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2 years ago

Which of the following does Bernoulli's principle help to explain?

Physics

2 answers:

Kryger [21]2 years ago

7 0

B. Surface tension

Hope this helps

Anon25 [30]2 years ago

6 0

I would say B surface tension

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A car with a mass of 3 Kg and velocity of 40 m/s collided with a truck of a velocity of 60 m/s, if the momentum is conserved wha

Luden [163]

Answer:

the mass of the truck is 2 kg.

Explanation:

Given;

mass of the car, m₁ = 3 kg

initial velocity of the car, u₁ = 40 m/s

initial velocity of the truck, u₂ = 60 m/s

let the mass of the truck = m₂

Apply the principle of conservation of linear momemtum;

m₁u₁ = m₂u₂

m₂ = (m₁u₁) / u₂

m₂ = (3 x 40) / (60)

m₂ = 2 kg

Therefore, the mass of the truck is 2 kg.

4 0

2 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

2 years ago

An object that is falling has the following type(s) of energy. Ignore air resistance.

Anton [14]

Potential and kinetic

6 0

2 years ago

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To minimize signal distortion, at each end of the J-1939 CAN-bus there is a(n)_____________resistor.

EastWind [94]

5-ohm
Extra
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120-ohm
Variable
Pg. 614

8 0

2 years ago

In which situation are both the speed and velocity of the car changing?

lubasha [3.4K]

Answer is A because the speed and velocity would change. Think of it as GTA, your going 30+ miles per hour and you take a left turn, the speed and velocity would change in an instant..
Hope this helped.

6 0

2 years ago

Read 2 more answers