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3 years ago

9

a woman is swimming across a cold lake. her body temperature is 98 degrees fahrenheit , and the lake water is at 60 degrees fahr

enheit. Which has more thermal energy, the woman or the lake? Explain.

1 answer:

vagabundo [1.1K]3 years ago

6 0

the water has more thermal energy - this is a key concept of thermodynamics

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object x and y fall from a same height and object x is heavier than y which object would fall faster qnd y

hjlf

Answer: They’d fall at the same speed, because air resistance is the only thing that makes an object fall faster than another. There’s a video somewhere on the internet of a bowling ball and a feather falling at the same speed in a vacuum, if you look for it. Hope this helps!

4 0

3 years ago

Who’s going faster in the attached image?

ioda

Answer:

Art

Explanation:

Polly's line is linear, while arts line is going up with constant velocity. There for art is going faster.

8 0

3 years ago

Read 2 more answers

A 2.40 kg ball is attached to an unknown spring and allowed to oscillate. The figure below shows a graph of the ball’s position

irga5000 [103]

(a) The period of the oscillation is 0.8 s.

(b) The frequency of the oscillation is 1.25 Hz.

(c) The angular frequency of the oscillation is 7.885 rad/s.

(d) The amplitude of the oscillation is 3 cm.

(e) The force constant of the spring is 148.1 N/m.

The given parameters:

<em>Mass of the ball, m = 2.4 kg</em>

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From the given graph, we can determine the missing parameters.

The amplitude of the wave is the maximum displacement, A = 3 cm

The period of the oscillation is the time taken to make one complete cycle.

T = 0.8 s

The frequency of the oscillation is calculated as follows;

$f = \frac{1}{T} \\\\f = \frac{1 }{0.8} \\\\f = 1.25 \ Hz$

The angular frequency of the oscillation is calculated as follows;

$\omega = 2\pi f\\\\\omega = 2\pi \times 1.25\\\\\omega = 7.855 \ rad/s$

The force constant of the spring is calculated as follows;

$\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\ k = \omega ^2 m\\\\k = (7.855)^2 \times 2.4\\\\k = 148.1 \ N/m$

Learn more about general wave equation here: brainly.com/question/25699025

4 0

3 years ago

48 You are given two amplifiers, A and B, to connect in cascade between a 10-mV, 100-k source and a 100- load. The amplifiers ha

Sedbober [7]

Answer:

SABL

Explanation:

The best amplifier will be the one that gives us a bigger gain. In each stage will be a load factor that will reduce the gain, that is defined as:

$Fp=\frac{R_{in}}{R_{in}+R_{out}}\\$

where Rin is the input resistance of the next stage and Rout the output resistance of the previous stage.

Analyzing SABL:

$Fp_1=\frac{100K}{100K+100K}=0.5\\\\Fp_2=\frac{10K}{10K+10K}=0.5\\\\Fp_3=\frac{100}{100+1K}=0.0909$

the total gain will be the total gain of each stage multiplied by the load factor.

$SABL_{gain}=0.5*100*0.5*10*0.0909\\SABL_{gain}=22.73\\SABL_{gain_{db}}=20*\log{22.73}=27.13db$

Analyzing SBAL:

$Fp_1=\frac{10K}{10K+100K}=0.0909\\\\Fp_2=\frac{100K}{100K+10K}=0.909\\\\Fp_3=\frac{100}{100+10K}=0.0099$

the total gain will be the total gain of each stage multiplied by the load factor.

$SABL_{gain}=0.0909*10*0.99*100*0.0099\\SABL_{gain}=0.89\\SABL_{gain_{db}}=20*\log{0.89}=-1.01db$

So the best amplifier arrangement is SABL.

4 0

3 years ago

You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m.

Gwar [14]

Answer:

0.95 seconds

Explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

a = Acceleration = 9.81 m/s² (downward positive, upward negative)

Time taken by the ball to reach the maximum height

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-15}{-9.81}\\\Rightarrow t=1.52\ s$

Maximum height

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-15^2}{2\times -9.81}\\\Rightarrow s=11.47\ m$

Distance between maximum height of the ball and the branch is 11.47-7 = 4.46 m

So, the distance that will be covered on the way down is 4.46 m

Now

u = 0

s = 4.46

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.46=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{4.46\times 2}{9.81}}\\\Rightarrow t=0.95\ s$

Time taken by the ball from the maximum height to the tree branch is 0.95 seconds.

Total time taken from the moment the ball is thrown to reach the tree branch is 1.52+0.95 = 2.47 seconds

7 0

3 years ago