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3 years ago

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A hot lump of 35.8 g of copper at an initial temperature of 70.7 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to

reach thermal equilibrium. What is the final temperature of the copper and water, given that the specific heat of copper is 0.385 J/(g·°C)? Assume no heat is lost to surroundings.

1 answer:

AURORKA [14]3 years ago

4 0

I upload the answer via Image because Brainly warned me that there were inappropriate links or words.

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During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi

geniusboy [140]

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

$a_t$ = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

$\omega_i$ = Initial angular velocity = 95 rad/s

$\theta$ = Angle of rotation

$\omega_f$ = Final angular velocity

t = Time taken

Angular acceleration is given by

$\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2$

The angular acceleration is -24.28571 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev$

The number of revolutions is 29.57239

$\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-24.28571}\\\Rightarrow t=3.91176\ s$

The time it takes for the car to stop is 3.91176 seconds

Linear distance

$s=r\theta\\\Rightarrow s=0.28\times 185.80885\\\Rightarrow s=52.026478\ m$

The distance the car travels is 52.026478 m

8 0

3 years ago

An object is dropped from the top of a building and is observed to take 7. 2s to hit the ground. How tall is the building?.

maw [93]

the height of the building is H=36 m.

<h3>What is The Law of Gravity?</h3>

According to Newton's law of gravity, every particle of matter in the universe is attracted to every other particle with a force that varies directly as the product of their masses and inversely as their distance from one another.

Properties of Gravity -

It is a universal attractive force. It is directly proportional to the product of the masses of the two bodies.
It obey inverse square law.
It is the weakest force known in nature.

Examples of Gravity -

The force that holds the gases in the sun.
The force that causes a ball you throw in the air to come down again.
The force that causes a car to coast downhill even when you aren't stepping on the gas.

v₀=0 m/s

H₀=0 m

g=10 m/s²

t=7,2 s

H - ?

$H= H_{0} + v_{0}t+\frac{gt^{2} }{2} \\$

H = 0 +0 × 7.2 + 10(7.2)²/2

H = 36m

to learn more about Gravity go to - brainly.com/question/12528243

#SPJ4

8 0

2 years ago

you're reading from the journal of a European explorer from the early 1600s. In one passage, the explorer describes itting on th

Zepler [3.9K]

Answer: horse latitudes

Explanation:

8 0

3 years ago

Read 2 more answers

Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy

8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

$E_n=-13.6\times \frac{Z^2}{n^2}ev$

Formula used for the radius of the $n^{th}$ orbit will be,

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

$E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV$

Energy of the second orbit in H atom .

$E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV$

Energy of the third orbit in H atom .

$E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV$

Energy of the fifth orbit in H atom .

$E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV$

Energy of the sixth orbit in H atom .

$E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV$

Energy of the seventh orbit in H atom .

$E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV$

During an absorption of energy electron jumps from lower state to higher state.So, absorption will take place in :

1) n = 2 → n = 3

2) n= 5 → n = 6

Energy absorbed when: n = 2 → n = 3

$E=E_3-E_2$

$E=(-1.51 eV) -(-3.40 eV)=1.89 eV$

Energy absorbed when: n = 5 → n = 6

$E'=E_6-E_5$

$E'=(-0.378 eV)-(-0.544 eV) =0.166 eV$

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0

3 years ago

The graphs display velocity data Velocity is on the y-axis (m/s), while time is on the x-axis (s). Based on the graphs, which da

RUDIKE [14]

Answer:

The first graph is showing the constant acceleration (1 m/s)

Explanation:

The second graph showing the flexible velocity therefore a in the graph is different at t1, t2, t3, t4

The last graph is showing constant velocity therefore there is no acceleration (a = 0)

5 0

3 years ago

Read 2 more answers