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3 years ago

The best material to connect a power source to a light bulb is one that

Physics

2 answers:

Sergeu [11.5K]3 years ago

5 0

Will allow electrons to flow freely.

Naddik [55]3 years ago

4 0

Answer: the answer is d

i took the test and can confirm that it's right :)

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Thomas needs to move an 80 kg rock, but cannot lift it. He decides to use a

ArbitrLikvidat [17]

Answer:

Explanation:

The weight of the rock is W = mg = (80 kg) (10 m/s²) = 800 N.

The mechanical advantage is therefore 800 N / 200 N = 4.

5 0

3 years ago

What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha

larisa86 [58]

Explanation:

Below is an attachment containing the solution.

7 0

3 years ago

A skater slides across the ice with an initial velocity of 5.0 m/s. She slows 10 points

zvonat [6]

Explanation:

Given that,

The initial velocity of a skater is, u = 5 m/s

She slows to a velocity of 2 m/s over a distance of 20 m.

We can find the acceleration of skater. It is equal to the rate of change of velocity. So, it can be calculated using third equation of motion as follows :

$v^2-u^2=2as$

a = acceleration

$a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{(2)^2-(5)^2}{2\times 20}\\\\a=-0.525\ m/s^2$

So, her acceleration is $0.525\ m/s^2$ and she is deaccelerating. Also, her initial velocity is given i.e. 5 m/s.

7 0

3 years ago

A dragster starts with zero velocity and completes a 404.5 m (0.2528 mile) run in 4.922 s. If the car had a constant acceleratio

KIM [24]

Answer:

a. a=33.34ms⁻², V=164.4m/s

Explanation:

Since the dragster started with zero velocity, de determine the acceleration using of the equations of motion.

Below are the data given

Distance, s=404.5m,

time taken,t=4.922secs

Using the equation

S=ut+1/2at²

where u is the initial velocity and u=0

Making the acceleration the subject of the formula, we arrive at

a=2s/t²

a=(2*404.5)/4.922²

a=33.34ms⁻².

To determine the velocity, we use

V=u+at

V=0+33.34ms⁻² *4.922sec

V=164.4m/s

5 0

3 years ago

Read 2 more answers

A series of optical telescopes produced an image that has a resolution of about 0.00350 arc second.

Mila [183]

Answer:

The smallest diameter is $D =122 \ m$

Explanation:

From the question we are told that

The resolution of the telescope is $\theta = 0.00350 \ arc \ second$

The wavelength is $\lambda = 1.70 \mu m = 1.70 *10^{-6} \ m$

From the question we are told that

$1 arc \ sec = \frac{1}{3600^o}$

So $0.00350 \ arc \ second = x$

Therefore

$x = 0.00350 * \frac{1}{3600 }$

$x = ( 9.722*10^{-7} )^o$

Now $1^o = \frac{\pi}{180}$

So $(9.722*10^{-7})^o = \theta$

=> $\theta = (9.722*10^{-7}) * \frac{\pi}{180}$

$\theta = 1.69*10^{-8} rad$

The smallest diameter is mathematically represented as

$D = \frac{1.22 \lambda }{\theta }$

substituting values

$D = \frac{1.22 * 1.7 *10^{-6}} {1.69 *10^{-8} }$

$D =122 \ m$

6 0

4 years ago