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3 years ago

How many milligrams are equivalent to 150 dekagrams?

Physics

2 answers:

riadik2000 [5.3K]3 years ago

5 0

Hello,
150g=150×10000=15000000 dekg=1,5e^6dekg

Bye:-)

dybincka [34]3 years ago

4 0

i believe the answer is 1.5e+6

hope this helps!

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Light travels at a speed of close to 3 x 10^5 km/s in vacuum. Given that it takes light 8 min and 19 s to travel the distance fr

never [62]

Answer:

1497×10⁵ km

Explanation:

Speed of light in vacuum = 3×10⁵ km/s

Time taken by the light of the Sun to reach the Earth = 8 min and 19 s

Converting to seconds we get

8×60+19 = 499 seconds

Distance = Speed × Time

$\text{Distance}=3\times 10^5\times 499\\\Rightarrow \text{Distance}=1497\times 10^5\ km$

1 AU = 1497×10⁵ km

The Sun is 1497×10⁵ km from Earth

8 0

3 years ago

If we double the frequency of a system undergoing simple harmonic motion, which of the following statements about that system ar

AnnyKZ [126]

Answer:

a. The angular frequency is doubled.

e. The period is reduced to one-half of what it was.

Explanation:

Angular frequency is given as;

ω = 2πf

$\frac{\omega _1}{f_1} = \frac{\omega _2}{f_2}$

when the frequency is doubled

$\frac{\omega _1}{f_1} = \frac{\omega _2}{(2f_1)} \\\\\omega _1 = \frac{\omega _2}{2}\\\\\omega _2 = 2\omega _1$

Thus, the angular frequency will be doubled.

Amplitude in simple harmonic motion is the maximum displacement.

Frequency is related to period in simple harmonic motion as given in the equation below;

$f = \frac{1}{T} \\\\f_1T_1= f_2T_2\\\\T_2 = \frac{f_1T_1}{f_2}$

when the frequency is doubled;

$T_2 = \frac{f_1T_1}{2f_1} \\\\T_2 = \frac{T_1}{2}$

Thus, the period will be reduced to one-half of what it was.

5 0

3 years ago

In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo

irinina [24]

Answer:

The torque about the origin is $2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}$

Explanation:

Torque $\overrightarrow{\tau}$ is the cross product between force $\overrightarrow{F}$ and vector position $\overrightarrow{r}$ respect a fixed point (in our case the origin):

$\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}$

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

$\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]$

$\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})$

$\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})$

$\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}$

4 0

2 years ago

How does black paint interact with a light wave?

KATRIN_1 [288]

Answer:

Light rays that come from a source such as the sun reflect off items and enter our eye. ... While black objects absorb the energy from all colors and become hot, the objects gradually release some of that energy back into the air around it

Explanation:

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8 0

2 years ago

An AC source operating at 59 Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 kΩ) and an induc

Alexxandr [17]

I = V/Z

V = voltage, I = current, Z = impedance

First let's find the total impedance of the circuit.

The impedance of the resistor is:

$Z_{R}$ = R

R = resistance

Given values:

R = 1200Ω

Plug in:

$Z_{R}$ = 1200Ω

The impedance of the inductor is:

$Z_{L}$ = j2πfL

f = source frequency, L = inductance

Given values:

f = 59Hz, L = 2.4H

Plug in:

$Z_{L}$ = j2π(59)(2.4) = j889.7Ω

Add up the individual impedances to get the Z, and convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$

Z = 1200 + j889.7

Z = 1494∠36.55°Ω

I = V/Z

Given values:

V = 170∠0°V (assume 0 initial phase)

Z = 1494∠36.55°Ω

I = 170∠0°/1494∠36.55°Ω

I = 0.1138∠-36.55°A

Round the magnitude of I to 2 significant figures and now you have your maximum current:

I = 0.11A

5 0

3 years ago