I truly believe so, that’s a definite yes
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:
You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.
Explanation:
Given Values:
L = 50 cm = 0.5 m
H = 170 j/s
To find the diameter of the rod, we have to find the area of the rod using the following formula.
Here Tc = 100.0° C
k = 50.2
H = k × A × ![\frac{[T_{H -}T_{C} ] }{L}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BT_%7BH%20-%7DT_%7BC%7D%20%5D%20%7D%7BL%7D)
Solving for A
A = ![\frac{H * L }{k * [ T_{H}- T_{C} ] }](https://tex.z-dn.net/?f=%5Cfrac%7BH%20%2A%20L%20%7D%7Bk%20%2A%20%5B%20T_%7BH%7D-%20T_%7BC%7D%20%5D%20%7D)
A = ![\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}](https://tex.z-dn.net/?f=%5Cfrac%7B170%20%2A%200.5%7D%7B50.2%20%2A%20%5B%20350%20-%20100%20%5D%7D)
A =
= 6.77 ×
m²
Now Area of cylinder is :
A =
d²
solving for d:
d = 
d = 9.28 cm
Answer:
Height, h = 16.67 m
Explanation:
We have,
Mass of a squirrel is 0.765 kg.
He jumps off the tree and hits the ground with 125 joules of energy.
It is required to find the height up on the tree the squirrel was when it jumped.
The energy possessed by the squirrel is called its gravitational potential energy. It can be given by :

h is height up on the tree the squirrel was when it jumped

So, the squirrel will go to a height of 16.67 m.