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Leno4ka [110]
3 years ago
8

If a total force exerted by water in a container with a bottom area of 2 square meters is 450 newtons, what's the water pressure

at the bottom of the container?        A. 0.900 kPa   B. 0.225 kPa   C. 0.575 kPa   D. 0.300 kPa
Physics
2 answers:
m_a_m_a [10]3 years ago
7 0

its B 0.225kPa using the formula p=f/A then change the pascals into kpa

Ipatiy [6.2K]3 years ago
7 0

The Correct answer to this question for Penn Foster Students is: 0.225 kPa

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^^^^^^^^^^^^^^^^^^^^^^^^^^^ is correct
5 0
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The centrifugal force which tends to throw the electrons out it's orbit as stated with mass moving velocity in an orbit of radiu
sammy [17]

Answer:

In Motion in Two and Three Dimensions, we examined the basic concepts of circular motion. An object undergoing circular motion, like one of the race cars shown at the beginning of this chapter, must be accelerating because it is changing the direction of its velocity. We proved that this centrally directed acceleration, called centripetal acceleration, is given by the formula

\[{a}_{\text{c}}=\frac{{v}^{2}}{r}\]

3 0
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A centripetal-acceleration addict rides in uniform circular motion with period T = 2.0 s and radius r = 3.00 m. At t1 his accele
Dmitry_Shevchenko [17]

centripetal acceleration always points  towards the center of the circular path and velocity of object in circular motion always points towards the tangent on the the circle in this way centripetal acceleration and velocity are perpendicular to each other and the dot product of  perpendicular vectors is always zero,

therefore v•a=vacosα...........(1)

here α is angle between centripetal acceleration and velocity which is 90

therfore,

From equation (1)

v.a= vacos90

v.a=vax0..............(because cos90=0)

v.a=0 m^2

centripetal acceleration vector points towards center  it means it point towards inwards direction, so it lies along the radius vector,and radius vector points towards outward direction of the circle in this way centripetal acceleration and radius vector are in exact opposite direction so angle between them is 180 degree,

therefore r x a = rasin180

rxa=rax0            (because sin180=0)

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6 0
3 years ago
1. Earth is approximately a sphere of radius 6.37х106 m. what are (a) its circumference in kilometers, (b) its surface are in sq
Minchanka [31]

As we know that circumference of the sphere is given as

C = 4\pi R

here we know that

R = 6.37 \times 10^6 m

now we have

C = 4\pi (6.37 \times 10^6)

C = 8\times 10^7 m

C = 8 \times 10^4 km

PART B)

surface area of the sphere is given as

A = 4\pi R^2

R = 6.37 \times 10^3 km

A = 4\pi (6.37\times 10^3)^2

A = 5.1 \times 10^8 km^2

PART C)

Volume of the sphere is given as

V = \frac{4}{3}\pi R^3

here we have

V = \frac{4}{3}\pi(6.37 \times 10^3)^3

V = 1.1 \times 10^{12} km^3

5 0
4 years ago
Read 2 more answers
A 0.09 g honeybee acquires a charge of +23 pc while flying. The earth's electric field near the surface is typically (100 N/C, d
hodyreva [135]

Answer:

2.6\times 10^{-5}

Explanation:

<u>Given:</u>

  • m = mass of the honeybee = 0.09 g = 9\times 10^{-5}\ kg
  • q = charge on the honeybee = 23 pC = 2.3\times 10^{-11}\ C
  • E = electric field near the surface of earth = 100 N/C

<u>Assume:</u>

  • g = acceleration due to gravity = 9.8\ m/s^2
  • W = weight of the honeybee
  • F = electric force on the honeybee
  • R = ratio of the electric force and the weight of the honeybee

We know that

F = qE\,\,\, and\,\,\, W = mg\\\therefore R = \dfrac{F}{W}\\\Rightarrow R = \dfrac{qE}{mg}\\\Rightarrow R = \dfrac{2.3\times 10^{-11}\ C\times 100 N/C}{9\times 10^{-5}\ kg\times 9.8\ m/s^2}\\\Rightarrow R = 2.6\times 10^{-6}

So, the ration of the electric force on the bee to its weight is 2.6\times 10^{-6}.

On multiplying this ration by 10, the ratio becomes 2.6\times 10^{-5}.

8 0
4 years ago
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