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2 years ago

3500kg of water are poured into a water tank and fill a volume equal to 3.5m^3. a. What is the density of the water?

Physics

1 answer:

Vaselesa [24]2 years ago

8 0

Answer:

a )

Density = mass / volume

= 3500 / 3.5

= 1000 kg m⁻³

b) gauge pressure = h d g

h is height of water column

d is density

g is acceleration due to gravity

Put the values in the equation above

Gauge pressure = .15 x 1000 x 9.8

= 1470 Nm⁻²

Atmospheric pressure = 100000 Nm⁻²

Total pressure

= 101470 N m⁻²

c ) Volume of water displaced = volume of object

= 55 x 10⁻⁶ m³

Buoyant Force = weight of displaced water

= 55 x 10⁻⁶ x 1000 x 9.8

= .539 N.

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If it takes 200 joules of energy to lift a bucket of water 3 meters in 2 seconds, how much power would be required to do the sam

Harlamova29_29 [7]

200 joules of work energy are involved. That's all we need to know to answer the question. Once we know that 200 joules of work energy are involved, we don't care what was lifted, or how far, or how long it took, or how many people worked on it, or how much they were paid, or what was the distribution of their gender identities, or the ethnic diversity among the team. or what day each of them celebrates as their sabbath. Any other information besides the 200 joules is only there to distract us, and see whether we're paying attention.

Power = (work or energy) / (time to do the work or move the energy)

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3 years ago

The 1.5kg ball is launched straight upward with an initial velocity of 7m/s. What is the maximum height it will reach?

Temka [501]

Answer:

h = 2.5 m

Explanation:

Given that,

Mass of a ball, m = 1.5 kg

Initial velocity of the ball, u = 7 m/s

We need to find the maximum height reached by the ball. Let it is be h. Using the conservation of energy to find it such that,

$mgh=\dfrac{1}{2}mu^2\\\\h=\dfrac{u^2}{2g}$

Put all the values,

$h=\dfrac{7^2}{2\times 9.8}\\\\=2.5\ m$

So, it will reach to a height of 2.5 m.

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2 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

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2 years ago

Read 2 more answers