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anyanavicka [17]
3 years ago
14

In the symbol 42He, the subscript 2 is the _____ for helium and the superscript 4 is the _____ for helium. atomic mass; atomic n

umber atomic number; atomic mass mass number; atomic number atomic number; mass number
Physics
2 answers:
aleksklad [387]3 years ago
6 0
Mass number ; atomic number
Scrat [10]3 years ago
3 0

Answer:

2 is the atomic number while 4 is the mass number.

Explanation:

The symbol of an element is Z X A. Where Z is the number of protons or called atomic number of the element and A be the sum of number of protons and neutrons or called mass number of that element,

Here 42He,

Z = 2, it means atomic number is 2, which shows that number of proton are 2.

A = 4, it means mass number, which shows that the sum of number of neutrons and protons are 4.

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Kg. m
Irina18 [472]

Answer: 0.00211

Explanation:

7 0
3 years ago
Free body diagram definition <br>in physics​
kirill115 [55]

Answer:

In physics and engineering, a free body diagram (force diagram, or FBD) is a graphical illustration used to visualize the applied forces, moments, and resulting reactions on a body in a given condition.

4 0
3 years ago
A jeweler is determining the optical properties of an unknown blue gemstone. She uses an angle of incidence of 62°, and measures
Juliette [100K]

Answer:

n = 1.76

Explanation:

According to the rule of ( n1 sin theta1 = n2 sin theta2 )

we know both angles so we insert them to the law and apply n1 = 1

so 1/2 = n2 sin 62 and we get the final answer

3 0
4 years ago
A circular loop of flexible iron wire has an initial circumference of 164cm , but its circumference is decreasing at a constant
Travka [436]

Answer:

emf = 0.02525 V

induced current with a counterclockwise direction

Explanation:

The emf is given by the following formula:

emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}\ \ =-B\frac{A_2-A_1}{t_2-t_1}   (1)

ФB: magnetic flux =  BA

B: magnitude of the magnetic field = 1.00T

A2: final area of the loop; A1: initial area

t2: final time, t1: initial time

You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.

In t = 4 s the final circumference will be:

c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm

To find the areas A1 and A2 you calculate the radius:

r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm

r1 = 0.261 m

r2 = 0.190 m

Then, the areas A1 and A2 are:

A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2

Finally, the emf induced, by using the equation (1), is:

emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV

The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.

4 0
3 years ago
The bar ab has a velocity at point a of va = [20,9] m/s and an angular velocity w = 23 rad/s. If the distance between a and b is
DiKsa [7]

speed of point A is given as

v_a = 20.9 m/s

v_a = r*\omega

here r = distance from the axis

so here we have

20.9 = r*23

r  = 0.91 m

now the distance of point A and B is 0.71 m

while the distance of axis from point A is 0.91 m

so distance of axis from point B will be given as

r = 0.91 + 0.71 = 1.62 m

now for the speed of point b is given as

v = r \omega

v = 1.62 * 23

v = 37.26 m/s

so end point B will move with speed 37.26 m/s

8 0
3 years ago
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