Answer:
19.62 ms
Explanation:
t = Time taken = 2 s
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive)
Equation of motion
The speed of the pebble when it hit the water is 19.62 ms
Answer:
sorry mate i dont know
Explanation:
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Option (ii) B is the correct option. The object on the moon has greater mass.
To resolve this, utilize the formulas Force = Mass * Acceleration.
The equation can be used to find the mass given the force in Newtons, using 9.8 m/s² for the acceleration of gravity of the earth and 1.6 m/s² for the moon.
Calculating the mass on earth:
30 N = 9.8 m/s² * mass
This results in a mass of 3.0 kg for the object on Earth.
Calculating the mass of the moon:
30 N = 1.6 m/s²2 * mass
Thus, the moon's object has a mass of 19. kg.
This can be explained by the fact that the earth has a stronger gravitational pull than the moon, producing more force per kilogram of mass. As a result, the moon's mass must be bigger to produce the same amount of force at a lower acceleration from gravity (1.6 m/s² vs. 9.8 m/s²).
To know more about Mass, refer to this link :
brainly.com/question/13386792
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The imaginary second level is 60 dB more intense than the real level of the caller.
60 dB means a multiplication of 10⁶ = <em>1 million.</em>
That's how many equally-talented callers it would take to be 60 dB louder than him.
Answer:
1.dr/dt=0.0096cm/s
2. dA/dt=2.19cm^2/s
Explanation:
A spherical balloon is deflating at 10 cm3/s. At what rate is the radius changing when the volume is 1000π cm3 ? What is the rate of change of surface area at this moment?
for this question, we need to analyze the parameters we know
V=volume of the spherical balloon 1000π cm3
volume of the sphere=
1000π=4/3πr^3
dividing both sides by 4
250*3=r^3
r=9.08cm, the radius of the balloon
dv/dt=dv/dr*dr/dt...................................1
dv/dr ,means
V=
dv/dr=4*pi*r^2
dv/dt=10 cm3/s
from equ 1
10=4*pi*9.08^2*dr/dt
10=1036 dr/dt
dr/dt=10/1036
dr/dt=0.0096cm/s
2. to find the rate at which the area is changing we have,
dA/dt=dA/dr*dr/dt
area of a sphere is 4πr^2
differentiate a with respect to r, radius
dA/dr=8πr
dA/dt=8πr*0.0096
dA/dt=8*pi*9.08*0.0096
dA/dt=2.19cm^2/s
is the rate of change of the surface area
