First we need to find the time in the air
d = 1/2gt^2 + vt
Since there is no inital vertical velocity
4.1 = 1/2 (9.8) t^2
8.2/9.8 = t^2
0.8367 = t^2
t = 0.9147s
Now we multiply by speed
7m/s x 0.9147s = 6.4m
Power = (voltage) x (current)
Divide each side by 'voltage': Current = power / voltage
= (60 watts) / (120 volts) = 0.5 ampere
Answer Explanation :
Poiseuille equation: this equation is used for non ideal flow this is used for the calculation of pressure in laminar flow it is physical law we know that fluid in laminar flow, flows across the pipe whose diameter is larger than the length of pipe
in mathematical form the equation can be expressed as
Q =
where η is the cofficient of viscosity
now if we assume a small sphere of radius a is suspended freely in the plane of the laminar flow then for assuring that the sphere does not migrate with the flow we have to calculate the rate of flow of the liquid
1. To solve this problem, you must apply formula of Universal Gravitation Law, which is shown below:
F=Gm1m2/r²
F=2.75x10^-12
G=6.7x10^-11
r=2.6 m
m2=2m1
2. You must clear m1, as below:
F=G(m1)(2m1)/r²
F=G(2m1)²/r²
m1=√(Fr²/2G)
3. When you susbtitute the values into the formula m1=√(Fr²/2G), you obtain:
m1=√(Fr²/2G)
m1√0.13
m1=0.37
m2=2m1
m2=2(0.37)
m2=0.74
4. Therefore, the answer is:
m1=0.37
m2=0.74
Taking into account the Newton's first Law, the correct answer is option C. To overcome an object's inertia, it must be acted upon by a force.
Newton's First Law, also called the Law of inertia, indicates that "Every body perseveres in its state of rest or of uniform rectilinear motion unless it is forced to change its state by forces impressed on it." This means that for a body to come out of its state of rest or of uniform rectilinear motion, it is necessary for a force to act on it.
In other words, it is not possible for a body to change its initial state (be it rest or motion) unless one or more forces intervene.
Finally, the correct answer is option C. To overcome an object's inertia, it must be acted upon by a force.
Learn more: