propto x^{1.25}$

$y=kx^{1.25}$

$k=\dfrac{y}{x^{1.25}}$

Put the value into the formula

$k=\dfrac{1.8}{1.1^{1.25}}$

$k=1.5978$

We need to calculate the crab weight

$y=kx^{1.25}$

Here, x = 0.85 g

Put the value into the formula

$y=1.5978\times(0.85)^{1.25}$

$y=1.3040\ gram$

Hence, The weight of body is 1.3040 gram.

7 0

4 years ago

A planet is orbiting a nearby star. Suppose that the star’s gravity increases over time. What effect will this increase most lik

zhenek [66]

The planet would stay in the same orbit but start revolving faster.
(Its year would get shorter.)

7 0

4 years ago

A negative charge of -0.00067 C and a positive charge of 0.00096 C are separated by 0.7 m. What is the force between the two cha

koban [17]

Answer:

Explanation:

According to coulombs law, the Force between the two charges is expressed as;

F = kq1q2/d²

k is the coulombs constant = 9*10⁹kg⋅m³⋅s⁻²⋅C⁻².

q1 = -0.00067 C

q2 = 0.00096 C

d = 0.7m

Substitute into the formula:

F = 9*10^9 * -0.00067 * 0.00096/0.7²

F = 9*10⁹*-6.7*10⁻⁴*9.6*10⁻⁴/0.49

F = -578.88*10⁹⁻⁸/0.49

F = -578.88*10/0.49

F = -5788.8/0.49

F = -11,813.87N

<em>Hence the force between the two charges is -11,813.87N </em>

6 0

3 years ago

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi

Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

$m_{m}$ = mass of metal = 400 g

$c_{m}$ = specific heat of metal = ?

$T_{mi}$ = initial temperature of metal = 100 °C

$m_{a}$ = mass of aluminum cup = 100 g

$c_{a}$ = specific heat of aluminum cup = 900.0 J/kg ∙ K

$T_{ai}$ = initial temperature of aluminum cup = 15 °C

$m_{w}$ = mass of water = 500 g

$c_{w}$ = specific heat of water = 4186 J/kg ∙ K

$T_{wi}$ = initial temperature of water = 15 °C

$T$ = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

$m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}$

7 0

4 years ago

Use the information and diagram to answer the following question.

Mumz [18]

Answer:b

Explanation:

Because I don’t know

8 0

3 years ago