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3 years ago

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9. Kokio dydžo Archimedo jėga veikia 0,5 m² tūrio medinį rastą vandenyje?

1 answer:

ICE Princess25 [194]3 years ago

6 0

Answer:

i dont knowwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwww

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Who reported four “element” classifications, but included some substances that were combinations of elements rather than true el

marishachu [46]

Explanation :

Antoine-Laurent Lavoisier reported four "element" classifications but included some substances that were combinations of elements rather than true elements in his listing.

He is also known as " father of modern chemistry". He gives the modern system of naming chemical substances. He also gives a theory for chemical reactivity of the oxygen.

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3 years ago

Select the correct answer.

VLD [36.1K]

Answer:O – H

Explanation:

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2 years ago

A 3kg object has an initial velocity (6i - 2j) m/s (a ) what is its kinetic energy at this time? (b) Find total work done on the

guapka [62]

Answer:

$K.E = \frac{1}{2} m {v}^{2} \\ {v}^{2}_i = {v}^{2} _x + {v}^{2} _y \\ = {(6)}^{2} + {( - 2)}^{2} = 36 + 4 = 40m. {s}^{ - 1} \\ K.E_i = \frac{1}{2} (3) (40) = 60J \\ \\ {v}^{2}_f = {v}^{2} _x + {v}^{2} _y \\ = {(8)}^{2} + {(4)}^{2} = 80m. {s}^{ - 1} \\ K.E_i = \frac{1}{2} (3) (80) = 120J \\ W_{net}=K.E_f-K.E_i \\ = 120J - 60J \\ = 60J$

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3 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

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3 years ago

a typical cmall flashlight contains two batteries each having na emf of 2.0 v connected in series with a bulb havin ga resistanc

Helen [10]

Answer:

P = 0.25 W

Explanation:

Given that,

The emf of the battry, E = 2 V

The resistance of a bulb, R = 16 ohms

We need to find the power delivered to the bulb. We know that, the formula for the power delivered is given by :

$P=\dfrac{V^2}{R}\\\\P=\dfrac{2^2}{16}\\\\=0.25\ W$

So, 0.25 W power is delivered to the bulb.

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3 years ago