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3 years ago

Two cars collide and come to a complete stop. where did all of their energy go?

1 answer:

3 0

Most of the energy will be absorbed by the materials that make up the cars, causing them to deform. The energy will also be converted into sound energy, causing a loud bang upon collision. Also, some energy will be converted to thermal energy, which will cause the cars to heat up slightly.

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A basketball has a coefficient of restitution of 0.821 in collisions with the wood floor of a basketball court. The ball is drop

Tanya [424]

Answer: The height of its fourth bounce = 0.43m

Explanation:

The coefficient of restitution denoted by (e), is the ratio that shows the final velocity to initial relative velocity between two objects after collision

IT is given by the formula in terms of height as

Coefficient of Restitution, e = √(2gh))/√(2gH) = √(h/H)

Where

Coefficient of Restitution, e= 0.821

H = 2.07 m

At fourth bounce , we have that

Coefficient of Restitution, e⁴ =√(h₄/H)

Putting the given values and solving , we have,

e⁴ =√(h₄/H)

= 0.821⁴ = √(h₄/2.07)

(0.821⁴ )² =h₄/2.07

0.2064 x 2.07 = 0.427 = 0.43

At fourth bounce, h₄ height = 0.43m

7 0

3 years ago

Can someone tell me the answer for both of these pleaseeee i need it asap!!! i will give brainlist!!

inessss [21]

Answer:

1/ decreases

2/ yes;

In general there is no relationship, and any combinations of frequencies and amplitudes are allowed.

Explanation:

Explanation for 1/ - please see the picture attached below

6 0

3 years ago

How much heat is released when 35kg of water freezes?

inn [45]

You need an additional point of data here: the enthalpy of fusion, or conversely the enthalpy of melting (they differ only by their sign). For water (or ice) that value is gotten from sources such as the internet

ΔH°(fus) = 6.01 kJ/mole 

Since you have 35 000g, how many moles do you have? 
Moles H2O = 35000 g/(18.015 g/mole) = 1942.8 moles

So, take that ΔH°(fus) in kJ/mole, multiply by the number of moles, and there ya go!

6.01 x 1942.8 = 11,676 kJ of energy is released

Hope I helped!! xx

3 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

B c _ - _ a b ( a=3 b=-5 c=6)

Pachacha [2.7K]

(-5)/3 - 6/(-5)
You can solve it now :)

7 0

2 years ago