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3 years ago

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Suppose Person A runs off the edge of the cliff at 2 m/s and Person B runs off the edge at 1 m/s. Which will hit the ground fart

her from the base of the cliff?

1 answer:

djverab [1.8K]3 years ago

3 0

Answer:

<h2>Person A will hit a distance father</h2>

Explanation:

Based on the fact that the velocity of person A is more than that of person B, that is from the question, person A has a velocity of 2m/s and person B has a velocity of 1m/s, this clearly shows that person A has the tendency to hit a distance farther from the cliff than person B.

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Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = $\pi d^{2} /4$ = (3.142 x $(2*10^{-3} )^{2}$ )/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = $\sqrt{2I/ce_{0} }$

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

$e_{0}$ = 8.85 x 10^9 m kg s^-2 A^-2

E = $\sqrt{2*318.2/3*10^8*8.85*10^9}$ = <em>1.55 x 10^-8 v/m</em>

6 0

3 years ago

A ball traveling at a speed ν0 rolls off a desk and lands at a horizontal distance x0 away from the desk, as shown in the figure

klasskru [66]

Answer:

$3x_0$

Explanation:

The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by

$d=v_x t$

where

$v_x$ is the horizontal velocity

t is the time of flight

The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.

In the first situation, the horizontal distance covered is

$d=v_0 t = x_0$

in the second case, the horizontal velocity is increased to

$v_x' = 3v_0$

And so the new distance travelled will be

$d' = v_x' t = 3 v_0 t = 3 x_0$

So, the distance increases linearly with the horizontal velocity.

5 0

3 years ago

The resistance of a very fine aluminum wire with a 20 μm × 20 μm square cross section is 1200 Ω . A 1200 Ω resistor is made by w

notsponge [240]

Explanation:

The given data is as follows.

Resistance (R) = 1200 ohm, Area (A) = $20 \times 10^{-6}$ m (as $1 \mu m = 10^{-6} m$ )

Diameter (d) = 2.3 mm = $2.3 \times 10^{-3}$ m

First, we will calculate the length as follows.

R = $\rho \frac{L}{A}$

Here, $\rho$ = resistivity of aluminium = $2.65 \times 10^{-8}$

Putting the given values above and we will calculate the value of length as follows.

R = $\rho \frac{L}{A}$

1200 = $2.65 \times 10^{-8} \times \frac{L}{20 \times 10^{-6}}$

L = $9.056 \times 10^{5}$

As the circumference of circular wire = $2 \pi r$

or, = $2 \times \pi \times \frac{d}{2}$

= $\pi \times d$

And, number of turns will be calculated as follows.

No. of turns × Circumference = Length of wire

No. of turns × $3.14 \times 2.3 \times 10^{-3} = 9.056 \times 10^{5}$

= $1.25 \times 10^{8}$

Thus, we can conclude that $1.25 \times 10^{8}$ turns of wire are needed.

6 0

3 years ago

Now assume that Alice and Bob are twins, and Alice left Earth and Bob stayed behind fixing his spaceship. If Alice spent some ti

alexgriva [62]

我們的確認為我可以幫你們解決問題，我們要不要買不起房屋貸款土地

8 0

3 years ago

How much work output can a 150hp snowmobile do in 15 seconds if it is 18 % efficiency?

Sav [38]

The work output is $3.02\cdot 10^5 J$

Explanation:

The power of the snowmobile is

$P=150 hp$

Keeping in mind that

$1 hp = 746 W$

We can convert it into Watts:

$P=(150)(746)=111,900 W$

The energy used by the snowmobile can be found as follows:

$E=Pt$

where

P = 111,900 W is the power used

t = 15 s is the time interval

Substituting,

$E=(111,900)(15)=1.68\cdot 10^6 J$

However, the snowmobile is 18% efficient: this means that only 18% of this energy is converted into useful work. Therefore, the work output is

$W=0.18E=(0.18)(1.68\cdot 10^6)=3.02\cdot 10^5 J$

Learn more about power and work:

brainly.com/question/7956557

brainly.com/question/6763771

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#LearnwithBrainly

3 0

3 years ago