Answer:
The x-component of
is 56.148 newtons.
Explanation:
From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:
(1)
Where:
,
,
- External forces exerted on the ring, measured in newtons.
- Vector zero, measured in newtons.
If we know that
,
,
and
, then we construct the following system of linear equations:
(2)
(3)
The solution of this system is:
, 
The x-component of
is 56.148 newtons.
The data convincingly show that wave frequency does not affect wave speed. An increase in wave frequency caused a decrease in wavelength while the wave speed remained constant. The last three trials involved the same procedure with a different rope tension.
The functions that would be performed both by the placenta and the hatchery so that the embryos will survive is to maintain the temperature of the embryos. The temperature should also be at the temperature where the embryos would thrive and develop.
Answer:
Rotational kinetic energy = 0.099 J
Translational kinetic energy = 200 J
The moment of inertia of a solid sphere is
.
Explanation:
Rotational kinetic energy is given by

where <em>I</em> is the moment of inertia and <em>ω</em> is the angular speed.
For a solid sphere,

where <em>m</em> is its mass and <em>r</em> is its radius.
From the question,
<em>ω</em> = 49 rad/s
<em>m</em> = 0.15 kg
<em>r</em> = 3.7 cm = 0.037 m


Translational kinetic energy is given by

where <em>v</em> is the linear speed.
