Answer:
a) The magnitude of the thrust provided by the jet's engines is 4840 newtons.
b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.
Explanation:
a) By Newton's laws we construct the following equations of equilibrium. Please notice that both the glider and the jet experiments has the same acceleration:
Jet
(1)
Glider
(2)
Where:
- Thrust of jet engines, measured in newtons.
- Tension in the cable connecting the jet and glider, measured in newtons.
, 
- Masses of the glider and the jet, measured in kilograms.
- Acceleration of the glider-jet system, measured in meters per square second.
If we know that 
, 
and 
, then the solution of this system of equations:
By (2):
By (1):
The magnitude of the thrust provided by the jet's engines is 4840 newtons.
b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.
To find:
The equation to find the period of oscillation.
Explanation:
The period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity.
Thus the period of a pendulum is given by the equation,
Where L is the length of the pendulum and g is the acceleration due to gravity.
On substituting the values of the length of the pendulum and the acceleration due to gravity at the point where the period of the pendulum is being measured, the above equation yields the value of the period of the pendulum.
Final answer:
The period of oscillation of a pendulum can be calculated using the equation,
<span>Earth's rotation is the rotation of the planet Earth around its own axis. The Earth rotates from the west towards east. As viewed from North Star or polestar Polaris, the Earth turns counter-clockwise.</span>
Answer:
True
Explanation:
It Depends on the order in which each object was plotted, if two connected points objects pass through the same set of three points, the shapes created by each point may be different
Answer:
Explanation:
We shall consider a Gaussian surface inside the insulation in the form of curved wall of a cylinder having radius equal to 3mm and unit length , length being parallel to the axis of wire .
Charge inside the cylinder = 250 x 10⁻⁹ C .
Let E be electric field at the curved surface , perpendicular to surface .
Total electric flux coming out of curved surface
= 2π r x 1 x E
= 2 x 3.14 x 3 x 10⁻³ E
According to Gauss's theorem , total flux coming out
= charge inside / ε ( 250 x 10⁻⁹C charge will lie inside cylinder )
= 250 x 10⁻⁹ / 2.5 x 8.85 x 10⁻¹² ( ε = 2.5 ε₀ = 2.5 x 8.85 x 10⁻¹² )
= 11.3 x 10³ weber .
so ,
2 x 3.14 x 3 x 10⁻³ E = 11.3 x 10³
E = 11.3 x 10³ / 2 x 3.14 x 3 x 10⁻³
= .599 x 10⁶ N /C .
