The inaccurate measurements must be similar to the other two measurements (ex; 590, 589, 599), but different from the actual volume of water. (Ex; the actual volume is let say.. 100, but you measured 50, 49, 40)
They can't hear an echo in small room because in it the sound can't be reflected back. For an echo of a sound to be heard,the minimum distance between the source of sound and the walls of the roomshould be 17.2 m. Obviously,in asmall room echoes cannot beheard.
Answer:
0.0768 revolutions per day
Explanation:
R = Radius
= Angular velocity
As the mass is conserved the angular momentum is conserved
Moment of intertia for solid sphere
Moment of intertia for hollow sphere
Dividing the moment of inertia
From the first equation
The angular velocity, in revolutions per day, of the expanding supernova shell is 0.0768 revolutions per day
Answer:
9.81 m/s² constant
Explanation:
Any object which is falling has only the acceleration due to gravity acting on it. The value of acceleration due to gravity is 9.81 m/s² which is constant. This is the case if air resistance is not taken into consideration.
The air resistance is a result of the surface area of the object which is falling. This will slow down the object and the velocity reached is called the terminal velocity.
Answer:
(a) x = 0.25 m
(b) v = 1.46 m/s
(c) v = 2.4 m/s
Explanation:
mass (m) = 10.3 kg
force from thrust (F) = 240 N
spring constant (k) = 400 N/m
stretch distance from thrust (y) = 30 cm = 0.3 m
acceleration due to gravity (g) = 9.8 m/s^{2}
(A) from mg = kx
compression (x) = mg/ k
x = 
x = 0.25 m
(B) from the conservation of forces
(Fy) + (0.5k
) = (0.5k
) + mgh + (0.5m
)
v = ![\sqrt{[tex]\frac{(Fy) + (0.5k[tex]x^{2}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Btex%5D%5Cfrac%7B%28Fy%29%20%2B%20%280.5k%5Btex%5Dx%5E%7B2%7D)
) - (0.5k) - mgh }{0.5m}[/tex]}[/tex]
v = ![\sqrt{[tex]\frac{(240 x 0.3) + (0.5 x 400 x [tex]0.25^{2}](https://tex.z-dn.net/?f=%5Csqrt%7B%5Btex%5D%5Cfrac%7B%28240%20x%200.3%29%20%2B%20%280.5%20x%20400%20x%20%5Btex%5D0.25%5E%7B2%7D)
) - (0.5 x 400 x 
) - (10.3 x 9.8 x (0.25 + 0.3)) }{0.5 x 10.3}[/tex]}[/tex]
v = 1.46 m/s
(C) if the rocket weren't attached to the spring, the conservation of energy equation becomes
(Fy) + (0.5k) = mgh + (0.5m)
v = ) - mgh }{0.5m}[/tex]}[/tex]
v = ) - (10.3 x 9.8 x (0.25 + 0.3)) }{0.5 x 10.3}[/tex]}[/tex]
v = 2.4 m/s
