Which path would the laser take when entering the water from air? <br> brown or black

Galileo and Newton both studied the night sky. Galileo studied Jupiter and its moons. Newton studied the gravitational force tha

denis23 [38]

Both had different interests

7 0

4 years ago

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A 2.50 gram rectangular object has measurements of 22.0 mm, 13.5 mm, and 12.5 mm. what is the objectâs density in units of g/ml?

navik [9.2K]

<span>To calculate the density, we must first calculate the volume of the object. For a rectangle, the volume is equal to V=l*w*h. In this case, V=22*13.5*12.5=3712.5 mm^3. Next we have to convert the mm^3 to mL using the conversion factor of 1 mm^3=.001 mL. To get mL, we set up the expression 3712.5 mm^3 * (.001mL/1mm^3) = 3.7125 mL. Finally, to get the density, we know density is mass over volume, so we set the expression p=m/V=2.5g/3.7125mL=.6734 which rounded to 3 significant figures is .673. The answer will be .673</span>

8 0

3 years ago

What is the mass of 1.5 in newtons?

Thepotemich [5.8K]

Answer:

0.15 kg

Explanation:

F = mg

1.5N = m(9.8 m/s^2)

m = 0.15kg

8 0

3 years ago

The earth's hydrosphere includes which of the following?

Oduvanchick [21]

The answer is all of these

5 0

3 years ago

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A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial

Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

$t=\frac{2usin\theta}{g}$

Here θ = 30° and t = 20 s

a) $t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s$

Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

$H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m$

Maximum altitude = 490.5 m

c) Range of projectile is given by

$R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m$

Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

We have equation of motion s = ut + 0.5 at²

Horizontal motion

u = 169.91 m/s

a = 0 m/s²

t = 15 s

Substituting

s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

Vertical motion

u = 98.1 m/s

a = -9.81 m/s²

t = 15 s

Substituting

s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

$\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m$

Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0

4 years ago

Which path would the laser take when entering the water from air? brown or black