Ask question

Ask question

All categories

3 years ago

13

A laserbeam takes 24ms to travel from a rocket to a reflective surface of a planet and back to the rocket. How far, in miles, is

the rocket from this planet’s surface?
A)2200 miles
B)3600 miles
C)2400 miles

please show work, thank you!

2 answers:

aksik [14]3 years ago

8 0

Answer:B

Explanation:

finlep [7]3 years ago

5 0

It’s a virus don’t believe it

You might be interested in

A 3-m-high, 7-m-wide rectangular gate is hinged at the top edge and is restrained by a fixed ridge. Determine the hydrostatic fo

Shalnov [3]

Answer:

The Hydrostatic force is $F = 137.2 kN$

The location of pressure center is $Z = 1.333 \ m$

Explanation:

From the question we are told that

The height of the gate is $h = 3 \ m$

The weight of the gate is $w = 7 \ m$

The height of the water is $h_w = 2 \ m$

The density of water is $\rho_w = 1000 \ kg/m^3$

Note used $h_w$ for height of water and height of gate immersed by water since both have the same value

The area of the gate immersed in water is mathematically represented as

$A = h_w * w$

substituting values

$A = 2* 7$

$A = 14 \ m^2$

The hydrostatic force is mathematically represented as

$F = \rho_w * g * h_f * A$

Where

$h_f =h- h_w$

$h_f =3 -2$

$h_f = 1\ m$

So

$F = 1000 * 9.8 * 1 * 14$

$F = 137.2 kN$

The center of pressure is mathematically represented as

$Z = h_f + \frac{I_g}{h_f * A}$

Where $I_g$ is the moment of inertia of the gate which mathematically represented as

$I_g = \frac{w * h_w^2}{12}$

The $h_w$ is the height of gate immersed in water

$I_g = \frac{7 * 2^2 }{12}$

$I_g = 4.667\ kg m^2$

Thus

$Z = 1 + \frac{4.66}{1 * 14}$

$Z = 1.333 \ m$

3 0

3 years ago

A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th

melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore, the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F = 624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

5 0

3 years ago

What compounds are you made of?

Vanyuwa [196]

I am made of literally THOUSANDS of compounds ... too many to list here.
But the one compound that's most abundant, and actually comprises almost
80% of my entire beautiful body, is the compound DiHydrogen Oxide, with
the molecular formula H₂O . This compound is commonly known as "water".

6 0

3 years ago

Read 2 more answers

The wind increases to 14 mph from the north. Now what is your airspeed and what direction are you flying? If your destination is

ladessa [460]

Answer:

I-

Explanation:

...

7 0

4 years ago

A transmission diffraction grating with 420 lines/mm is used to study the light intensity of di event orders (n). A screen is lo

Goshia [24]

Answer:

Explanation:

Diffraction grating is used to form interference pattern of dark and bright band.

Distance between adjacent slits (a ) = 1 / 420 mm

= 2.38 x 10⁻³ mm

2.38 x 10⁻⁶ m

wave length of red light

= 680 x 10⁻⁹ m

For bright red band

position x on the screen

= n λD / a , n = 0,1,2,3 etc

D = distance of screen

putting n = 1 , 2 and 3 , we can get three locations of bright red band.

x₁ = λD / a

= 680 x 10⁻⁹ x 2.8 / 2.38 x 10⁻⁶

= .8 m

= 80 cm

Position of second bright band

= 2 λD / a

= 2 x 80

= 160 cm

Position of third bright band

= 3 λD / a

= 3 x 80

= 240 cm

5 0

3 years ago