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4 years ago

7

A descriptive investigation is a type of _____.

2 answers:

denpristay [2]4 years ago

8 0

It’s D or B Because investigation is some tho ur studying or experiment to know the final result Most likely D it’s not a or c

Lesechka [4]4 years ago

6 0

D because i took this test already and i got it correct so you should too

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The wattage marked on a lightbulb is not an inherent property of the bulb but depends on the voltage to which it is connected, u

Mashutka [201]

Answer:

The current that flows through the lamp is 0.5 A.

Explanation:

A lamp functions like a resistor and the real power absorbed by a resistor is given by the product of the voltage drop across it's terminals and the current that flows through it. If we wish to find the current that this lamp draws we should divide the wattage given (60 W) by the voltage drop provided (120 V). We then have:

i = P/V = 60/120 = 0.5 A.

5 0

3 years ago

How long does it take for the Earth to make a complete revolution around the sun?

timurjin [86]

365 days so a year basically

4 0

3 years ago

Read 2 more answers

PLEASE ANYONE CAN HELP ME !E.x/A block of metal has a volume of 0.09 m3

LuckyWell [14K]

Answer:

B = 1058.4 N

Explanation:

Given that,

The volume of a metal block, V = 0.09 m³

The density of fluid, d = 1200 kg/m³

We need to find the buoyant force when it's Completely immersed in brine. The formula for the buoyant force is given by :

$B=\rho gV$

g is acceleration due to gravity

$B=1200\times 9.8\times 0.09\\\\B=1058.4\ N$

So, the required buoyant force is 1058.4 N.

3 0

3 years ago

The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic

Drupady [299]

a) Density at 100 degrees: $1.34\cdot 10^4 kg/m^3$

Explanation:

The density of mercury at 0 degrees is $d=1.36\cdot 10^4 kg/m^3$

Let's take 1 kg of mercury. Its volume at 0 degrees is

$V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3$

The formula to calculate the volumetric expansion of the mercury is:

$\Delta V= \alpha V \Delta T$

where

$\alpha=180\cdot 10^{-6} K^{-1}$ is the cubic expansivity of mercury

V is the initial volume

$\Delta T$ is the increase in temperature

In this part of the problem, $\Delta T=100 C-0 C=100 C=100 K$

So, the expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3$

b) Density at 22 degrees: $1.355\cdot 10^4 kg/m^3$

We can apply the same formula we used before, the only difference here is that the increase in temperature is

$\Delta T=22 C-0 C=22 C=22 K$

And the volumetric expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3$

8 0

3 years ago

2. Which is NOT a unit of speed?

Ratling [72]

The answer is D) second/meter

7 0

2 years ago