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3 years ago

The beginning of the Phanerozoic is marked by what occurrence

Physics

1 answer:

Sergio039 [100]3 years ago

3 0

The beginning of the Phanerozoic is marked by the development of hard body parts, such as shells and bones.

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A person is lying on a diving board 2.00 m above the surface of the water in a swimming pool. She looks at a penny that is on th

wlad13 [49]

Answer:

7.98 m

Explanation:

In the given question,

distance above surface= 2 m

Distance penny from person = 8 m

Since the swimming pool is filled with water and atmosphere has air therefore the refractive index phenomenon will occur.

The refractive index of water: air is 4/3 (1.33).

Using the formula, 4/3 = real depth, apparent depth

real depth= 4/3 x apparent depth

Now, calculating apparent depth = 8 - 2

= 6 m

therefore, real depth = 4/3 x apparent depth

= 1.33 x 6

= 7.98

thus, 7.98 m is the real depth of water.

8 0

2 years ago

Is work required to pull a nucleon out of an atomic nucleus? Does the nucleon, once outside the nucleus, hove more mass than it

olga2289 [7]

<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>

6 0

3 years ago

The social interaction theory explains that the purpose of aggression is to control another person’s behavior.

meriva

Sounds like true. ....

5 0

2 years ago

Read 2 more answers

Calculate the force of gravity between a comet with a mass of 500kg and a small asteroid with a mass of 20kg that is separated b

givi [52]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The equivalent gravitational force is ~

$F \approx1.48\times 10 {}^{ - 7} \: \: N$

$\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}$

We know that ~

$\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}$

where,

F = gravitational force

$m_1$ = mass of 1st object = 500 kg

$m_2$ = mass of 2nd object = 20kg

G = gravitational constant = $6.674 × {10}^ {-11}$

r = distance between the objects = 2.12 m

Let's calculate the force ~

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 500 \times 20}{(2.12) {}^{2} }$

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 10 {}^{4} }{4.4944}$

$F = \dfrac{6.674}{4.4944} \times 10 {}^{ - 7}$

$F =1.484 \times 10 {}^{ - 7} \: \: newtons$

7 0

2 years ago

The pressure of a gas is reduced from 1200.0 mm hg to 850.0 mm hg as the volume of its container is increased by moving a piston

Kamila [148]

From gas laws:
$\frac{PV}{T} = Constant$

Therefore,
$\frac{ P_{1} V_{1} }{ T_{1} } = \frac{ P_{2} V_{2} }{ T_{2} }$

P1 = 1200 mm
V1 = 85 ml
T1 = 90°C = 363.15 K
P2 = 850 mm
V2 = 350 ml
T2 = ?

Substituting;
$T_{2} = \frac{ P_{2} V_{2} T_{1} }{ P^{1} V^{1} } = \frac{850*350*363.15}{1200*85} = 1059.19 K$ = 786.04 °C

8 0

3 years ago

Read 2 more answers