Answer: 1160 m
Explanation:
Speed = distance / time. Plug in 40 m/s for speed and 29 s for time in order to get the distance, 1160 m.
Answer: 0.0146m
Explanation: The formula that defines the velocity of a simple harmonic motion is given as
v = ω√A² - x²
Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.
The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA
One half of maximum speed = speed of motion
3ωA/2 = ω√A² - x²
ω cancels out on both sides of the equation, hence we have that
A/2 = √A² - x²
(0.0169)/2 = √(0.0169² - x²)
0.00845 = √(0.0169² - x²)
By squaring both sides, we have that
0.00845² = 0.0169² - x²
x² = 0.0169² - 0.00845²
x² = 0.0002142
x = √0.0002142
x = 0.0146m
Answer:
<h2> 4kg</h2>
Explanation:
Step one:
given
length of rod=2m
mass of object 1 m1=1kg
let the unknown mass be x
center of mass<em> c.m</em>= 1.6m
hence 1kg is 1.6m from the <em>c.m</em>
and x is 0.4m from the <em>c.m</em>
Taking moment about the <em>c.m</em>
<em>clockwise moment equals anticlockwise moments</em>
1*1.6=x*0.4
1.6=0.4x
divide both sides by 0.4 we have
x=1.6/0.4
x=4kg
The mass of the other object is 4kg
Answer:
Option A. 1 bar = 1 atm
Explanation:
Pressure has various units of measurement. Each unit of measurement can be converted to other units of measurement. For example:
1 atm = 1 bar
1 atm = 760 mmHg
1 atm = 760 torr
1 atm = 1×10⁵ N/m²
1 atm = 1×10⁵ Pa
With the above conversion scale we can convert from one unit to the other.
Considering the question given above, it is evident from the coversion scale illustrated above that only option A is correct.
Thus,
1 bar = 1 atm
Answer:
twice
Explanation:
From magnification = height of image / height of object
Distance of image/ distance of object = magnification
If the distance and height of the object represents the initial light distance and the exposed surface respectively.
And similarly the distance and height of the image represents the final light distance and the exposed surface respectively.
Hence the new image exposure would be twice as large.
If we use the formula our point of investigation is Height of image,
H2= D2/D1× H1
H2 = 2D2/D1 × H1
H2 = 2H1
