Answer:
0.000000002 m=2.0*10⁻⁹ m
Explanation:
Scientific notation allows us to write very large or very small numbers in abbreviated form. This notation simply consists of multiplying by a power of base 10 with a positive or negative exponent.
A number written in scientific notation has the form:
a*10ⁿ
where:
- the coefficient a has a value such that 1 ≤ a <10
- n is an integer. Represents the number of times the decimal point is shifted. It is always a whole number, positive if it is shifted to the left, negative if it is shifted to the right.
So to write the number 0.000000002 in scientific notation, the following steps are performed:
- The decimal point is moved to the right as many spaces until it reaches the right of the first digit.
- This number is then written, which will be the coefficient a in the expression of the previous product. So a=2.0
- The base 10 is written with the exponent equal to the number of spaces that the comma moves. So n=9. But this is a negative number because the comma shifts to the right.
So, you get: <u><em>0.000000002 m=2.0*10⁻⁹ m</em></u>
Answer: D) All of the above
Explanation:
Answer:
The answer to your question is m₂ = 38.5 kg
Explanation:
Data
distance = d = 2.1 x 10⁻¹ m
Force = 3.2 x 10⁻⁶ N
m₁ = 55 kg
m₂ = ?
G = 6.67 x 10 ⁻¹¹ Nm²/kg²
Process
1.- To solve this problem use Newton's law of Universal Gravitation.
F = G m₁m₂ / r²
-Solve for m₂
m₂ = Fr² / Gm₁
2.- Substitution
m₂ = (3.2 x 10⁻⁶)(2.1 x 10⁻¹)² / (6.67 x 10⁻¹¹)(55)
3.- Simplification
m₂ = 1.411 x 10⁻⁷ / 3.669 x 10⁻⁹
4.- Result
m₂ = 38.5 kg
Answer:
-10.8°, or 10.8° below the +x axis
Explanation:
The x component of the resultant vector is:
x = 3.14 cos(30.0°) + 2.71 cos(-60.0°)
x = 4.07
The y component of the resultant vector is:
y = 3.14 sin(30.0°) + 2.71 sin(-60.0°)
y = -0.777
Therefore, the angle between the resultant vector and the +x axis is:
θ = atan(y / x)
θ = atan(-0.777 / 4.07)
θ = -10.8°
The angle is -10.8°, or 10.8° below the +x axis.
Answers:
a) -171.402 m/s
b) 17.49 s
c) 1700.99 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
(3)
Where:
is the bomb's final jeight
is the bomb'e initial height
is the bomb's initial vertical velocity, since the airplane was moving horizontally
is the time
is the acceleration due gravity
is the bomb's range
is the bomb's initial horizontal velocity
is the bomb's fina velocity
Knowing this, let's begin with the answers:
<h3>b) Time</h3>
With the conditions given above, equation (1) is now written as:
(4)
Isolating :
(5)
(6)
(7)
<h3>a) Final velocity</h3>
Since , equation (3) is written as:
(8)
(9)
(10) The negative sign ony indicates the direction is downwards
<h3>c) Range</h3>
Substituting (7) in (2):
(11)
(12)
