Hot air balloon ( look at pic)

ser-zykov [4K]

Answer:

A force pump can be used to raise water by a height of more than 10m, the maximum height allowed by atmospheric pressure using a common lift pump.

In a force pump, the upstroke of the piston draws water, through an inlet valve, into the cylinder. On the downstroke, the water is discharged, through an outlet valve, into the outlet pipe.

8 0

2 years ago

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The slowest animal ever discovered was a crab found in the Red Sea. It traveled with an average speed of 5.70km/yr. how long wou

Vaselesa [24]

If it takes 1 year for this crab to travel 5,70km

Then, it will takes approximatly 1000/5,70 = 175 years to travel 1000 km

8 0

2 years ago

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The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

2 years ago

The velocity of sound on a particular day outside is 331 meters/second. What is the frequency of a tone if it has a wavelength o

Leya [2.2K]

Answer:
frequency = 5.52 * 10² Hz

Explanation:
the equation that relates velocity, frequency and wavelength is:
velocity = frequency * wavelength

We are given that:
velocity = 331 m/sec
wavelength = 0.6 m

Substitute with the givens in the equation to get the frequency as follows:
velocity = frequency * wavelength
331 = frequency * 0.6
frequency = 331 / 0.6
frequency = 5.52 * 10² Hz

Hope this helps :)

4 0

3 years ago

What from the following list of statements about vectors is definitely true? (section 3.3) The magnitude of a vector can be smal

Mashcka [7]

Answer:

"Magnitude of a vector can be zero only if all components of a vector are zero."

Explanation:

"The magnitude of a vector can be smaller than length of one of its components."

Wrong, the magnitude of a vector is at least equal to the length of a component. This is because of the Pythagoras theorem. It can never be smaller.

"Magnitude of a vector is positive if it is directed in +x and negative if is is directed in -X direction."

False. Magnitude of a vector is always positive.

"Magnitude of a vector can be zero if only one of components is zero."

Wrong. For the magnitude of a vector to be zero, all components must be zero.

"If vector A has bigger component along x direction than vector B, it immediately means, the vector A has bigger magnitude than vector B."

Wrong. The magnitude of a vector depends on all components, not only the X component.

"Magnitude of a vector can be zero only if all components of a vector are zero."

True.

6 0

3 years ago