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3 years ago

Determine the density of a liquid from the following data .

2 answers:

6 0

Density = mass of liquid / Volume of liquid

Density = 33.79 / 13.3

Density = 2.54

revise the process for the second one also.

jolli1 [7]3 years ago

4 0

Mass = 40.14 - 33.79= 6.35 gram
vol = 13.3 ml
density = m/v = 6.35/13.3 = 0.477 g/ml

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What is a partial and total lunar eclipse?

Minchanka [31]

Partial Lunar Eclipse:

A partial lunar eclipse is when the earth gets between the Sun and Moon. However, all three bodies are not in alignment meaning we are able to see some more like part of the moon's surface as it moves in route of the Earth's shadow.

Total Lunar Eclipse:

The three celestial bodies are perfectly aligned which allows for the earth to completely block the sun's rays from hitting/reaching the moon. The sun is positions is in back of the Earth which then causes the shadow of the earth to be cast on the Moon covering the moon completely. When that happens you get the phenomenon called a total lunar eclipse.

Hopefully this helped and good luck.

7 0

3 years ago

A projectile is fired vertically from Earth's surface with

scoray [572]

Answer:

$h=25.52\times10^6 m$

Explanation:

Initial speed, v = 10 x 10^3 m/s

Mass of the earth, M = 6 x 10^24 kg

Radius of the earth, R = 6.4 x 10^6 m

Maximum from the surface of earth, h = ?

Let m = Mass of the projectile

Solution:

Potential energy at maximum height = ( Potential + Kinetic energy ) at the surface

$-G M m / ( R + h )=- G M m / R + (1/2) m v^2$

$- G M / ( R + h ) = - G M / R + (1/2) v^2$

$-2\times G M / ( R + h ) = ( - 2 G M / R ) + v^2$

$-2\times6.67\times10^{-11}\times6\times10^{24}/ ( R + h )$

= $( (- 2\times 6.67\times10^{-11}\times6\times10^{24}) /(6.4\times10^6)} +10000^2$

= $-2.50625\times10^7 J$

$=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7$

$R+h=31.92\times10^{6}$

$h=31.92\times10^{6}-6.4\times10^6$

$h=25.52\times10^6 m$

5 0

3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim

Bingel [31]

Answer:

given,

speed of the car = 20 m/s

final speed of car = 0 m/s

distance between car and the deer = 38 m

reaction time, t = 0.5 s

deceleration of the car = 10 m/s².

a) distance between deer and car

distance travel in the reaction time

d₁ = v x t

d₁ = 20 x 0.5 = 10 m

distance travel after you apply brake

using equation of motion

v² = u² + 2 a s

0 = 20² - 2 x 10 x s

s = 20 m

total distance traveled by the car

D = d₁ + d₂

D = 20 + 10 = 30 m

distance between car and the deer = 38 m - 30 m

= 8 m

b) now, maximum speed car.

distance travel in reaction time

d₁ = s x t

d₁ = 0.5 V

distance left between them

d₂ = 38 - d₁

d₂ = 38 - 0.5 V

distance travel after you apply brake

using equation of motion

v² = u² + 2 a d₂

0 = (V)² - 2 x 10 x (38 - 0.5 V)

V² + 10 V - 760 = 0

now, solving the quadratic equation

$x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}$

V = 23.01 , -33.01

rejecting the negative term.

hence, maximum speed of the car could be V = 23.01 m/s

7 0

3 years ago

A space probe is launched from Earth headed for deep space. At a distance of 10,000 miles from Earth's center, the gravitational

Pavlova-9 [17]

<span> gravitational force varies based on 1/r^2
when you're double the distance =10,000 to 20,000, the force is 4 times smaller so on and so forth.
</span><span>As force is proportional to 1 / {distance squared}, the force will be 1 / 2^2 (i.e. 1/4) of the force at the reference distance (i.e. 1/4 * 600 = 150 lb)
</span>hope this helps

5 0

4 years ago

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Aloiza [94]

Answer:

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3 years ago