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Marta_Voda [28]

3 years ago

8

A certain part of a computer chip has a length of 1.5 micrometers. How many meters is this ?

1 answer:

Alchen [17]3 years ago

4 0

When you're working with units of measure, "micro" almost always means "millionth", or 1 x 10⁻⁶ . And that's exactly the case in this example.

"1.5 micrometer" = 1.5 millionths of a meter, or <em>1.5 x 10⁻⁶ meter.</em>

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If two swimmers compete in a race, does the faster swimmer develop more power?

valkas [14]

Power is equal to energy per unit time. In this case, power is proportional to energy while is inversely proportional to time,on the other hand. Given the two swimmers exerts same amount of energy but the faster swimmer just does things in faster time, then the faster swimmer should develop more power from shorter time

7 0

3 years ago

A mass of 2 kg traveling at 3 m/s undergoes a one-dimensional elastic collision with a group of four 1kg masses that are at rest

alukav5142 [94]

2kg + 3 kg = 5 kg as result of the conclusion

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3 years ago

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

<u>Now, gravitational force on particle 3 due to particle 1:</u>

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

<u>Now the net force</u>

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

<em>For angle in counterclockwise direction from the +x-axis</em>

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

4 0

3 years ago

A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth

Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

$\tan\theta=\dfrac{8}{3}$

$\thta=\tan^{-1}\dfrac{8}{3}$

$\theta=69.44^{\circ}$

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

$mg=2T\cos\theta$

Put the value into the formula

$3\times9.8=2T\times\cos69.44$

$T=\dfrac{3\times9.8}{2\times\cos69.44}$

$T=41.85\ N$

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

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3 years ago

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Select the statement that correctly describes how light travels? light can travel in a vacuum, and its speed is constant even if

Irina18 [472]

<span>Light can travel in a vacuum, and ... strange as it may seem ...
its speed is always the same, even if the light source is moving. </span>

3 0

3 years ago

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