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2 years ago

Which part of the electromagnetic spectrum generally gives us our best views of stars forming in dusty clouds?

1 answer:

3 0

We can see stars forming in dusty clouds because of the infrared portion of the electromagnetic spectrum. Infrared waves pass through dust in outer space more easily than visible light. Interstellar dust absorbs visible light, a phenomena known as dust extinction.

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A nucleus with a mass number of 64 has a mean radius of about: _________.

dangina [55]

Answer:

The correct option is A

Explanation:

From the question we are told that

The mass number is $A= 64$

Generally the mean radius is mathematically evaluated as

$R = R_o A^{\frac{1}{3} }$

Here $R_o$ is a constant with a value $R_o =1.2*10^{-15}$

$R = 1.2*10^{-15} * 64^{\frac{1}{3} }$

$R = 4.8 *10^{-15}$

$R = 4.8\ fm$

5 0

2 years ago

Near the surface of Earth an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (

pentagon [3]

Answer:

$q=2.997\times 10^{-4}$ C

Sign-Negative

Explanation:

We are given that

Electric field = $E=100NC^{-1}$ (Radially downward)

Acceleration= $0.19 ms^{-2}$ (Upward)

Mass of charge=3 g= $3\times 10^{-3}$ kg

1kg=1000g

We have to find the magnitude and sign of charge would have to be placed on a penny .

By newton's second law

$\sum F_y=ma$

$\sum F_y=qE-mg$

Substitute the values then we get

$qE-mg=ma$

Substitute the values then we get

$q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)$

$100q-29.4\times 10^{-3}=0.57\times 10^{-3}$

$100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}$

$q=\frac{29.97\times 10^{-3}}{100}$

$q=2.997\times 10^{-4}$ C

Sign of charge =Negative

Because electric force acting in opposite direction of electric field therefore,charge on penny will be negative.

4 0

3 years ago

A ball is attached to a string and rotates in a circle. if it takes 1.00 s to complete one revolution, what is the angular veloc

Paha777 [63]

An object roating at one revokution per second has an angular velocity of 360 degrees per second or 2pi radians per second. This is found by taking the number of revolutions over a period of time and than dividing by the chosen period of time to get the velocity. There are 360 degrees or 2pi radians in one revolution.

8 0

2 years ago

Define the unit of time and unit of length

garik1379 [7]

This is your perfect answer

The base unit for time is the second (the other SI units are: metre for length, kilogram for mass, ampere for electric current, kelvin for temperature, candela for luminous intensity, and mole for the amount of substance). The second can be abbreviated as s or sec.

7 0

2 years ago

Read 2 more answers

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

Known Data

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

Time of the dropped Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

Time of the Thrown Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$

Finally, we can find how much earlier does the thrown rock strike the ground, so $t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s$

6 0

3 years ago