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3 years ago

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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

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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it is about 384,750 kilometers from earth to the moon. it took the apollo astronauts about 2 days and 19.5 hours to fly to the m

Jet001 [13]

We know, speed = Distance / Time
d = 384,750 Km
t = 2 days, 19.5 hours = 48+19.5 = 67.5 hour

Substitute their values,
s = 384,750 / 67.5
s = 5700 Km/h

In short, Your Answer would be 5700 Km/h

Hope this helps!

7 0

3 years ago

Read 2 more answers

1. If an object of mass m collides and velocity v collides inelastically with an object of mass 3m that is initially at rest, th

Archy [21]

Answer:

1a). 4 times. 1b) 4. 1c) 1/4.

2a) 5 times. 1b) 5 1c) 1/5

3a) 7/3 times 3b) 7/3 3c) 3/7

4a) 8/5 times 4b) 8/5 4c) 5/8

Explanation:

1) Assuming no external forces acting during the collision, total momentum must be conserved, so the following general expression applies:

m₁*v₁₀ + m₂*v₂₀ = m₁*v₁f + m₂*v₂f (1)

If we assume that the collision is perfectly inelastic, this means that both masses stick together after the collision, so v₁f = v₂f.

If m₂ is initially at rest, ⇒ v₂₀ = 0.

Replacing in (1) we get the expression of vf as a function of v₁₀, as follows:

vf = v₁₀*(m₁/(m₁+m₂)

So, for the four cases we have the following:

1) initial mass = m

final mass = m+3m = 4 m

⇒final mass / initial mass = 4

vf = v₀* (m/4m) = v₀/4 ⇒v₀/vf = 4

So, the velocity of the system will decrease by a factor of 4. The new velocity will be vf= v₀/4.

2) Applying the same considerations, we get:

2a) final mass / initial mass = 5

2b) vf = v₀* (m/5m) = v₀/5 ⇒v₀/vf = 5

2c) vf = v₀/5

3) Applying the same considerations, we get:

3a) final mass / initial mass = 7/3

3b) vf = v₀* (3m/7m) =3/7* v₀ ⇒v₀/vf = 7/3

3c) vf = 3/7*v₀

4) Applying the same considerations, we get:

4a) final mass / initial mass = 8/5

4b) vf = v₀* (5m/8m) = 5/8*v₀ ⇒v₀/vf = 8/5

4c) vf = 5/8*v₀

3 0

3 years ago

a bullet moving with a velocity of 100m/s pierce a block of wood and moves out with a velocityof 10 m/s.if the thickness of the

erma4kov [3.2K]

The emerging velocity of the bullet is <u>71 m/s.</u>

The bullet of mass <em>m</em> moving with a velocity <em>u</em> has kinetic energy. When it pierces the block of wood, the block exerts a force of friction on the bullet. As the bullet passes through the block, work is done against the resistive forces exerted on the bullet by the block. This results in the reduction of the bullet's kinetic energy. The bullet has a speed <em>v</em> when it emerges from the block.

If the block exerts a resistive force <em>F</em> on the bullet and the thickness of the block is <em>x</em> then, the work done by the resistive force is given by,

$W=Fx$

This is equal to the change in the bullet's kinetic energy.

$W=Fx=\frac{1}{2} m(u^2-v^2)......(1)$

If the thickness of the block is reduced by one-half, the bullet emerges out with a velocity v<em>₁.</em>

Assuming the same resistive forces to act on the bullet,

$F(\frac{x}{2} )=\frac{1}{2} m(u^2-v_1^2)......(2)$

Divide equation (2) by equation (1) and simplify for v<em>₁.</em>

$\frac{\frac{Fx}{2} }{Fx} =\frac{(u^2-v_1^2)}{(u^2-v^2)} \\\frac{100^2-v_1^2}{100^2-10^2} =\frac{1}{2} \\v_1^2=5050\\v_1=71.06 m/s$

Thus the speed of the bullet is 71 m/s

3 0

3 years ago

11. A 3.8 kg object is lifted 12 meters. Approximately how much work is performed during the lifting?

Marianna [84]

I found the answer for you if u need any help ask anytime!

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3 years ago

If the tension of the cable is 25.0 N what is the mass of the ball

marysya [2.9K]

Answer:576

Explanation:

5 0

2 years ago