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2 years ago

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A researcher finds that blood alcohol levels cause Progressive damage the liver all of mice and his study were fed the same amou

nt and types of food different consideration of alcohol injected into different mice represents the

1 answer:

Juli2301 [7.4K]2 years ago

4 0

Answer:hmm

Explanation:

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Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at 2

igomit [66]

Answer

Applying Wein's displacement

$\lamda_{max}\ T = 2898 \mu_mK$

1) for sun T = 5800 K

$\lambda_{max} = \dfrac{2898}{5800}$

$\lambda_{max} = 0.5 \mu_m$

2) for tungsten T = 2500 K

$\lambda_{max} = \dfrac{2898}{2500}$

$\lambda_{max} = 1.16 \mu_m$

3) for heated metal T = 1500 K

$\lambda_{max} = \dfrac{2898}{1500}$

$\lambda_{max} = 1.93 \mu_m$

4) for human skin T = 305 K

$\lambda_{max} = \dfrac{2898}{305}$

$\lambda_{max} = 9.50 \mu_m$

5) for cryogenically cooled metal T = 60 K

$\lambda_{max} = \dfrac{2898}{60}$

$\lambda_{max} = 48.3 \mu_m$

range of different spectrum

UV ----0.01-0.4

visible----0.4-0.7

infrared------0.7-100

for sun T = 5800

λ 0.01 0.4 0.7 100

λT 58 2320 4060 5.8 x 10⁵

F 0 0.125 0.491 1

fractions

for UV = 0.125

for visible = 0.441-0.125 = 0.366

for infrared = 1 -0.491 = 0.509

8 0

3 years ago

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

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3 years ago

A box is being pulled to the right. The free body diagram is shown.

Vinvika [58]

Answer:

25N

Explanation:

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3 years ago

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Find the resultant of two components, 3km west and 4 km south. How do I solve this using the Pythagorean theorem?

Rasek [7]

Answer:5

Explanation:

a=3

b=4

c=√(a²+b²)

c=√9+16

c=√25

c=5

3 0

3 years ago

How do we find the average force in physics?

svlad2 [7]

Force is equal to mass multiplied by acceleration. Hope this helps!

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3 years ago

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