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3 years ago

7

FREE BRAINLIEST! if you can answer this correctly ill give you brainliest and answer some of the questions you have posted :) th

ank you very much!!! (22pts)

1 answer:

Gala2k [10]3 years ago

4 0

b) accelerate to the left as much more pressure is pulling it in that direction and on the right however , there is less force .

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How many photons will be required to raise the temperature of 1.8 g of water by 2.5 k ?'?

tatyana61 [14]

Missing part in the text of the problem:
"<span>Water is exposed to infrared radiation of wavelength 3.0×10^−6 m"</span>

First we can calculate the amount of energy needed to raise the temperature of the water, which is given by
$Q=m C_s \Delta T$
where
m=1.8 g is the mass of the water
$C_s = 4.18 J/(g K)$ is the specific heat capacity of the water
$\Delta T=2.5 K$ is the increase in temperature.

Substituting the data, we find
$Q=(1.8 g)(4.18 J/(gK))(2.5 K)=18.8 J=E$

We know that each photon carries an energy of
$E_1 = hf$
where h is the Planck constant and f the frequency of the photon. Using the wavelength, we can find the photon frequency:
$\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{3 \cdot 10^{-6} m}=1 \cdot 10^{14}Hz$

So, the energy of a single photon of this frequency is
$E_1 = hf =(6.6 \cdot 10^{-34} J)(1 \cdot 10^{14} Hz)=6.6 \cdot 10^{-20} J$

and the number of photons needed is the total energy needed divided by the energy of a single photon:
$N= \frac{E}{E_1}= \frac{18.8 J}{6.6 \cdot 10^{-20} J} =2.84 \cdot 10^{20} photons$

4 0

3 years ago

After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T

PIT_PIT [208]

Answer:

160 m

Explanation:

The intensity, I, of the sound is inversely proportional to the square of the distance, r, from the source.

$I\propto \dfrac{1}{r^2}$

Hence,

$I_1r_1^2 = I_2r_2^2$

$r_2 = r_1\sqrt{\dfrac{I_1}{I_2}}$

From the question, $I_2$ is half of $I_1$

$r_2 = r_1\sqrt{\dfrac{I_1}{0.5I_1}}$

$r_2 = r_1\sqrt{2}$

$r_2 = 113\text{ m}\sqrt{2} = 160 \text{ m}$

3 0

3 years ago

Read 2 more answers

A skier starts from rest down a slope 500.0 M long, the skier accelerates at a constant rate of 2.00 m/s/s, what's the velocity

nevsk [136]

We can use the kinematic equation
$(v_f)^2 = (v_i)^2 + 2*a*d$
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance

we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s [if you are rounding this by significant figures]

8 0

2 years ago

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

Burka [1]

Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .

At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .

At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .

7 0

2 years ago

Read 2 more answers

As pressure decreases, temperature

Pani-rosa [81]

Temperature doesn't do anything, the boiling point of stuff decreases. If you put water in a vacuum chainber then it will start to boil

5 0

3 years ago