Answer:
The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.
Explanation:
Answer:
A. 50 m/s
Explanation:
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 4 s
Find: v
v = at + v₀
v = (10 m/s²) (4 s) + 0 m/s
v = 40 m/s
In the x direction, the velocity is constant at 30 m/s.
The overall speed is:
v² = (30 m/s)² + (40 m/s)²
v = 50 m/s
Answer:
Thin, aluminium and buried underground.
Explanation:
When it comes to electrification of a state or province, some characteristics of the wire to use must be considered. This would help to minimize and avoid power loss and wire burns.
i. The wire to use should be thin, and a quite number can be twisted one against the other so as to increase the surface area for heat dissipation.
ii. Aluminium wire is more preferable for this project. It has a high melting point, and reduces energy loss.
iii. Burying the wire underground through an insulator is the best choice, though expensive but would preserve the wire from external influence.
Answer:
Explanation:
Use the following equation:
and solve for F:
and filling in:
F = 4.0 × 
N
