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OverLord2011 [107]
3 years ago
11

Science Help Please?? It is believed that solar nebular material came from which of the following:

Physics
2 answers:
kolbaska11 [484]3 years ago
7 0
B would be the answer to this
Ray Of Light [21]3 years ago
5 0
A is the answer, (I put B and got it wrong)
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Find the magnitude of the resultant forces 6N and 8N inclined at an angle 240° to each<br>other.​
Nana76 [90]

Answer:

F = 7.2N

Explanation:

The resultant of two forces acting at some angle is given by using the vector addition as given below

F =√F1^2+F2^2+2F1F2cosθ

Where F1 = 6N and F2 = 8N

θ = 240°

Substituting the values into the equation above

F = √ 6^2+8^2+ 2(6×8)cos240

F =√ 36+64+96cos240

F = √ 100+96 ×-0.5

F = √ 100-48

F = √ 52

F = 7.211

F = 7.2N

7 0
3 years ago
Question 1 of 10
Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field E at a distance R from a charge Q is given by

E = k\dfrac{Q}{R^2}

where k = 9*10^9Nm/C is the coulomb's constant.

Now, in our case

R = 0.0075m

Q = 0.0035C;

therefore,

E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}

\boxed{E = 5.6*10^{11}N/C.}

which is choice C from the options given<em> (at least it resembles it).</em>

6 0
3 years ago
In nuclear physics wht units are used to measure the radius of an atom ?
Alecsey [184]
Angstrom = 10^-10 m
for nucleus size are used fermi (femtometer  10^-15 m )
6 0
3 years ago
HElP I NEED TO turn this in
Alexeev081 [22]
B student 2 because you add
8 0
2 years ago
A rock thrown with speed 8.50 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 19.0 m bef
frez [133]
Draw a diagram to illustrate the problem as shown below.

The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s

Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s

The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
   = 21.675 m

Answer: The height is 21.7 m (nearest tenth)

4 0
3 years ago
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