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4 years ago

Personality tests are one of the most accurate ways to let you know exactly who you are

1 answer:

8 0

If this is a True/False question the answer is False. the most accurate ways to let you know exactly who you are is to... well, be exactly who you are! That's all I can really say here. Personality tests may be on target sometimes but only you can know who you truly are, you know?

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If the depth of water in a well is 10m, what is the pressure exerted by it the bottom of the well ? ( Use g = 10 m/s2)

Tasya [4]

Answer:

The precise answer depends on the density and therefore the temperature of the water, but we can obtain a reasonable approximation by assuming that the density of the water is 1000 kilograms per cubic meter (kg/m³).

Since the depth of the water in the well is 10 m, the volume of water directly above an area A of a square meters (m²) at the bottom of the well is 10×a m³.

Since the density of the water is 1,000 kg/m³, the mass of water directly above area A is (1,000 kg/m³) × (10×a m³) = (1000×10×a kg) = 10,000×a kg.

Since g = 9.8 m/s², the force of gravity acting on the water directly above area A is (9.8 m/s²) × (10,000×a kg) = 9.8×10,000×a N (newtons) = 98,000×a N.

So the pressure of water acting on area A is (98,000×a N)/(a m²) = (98,000×a)/a N/m² = 98,000 pascals (pa). And since A could be any given area at the bottom of the well, this is the pressure at any point at the bottom of the well.

So the pressure at the bottom of the well is 98,000 pascals (or 98,000/101,325 standard atmospheres = 560/579 atmospheres ~ 0.967 standard atmospheres).

Please comment below if you have any questions.

7 0

3 years ago

A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.13 s. W

vitfil [10]

Answer:

Part a)

T = 0.52 s

Part b)

$f = 1.92 Hz$

Part c)

$speed = 3.65 m/s$

Explanation:

As we know that the particle move from its maximum displacement to its mean position in t = 0.13 s

so total time period of the particle is given as

$T = 4\times 0.13 = 0.52 s$

now we have

Part a)

T = time to complete one oscillation

so here it will move to and fro for one complete oscillation

so T = 0.52 s

Part b)

As we know that frequency and time period related to each other as

$f = \frac{1}{T}$

$f = \frac{1}{0.52}$

$f = 1.92 Hz$

Part c)

As we know that

wavelength = 1.9 m

frequency = 1.92 Hz

so wave speed is given as

$speed = wavelength \times frequency$

$speed = 1.92 \times 1.9$

$speed = 3.65 m/s$

4 0

3 years ago

An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 2530 km. When the astronaut jumps upwa

Natali [406]

Answer:

1.38*10^18 kg

Explanation:

According to the Newton's law of universal gravitation:

$F=G*\frac{m_a*m_p}{r^2}$

where:

G= Gravitational constant (6.674×10−11 N · (m/kg)2)

ma= mass of the astronaut

mp= mass of the planet

$F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2$

so:

$m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg$

7 0

4 years ago

Photo shows all! Need help ASAP! Will mark brainlist

Goryan [66]

Answer:

The student is getting different info bc the students probable keeping track of the distance instead of the displacement.

Explanation:

5 0

3 years ago

A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro

tekilochka [14]

Answer:

distance between object and image = 18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm ......................3

so from equation 2 and 3

distance between object and image = 5.4 + 13.5

distance between object and image = 18.9 cm

6 0

3 years ago