Wow ;-; im not gonna even try

rewona [7]

4/3 pi (radius)^3
If the radius is in cm the volume will be in mL

6 0

3 years ago

Three metal fishing weights, each with a mass of 1.00x102 g and at a temperature of 100.0°C, are placed in 1.00x102 g of water a

worty [1.4K]

Answer:

Approximately $0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ assuming no heat exchange between the mixture and the surroundings.

Explanation:

Consider an object of specific heat capacity $c$ and mass $m$ . Increasing the temperature of this object by $\Delta T$ would require $Q = c\, m \, \Delta T$ .

Look up the specific heat of water: $c(\text{water}) = 4.182\; {\rm J \cdot g^{-1} \cdot K^{-1}}$ .

It is given that the mass of the water in this mixture is $m(\text{water}) = 1.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the water: $\Delta T(\text{water}) = (45 - 35)\; {\rm K} = 10\; {\rm K}$ .

Thus, the water in this mixture would have absorbed :

$\begin{aligned}Q &= c\, m\, \Delta T \\ &= 4.182\; {\rm J \cdot g^{-1}\cdot K^{-1}} \\ &\quad \times 1.00 \times 10^{2}\; {\rm g} \times 10\; {\rm K} \\ &= 4.182 \times 10^{3}\; {\rm J}\end{aligned}$ .

Thus, the energy that water absorbed was: $Q(\text{water}) = 4.182 \times 10^{3}\; {\rm J}$ .

Assuming that there was no heat exchange between the mixture and its surroundings. The energy that the water in this mixture absorbed, $Q(\text{water})$ , would be the opposite of the energy that the metal in this mixture released.

Thus: $Q(\text{metal}) = -Q(\text{water}) = -4.182 \times 10^{3}\; {\rm J}$ (negative because the metal in this mixture released energy rather than absorbing energy.)

Mass of the metal in this mixture: $m(\text{metal}) = 3 \times 1.00 \times 10^{2}\; {\rm g} = 3.00 \times 10^{2}\; {\rm g}$ .

Temperature change of the metal in this mixture: $\Delta T(\text{metal}) = (100 - 45)\; {\rm K} = 55\; {\rm K}$ .

Rearrange the equation $Q = c\, m \, \Delta T$ to obtain an expression for the specific heat capacity: $c = Q / (m\, \Delta T)$ . The (average) specific heat capacity of the metal pieces in this mixture would be:

$\begin{aligned}c &= \frac{Q}{m\, \Delta T} \\ &= \frac{-4.182 \times 10^{3}\; {\rm J}}{3.00 \times 10^{2}\; {\rm g} \times (-55\; {\rm K})} \\ &\approx 0.253\; {\rm J \cdot g^{-1} \cdot K^{-1}}\end{aligned}$ .

6 0

3 years ago

It is always a good idea to get some sense of the "size" of units. For example, the mass of an apple is about 100 g , whereas a

marta [7]

Answer and Explanation:

The estimation of the mass of the objects is as follows

Given that

The mass of an apple is about 100 g

Just like that, we do some estimation of different objects

A dime is around 1 g

A smartphone is about 100 g

An adult male is approx 100 kg

A college physics textbook is around 1 kg

A ripe banana is around 100 g

A small car is around 1,000 kg

4 0

4 years ago

What type of drawing is shown in the figure?

Rasek [7]

$Question$

What type of drawing is shown in the figure?

Answer:

Hi, There! My Answer would Be A. Block diagram Or C. panel Diagram!

Hope this helps! ♚♛♕♔ッ✨♚

$-xXxAnimexXx-$

6 0

3 years ago

Kevin jump straight up in the air to a height of 1 m at the top of his jump he has the potential energy of 1000 J answer the fol

NeTakaya

Answer:

1000N,4.48m/s

Explanation:

Height(h) = 1m

P. E = 1000J

P. E = mass x acceleration due to gravity x height

Acceleration due to gravity = 10m/s

P. E = mgh

Where weight = mg

P. E = weight x height

Weight = P. E / height

Weight = 1000 / 1

Weight. 1000N

b) as he leaves the ground P. E = K. E

P. E = 1/2 mv2

P. E = 1000 , mass = weight / 10

Mass =1000/10 = 100kg

P. E = 1/2 mv2

1000 = 1/2 x 100 x v2

Multiply both sides by 2

1000 x 2 = 100 x v2

V2 = 2000/ 100

V2 =20

Square root both sides

V = square root of 20

V = 4.48m/s

I hope this was helpful, Please mark as brainliest

4 0

3 years ago