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3 years ago

A lab cart is loaded with different masses and moved at various velocities

Physics

1 answer:

Sedaia [141]3 years ago

6 0

A lab cart is loaded with different masses and moved at various constant velocities? the anser should be

1.0m/s → 4kg

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A cylinder contains a gas at a temperature of 400K. Explain two ways that you can add energy to the gas and raise its temperatur

Lelechka [254]

Heat

You can usually warm something by adding energy. The added energy can be from light, electricity, friction, a chemical reaction, nuclear reaction, or any other kind of energy. When first added to a substance, energy might be concentrated in one atom, but this one will soon bump into others and spread the energy. Eventually, every atom or molecule in the substance will move a bit faster. When the added energy is spread throughout a substance, it is then called heat energy, thermal energy, or, simply heat. All three terms mean the same thing. Heat is a form of energy, so it has the units of energy. In the SI system, this is Joules. Many other units to measure thermal energy are in common use. Calories and BTU's are common heat units.

Temperature

You cannot measure heat directly, but you can detect its effect on a substance. Changes in heat can usually be detected as changes in temperature. Usually, when you add energy to a bunch of atoms they move faster and get hotter. Similarly, if you remove energy from a bunch of atoms, they usually move less and get cooler.

5 0

3 years ago

What fraction of an iceberg is submerged? (ρice = 917 kg/m3, ,ρsea = 1030 kg/m3.)

aksik [14]

Answer:

Choice d. Approximately $89\%$ of the volume of this iceberg would be submerged.

Explanation:

Let $V_\text{ice}$ denote the total volume of this iceberg. Let $V_\text{submerged}$ denote the volume of the portion that is under the liquid.

The mass of that iceberg would be $\rho_\text{ice} \cdot V_\text{ice}$ . Let $g$ denote the gravitational field strength ( $g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth.) The weight of that iceberg would be: $\rho_\text{ice} \cdot V_\text{ice} \cdot g$ .

If the iceberg is going to be lifted out of the sea, it would take water with volume $V_\text{submerged}$ to fill the space that the iceberg has previously taken. The mass of that much sea water would be $\rho_\text{sea} \cdot V_\text{submerged}$ .

Archimedes' Principle suggests that the weight of that much water will be exactly equal to the buoyancy on the iceberg. By Archimedes' Principle:

$\text{buoyancy} = \rho_\text{sea} \cdot V_\text{submerged} \cdot g$ .

The buoyancy on the iceberg should balance the weight of this iceberg. In other words:

$\underbrace{\rho_\text{ice} \cdot V_\text{ice} \cdot g}_\text{weight of iceberg} = \underbrace{\rho_\text{sea} \cdot V_\text{submerged} \cdot g}_\text{buoyancy on iceberg}$ .

Rearrange this equation to find the ratio between $V_\text{submerged}$ and $V_\text{ice}$ :

$\begin{aligned} &\frac{V_\text{submerged}}{V_\text{ice}} \\&= \frac{\rho_\text{ice} \cdot g}{\rho_\text{sea} \cdot g}\\ &= \frac{\rho_\text{ice}}{\rho_\text{sea}}\ = \frac{917\; \rm kg \cdot m^{-3}}{1030\; \rm kg \cdot m^{-3}} \approx 0.89 \end{aligned}$ .

In other words, $89\%$ of the volume of this iceberg would have been submerged for buoyancy to balance the weight of this iceberg.

8 0

2 years ago

Radioactivity was discovered by Albert Einstein. True or False

asambeis [7]

Hello there!
The answer is <span>False.
</span><span>Henri Becquerel was the one to discover radioactivity.
</span>Hope I helped!

6 0

3 years ago

Read 2 more answers

A light-weight potter’s wheel, having a moment of inertia of 24 kg m2 is spinning freely at 40.0 rpm. The potter drops a small b

mixer [17]

Answer:

m=4.03 kg

Explanation:

Given that

r= 1.2 m

I=24 kg.m²

N₁ = 40 rpm

$\omega_1=\dfrac{2\pi N_1}{60}\ rad/s$

$\omega_1=\dfrac{2\pi \times 40}{60}\ rad/s$

ω₁= 4.1 rad/s

N₂=32 rpm

$\omega_2=\dfrac{2\pi \times 32}{60}\ rad/s$

ω₂ = 3.3 rad/s

Lets take mass of clay = m kg

Initial angular momentum L₁

L₁ = Iω₁

Final linear momentum L₂

L₂= I₂ ω₂

I₂ = I + m r²

The is no any external torque that is why angular momentum will be conserve

L₁ = L₂

Iω₁ = I₂ ω₂

Iω₁ = ( I + m r²) ω₂

Now by putting the all values

Iω₁ = ( I + m r²) ω₂

24 x 4.1 = (24 + m x 1.2²) x 3.3

24 x 4.1 = (24 + 1.44 m) x 3.3

29.81 = 24 + 1.44 m

1.44 m = 5.81

m=4.03 kg

7 0

3 years ago

Help please !!! thanks

blsea [12.9K]

Answer:

3. a) The current in R1 is 0.5 A

The current in R₂ is 0.2 $\overline {27}$ A

b) The power dissipated in R₁ is 0.5 W

Explanation:

The given circuit parameters are;

The voltage in the circuit = 5 V

The resistances in the circuit are;

R1 = 3 Ω, R2 = 4 Ω, R3 = 8 Ω, R4 = 10 Ω, R5 = 4 Ω, Ra = 4 Ω

3. a) The equivalent resistance of the circuit, $R_{E}$ , is given as follows;

$R_{E} = R1 +\dfrac{\left( \dfrac{R5 \cdot Ra}{R5 + Ra} + R4 \right) \times (R2 + R3)}{\left( \dfrac{R5 \cdot Ra}{R5 + Ra} + R4 \right) + (R2 + R3)}$

Plugging in the values, we get;

$R_{E} = 4 +\dfrac{\left( \dfrac{4 \times 4}{4 +4} + 10 \right) \times (4 + 8)}{\left( \dfrac{4 \times 4}{4 +4} + 10 \right) + (4 + 8)} = 10$

The equivalent resistance of the circuit, $R_{E}$ = 10 Ω

The current in R1 = The current in the circuit, $I_E$ = V/ $R_{E}$

∴ I = 5 V/(10 Ω) = 0.5 A

The current in R1 = 0.5A

Let, 'I₂' represent the current flowing through R₂

By the current divider rule, we have;

$I_2 = \dfrac{R_{E}}{R2 + R3 + R_{E}} \times I_T$

Which gives;

$\therefore I_2 = \dfrac{10 \, \Omega}{4 \, \Omega + 8 \, \Omega + 10 \, \Omega} \times 0.5A = \dfrac{5}{22} \, A = 0.2 \overline {27} \, A$

The current flowing through R₂, I₂ = 0.2 $\overline {27}$ A

b) The power dissipated in R₁, P₁ = $I_E^2$ × R₁

∴ The power dissipated in R₁, P₁ = (0.5 A)² × 4 Ω = 0.5 W

6 0

3 years ago