Answer:
it will double because im right
The answer is C. Hope this helps.
Answer:
-2.26×10^-4 radians
Explanation:
The solution involves a right angle triangle
Length is z while the horizontal is the height x
X^2+ 100^2=z^2
Taking the derivatives
2x(dx/dt)=Z^2(dz/dt)
Specific moments = Z= 200 ,X= 100sqrt3 and dx/dt= 11
dz/dt= 1100sqrt3/200 = 9.53
Sin a= 100/a
Taking derivatives in terms of t
Cos a(da/dt)=100/z^2 dz/dt
a= 30°
Cos (30°)da/dt= (-100/40000×9.5)
a= -2.26×10^-4radians
Answer:
W =1562.53 N
Explanation:
It is given that,
Radius of the aluminium ball, r = 24 cm = 0.24 m
The density of Aluminium, 
We need to find the thrust and the force. The mass of the liquid displaced is given by :

V is volume
Weight of the displaced liquid
W = mg

So,

So, the thrust and the force is 1562.53 N.
Answer:
.
Explanation:
By Newton's Second Law, the acceleration
of an object is proportional to the net force
on it. In particular, if the mass of the object is
, then
.
Rewrite this equation to obtain:
.
In this case, the assumption is that the
force is the only force that is acting on the object. Hence, the net force
on the object would also be
Make sure that all values are in their standard units. Forces should be in Newtons (same as
, and the acceleration of the object should be in meters-per-second-squared (
). Apply the equation
to find the mass of the object.
.