Ask question

Ask question

All categories

2 years ago

7

A physicist is investigating a beam of laser light of wavelength 550 nm. The light strikes a target that is 189 meters away from

the source of the light. What is the travel time of the light from the source to the target? Let the speed of light c = 3.00 × 10^8 m/s.

1 answer:

Lena [83]2 years ago

8 0

Answer: here you go I was looking for this answer everywhere,I have it now so it’s 6.30 x 10^-7 s

Explanation:

I hope this helps☺️

You might be interested in

What is the scientific method in physics ? <br><br>grade 9

a_sh-v [17]

A method of investigation in which a problem is first identified and observations, experiments, or other relevant data are then used to construct or test hypotheses that purport to solve it.

5 0

2 years ago

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts help.

GrogVix [38]

Answer:

$H = 532 m$

Explanation:

When teacher falls into the cliff then she shout for Help at the same time

so here we know that sound will go down and reflect back up

so here in 3 s distance traveled by the sound

$d = vt$

$d = 340 \times 3$

$d = 1020 m$

now in the same time the distance that teacher will fall down is given as

$d_1 = \frac{1}{2}gt^2$

$d_1 = \frac{1}{2}(9.81)(3^2)$

$d_1 = 44.1 m$

now total distance traveled by teacher and sound in 3 s

$d_{total} = d + d_1$

$d_{total} = 1020 + 44.1$

this total distance must be equal to twice the height of the cliff

$2H = 1064.1 m$

$H = 532 m$

7 0

2 years ago

Read 2 more answers

Two inductors, L1 and L2, are in parallel. L1 has a value of 25 mH and L2 a value of 50 mH. The parallel combination is in serie

Bond [772]

Answer:

Explanation:

For parallel inductors ,

$\frac{1}{L_R} = \frac{1}{L_1} +\frac{1}{L_2}$

$\frac{1}{L_R} =\frac{1}{25} +\frac{1}{50}$

$L_R=16.67 mH.$

For series combination

Total inductance

= 16.67 + 20

= 36.67 mH .

reactance of total inductance at 300 kHz

= ω $L_{total}$ where ω is angular frequency

= 2πf $L_{total}$

= 2 x 3.14 x 300 x 10³ x 36.67 x 10⁻³

= 69.1 x 10³ ohm

Total rms current = Vrms / reactance

= 60 / 69.1 x 10³ A

= .87 x 10⁻³ A

= .87 mA

7 0

2 years ago

A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft

Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$ ,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

$V_2 = 109.37m/s$

The gain in velocity is 0.37m/s

4 0

2 years ago

Suppose that a spring with a larger spring constant was used in this same apparatus. If a given mass were rotated at the same ra

malfutka [58]

Answer:

The new period of rotation using the new spring would be less than the period of rotation using the original spring

Explanation:

Generally the period of rotation of the mass is mathematically represented as

$T = 2 \pi \sqrt{\frac{I}{k} }$

Here I is the moment of inertia of the mass about the rotation axis and k is the spring constant

Now looking at the equation we can tell that T is inversely proportional to the square root of the spring constant which means that for a larger spring constant the time period would be lesser

6 0

2 years ago