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kakasveta [241]

3 years ago

7

Which weighs more: A. A large bathtub filled to the brim with water B. A large bathtub filled to the brim with water with a batt

leship floating in it C. No difference: these two options weigh the same

1 answer:

Airida [17]3 years ago

7 0

Answer:

Explanation:yes

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An oscillator consists of a block attached to a spring (k = 500 N/m). At some time t, the position (measured from the system's e

Alex_Xolod [135]

Answer:

a) $\omega = 10.407\,\frac{rad}{s}$ , b) $m = 4.617\,kg$ , c) $A = 1.355\,m$

Explanation:

a) The system have a simple armonic motion, whose position function is:

$x(t) = A\cdot \cos (\omega\cdot t + \phi)$

The velocity function is determined by deriving the position function in terms of time:

$v(t) = -\omega \cdot A \cdot \sin(\omega\cdot t + \phi)$

The acceleration function is found by deriving again:

$a(t) = -\omega^{2} \cdot A \cdot \cos (\omega\cdot t + \phi)$

Let assume that $t = 0\,s$ . The following nonlinear system is built:

$A\cdot \cos \phi = 0.660\,m$

$-\omega \cdot A \cdot \sin \phi = -12.3\,\frac{m}{s}$

$-\omega^{2}\cdot A \cdot \sin \phi = -128\,\frac{m}{s^{2}}$

System can be reduced by divinding the second and third expressions by the first expression:

$\omega \cdot \tan \phi = 18.636\,\frac{1}{s}$

$\omega^{2}\cdot \tan \phi = 193.94\,\frac{1}{s^{2}}$

Now, the last expression is divided by the first one:

$\omega = 10.407\,\frac{rad}{s}$

b) The mass of the block is:

$m = \frac{k}{\omega^{2}}$

$m = \frac{500\,\frac{N}{m} }{(10.407\,\frac{rad}{s})^{2} }$

$m = 4.617\,kg$

c) The phase angle is:

$\phi = \tan^{-1} \left(\frac{18.636\,\frac{1}{s} }{\omega} \right)$

$\phi \approx 0.338\pi$

The amplitude is:

$A = \frac{0.660\,m}{\cos 0.338\pi}$

$A = 1.355\,m$

8 0

4 years ago

Read 2 more answers

A beam of protons is moving toward a target in a particle accelerator. This beam constitutes a current whose value is 0.50 μA. H

Nataly_w [17]

I = 5*10 - 7A

q = 5 *15*1- - 7 cuolomb
= 75 *10 -7 C in 15 sec is indecent

1 proton has q = 1.6 *10 - 19C

75 *10-7 / 16* 10-19

= 4.7 x 10^13

Hope this helps

3 0

3 years ago

A 60-kg runner raises his center of mass approximately 0.5 m with each step. Although his leg muscles act as a spring, recapturi

Darina [25.2K]

Answer: P = 36.75W

The additional power needed to account for the loss is 36.75W.

Explanation:

Given;

Mass of the runner m= 60 kg

Height of the centre of gravity h= 0.5m

Acceleration due to gravity g= 9.8m/s

The potential energy of the body for each step is;

P.E = mgh

P.E = 60 × 9.8 × 0.5

PE = 294J

Since the average loss per compression on the leg is 10%.

Energy loss = 10% (P.E)

E = 10% of 294J

E = 29.4J

To calculate the runner's additional power

given that time per stride is = 0.8s

Power P = Energy/time

P = E/t

P = 29.4J/0.8s

P = 36.75W

5 0

3 years ago

Arthur is testing how well various types of disinfectants can kill E. coli bacteria. He starts with a Petri dish that is covered

maw [93]

The variable is the disinfectant that is used, because there different ones being used

3 0

3 years ago

Read 2 more answers

What is 1 joule work?

larisa86 [58]

$\large\color{pink}{{★ƛƝƧƜЄƦ ↦}}$

The amount of work done when a the force of 1 Newton displaces a body through a distance of 1m in the direction of the force applied is said to be as 1joule work

Hope it helps ~

$\sf{\:мѕнαcкεя\: ♪...}$

6 0

2 years ago