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:
Explanation:
a rigid object in uniform rotation about a fixed axis does not satisfy both the condition of equilibrium .
First condition of equilibrium is that net force on the body should be zero.
or F net = 0
A body under uniform rotation is experiencing a centripetal force all the time so F net ≠ 0
So first condition of equilibrium is not satisfied.
Second condition is that , net torque acting on the body must be zero.
In case of a rigid object in uniform rotation , centripetal force is applied towards the centre ie towards the line joining the body under rotation with the axis .
F is along r
torque = r x F
= r F sinθ
θ = 0 degree
torque = 0
Hence 2nd condition is fulfilled.
Pretty sure it is weather :))
Answer:
6.003×10¯⁶ N
Explanation:
We'll begin by converting 1 cm to m. This can be obtained as follow:
100 cm = 1 m
Therefore,
1 cm = 1 cm × 1 m / 100 cm
1 cm = 0.01 m
Finally, we shall determine the gravitational attraction. This can be obtained as follow:
Mass 1 (M₁) = 3 Kg
Mass 2 (M₂) = 3 Kg
Distance apart (r) = 0.01 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Force of attraction (F) =?
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 3 × / 0.01²
F = 6.003×10¯¹⁰ / 1×10¯⁴
F = 6.003×10¯⁶ N
Thus the gravitational attraction is 6.003×10¯⁶ N
It will take 13
seconds for the golf ball to hit the ground. The correct answer between
all the choices given is the last choice or letter D. I am hoping that this
answer has satisfied your query and it will be able to help you in your
endeavor, and if you would like, feel free to ask another question.
