How would you find the average speed of a cyclist throughout an entire race

HELP ASAP!!!! Which of the following is an example of a hypothesis? A) What will happen if a sugar cube in boiling water? B) I t

Maurinko [17]

Answer:

C) If an ice cube is placed into a boiling water, then it will melt in less than 2 minutes.

Explanation:

7 0

3 years ago

What is the area under the curve for the histogram below

igor_vitrenko [27]

Answer: B srry if i'm wrong

3 0

2 years ago

Your friend comes across a good deal to purchase a gold ring. She asks you for advice and for you to test the ring. The ring has

Luden [163]

Answer:

She is not getting a good deal.

Explanation:

The equation that relates heat with mass, specific heat and temperature change of an object is $Q=mc\Delta T$ .

Always convert temperature to Kelvin, although in our case it's not necessary because the $\Delta T$ will be the same, and we will leave the mass in grams because we will be getting $J/g^{\circ}C$ units for specific heat, which we can compare to the one given for gold.

We then calculate the specific heat of the object in question:

$c=\frac{Q}{m\Delta T}=\frac{94.8J}{(4.54g)(47.5^{\circ}C-23^{\circ}C)}=0.8523 J/g^{\circ}C$

Which shows it's not gold.

5 0

3 years ago

A gold bar 20.0kg at 35.0°c is placed in a large insulated 0.8kg glass container at 15°c and 2.0kg of water at 25°c.. calculate

Oksanka [162]

Answer:

The final equilibrium temperature is approximately 26.69 °C

Explanation:

The heat transferred, ΔQ, from a hot body to a cold one is given by the following formula;

ΔQ = m·c·ΔT

Where;

m = The mass of the body

c = The specific heat capacity of the body

ΔT = The temperature change of the body

The given mass of the gold bar, m₁ = 20.0 kg

The initial temperature of the gold bar, T₁ = 35.0 °C

The specific heat capacity of gold, c₁ = 0.13 kJ/(kg·K)

The mass of the glass container, m₂ = 0.8 kg

The initial temperature of the glass container, T₂ = 15°C

The specific heat capacity of glass, c₂ = 0.792 kJ/(kg·K)

The mass of the added water, m₃ = 2.0 kg

The initial temperature of the added water, T₃ = 25°C

The specific heat capacity of water, c₃ = 4.2 kJ/(kg·K)

The heat lost by the gold = The heat gained by the glass and the water

Let 'T' represent the temperature at the final equilibrium, we have;

m₁·c₁·ΔT₁ = m₂·c₂·ΔT₂ + m₃·c₃·ΔT₃

Where;

ΔT₁ = T₁ - T

ΔT₂ = T - T₂

ΔT₃ = T - T₃

∴ 20.0 × 0.13 × (35 - T) = 0.8 × 0.792 × (T - 15) + 2.0 × 4.2 × (T - 25)

Expanding and collecting like terms (using a graphing calculator) gives;

91 - 2.6·T = 9.0336·T - 219.504

9.0336·T + 2.6·T = 219.504 + 91 = 310.504

11.6336·T = 310.504

T = 310.504/11.6336 ≈ 26.69

The final equilibrium temperature, T ≈ 26.69 °C.

4 0

3 years ago

A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar

finlep [7]

Answer:

Explanation:

<u>Electric Potential Due To Point Charges </u>

The electric potential produced from a point charge Q at a distance r from the charge is

$\displaystyle V=k\frac{Q}{r}$

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

$d=\sqrt2 a$

where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

$\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}$

$\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m$

The total potential is

$V_t=V_1+V_2+V_3+V_4$

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of $\pm 3\mu\ C$ . Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.

$\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}$

$V_1=V_2=50912\ V$

The total potential is

$V_t=50912\ V+50912\ V=1\times 10^5\ V$

$\boxed{V_t=1\times 10^5\ V}$

6 0

3 years ago