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2 years ago

A tank is 3 m long 1 m wide and 0.5 m deep is filled with oil which weighs 12000 N

1 answer:

5 0

Area of the bottom = L x W = 3 x 1 = 3 square meters. Pressure = force/area = 12000/3 = 4000 N/square meter = 4 kPa. Since we know the weight of the oil, we don't need to know the depth.

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Which of the following statements about iron filings placed upon glass resting on top of a bar magnet is false? A. The direction

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<span>C. The filings will be clustered more densely where the field is weakest.</span>

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2 years ago

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What type of galaxy is this?

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Answer:

Barred Spiral.

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1 year ago

ou have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,

Dimas [21]

So far, since you moved into the apartment until the end of this much of the story, you haven't done ANY work on the dresser yet.

I'll admit that you pushed, groaned and grunted, sweated and strained plenty. You're physically and mentally exhausted, you're not interested in the dresser at the moment, and right now you just want to snappa cappa brew, crash on the couch, and watch cartoons on TV. But if you've done your Physics homework, you know you haven't technically done any <u><em>work</em></u> yet.

In Physics, "Work" is the product of Force times Distance.

Since the dresser hasn't budged yet, the Distahce is zero. So no matter how great the Force may be, it's multiplied by zero, so the <em>Work is zero</em>.

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2 years ago

A woman drops a vibrating tuning fork, which is vibrating at 513 Hertz, from a tall building. Through what distance (in m) has t

umka21 [38]

Answer:

h = 15.34 m

Explanation:

given,

tuning fork vibration = 513 Hz

speed of sound = 343 m/s

frequency after deflection = 489 Hz

the source (the fork) moves away from the observer, its speed increases and hence the apparent frequency decreases

$f_{apparent} = \dfrac{v}{v+u}f_0$

$489 = \dfrac{343}{343+u}\times 513$

$0.953 = \dfrac{343}{343+u}$

$343+u = \dfrac{343}{0.953}$

$343+u = 359.92$

u = 16.92 m/s

height of the building

v² = u² + 2 g s

16.92² = 2 x 9.8 x h

h = 14.61 m

time taken by sound to reach observer

$t = \dfrac{14.61}{343}$

$t =0.0426\ s$

in this time tuning fork has fallen one more now,

$h' = u t + \dfrac{1}{2}gt^2$

$h' = 16.92\times 0.0426 + \dfrac{1}{2}\times 9.8 \times 0.0426^2$

h' = 0.7296 m = 0.73 m

total distance

h = 14.61 + 0.73

h = 15.34 m

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2 years ago

A)Calculate the effective value of g, the acceleration of gravity, at 7900 m above the Earth's surface.

arsen [322]

The solution for the acceleration of gravity is given as

$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$$

$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$$

This is further explained below.

<h3>What is the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.?</h3>

Generally,

Mass of earth $M=5.97 \times 10^{24} \mathrm{~kg}$

Radius of earth $R=6371 \mathrm{~km}$

Gravitational const. $G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$

height $h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$$$

$R+h_{1}=6371+7.9 &\\\\R+h_{1}=6378.9 \mathrm{~km} \\\\&R+h_{1}=6378.9 \times 10^{3} \mathrm{~m}\end{aligned}$

In conclusion, acceleration due to gravity at this point will be

$g_{1}=\frac{G M}{\left(\bar{R}+\overline{h_{1}}\right)^{2}}$\\\\$g_{1}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(6378.9 \times 10^{3}\right)^{2}}$\\\\$g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$$

for $h_{2}=7900 \mathrm{~km}$

$R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\$g_{2}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(14271 \times 10^{3}\right)^{2}}$\\\\$g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$$

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1 year ago