Here, you need to use your "Protractor" as it is given in the question, but we can calculate the value with the help of our mathematical calculation too:
[ Protractor can be use only in real life, not here ]

Draw an imaginary line from initial position to final position.
Now, In that triangle, tan x = P/B
tan x = 1.4 / 2
tan x = 0.70
x = tan⁻¹ (0.70)
x = 35 [ tan 35 = 0.70 ]

In short, Your Answer would be 35 degrees

Hope this helps!

3 0

3 years ago

A hockey puck of mass m1=165 g slides from left to right with an initial velocity of 15.5 m/s. It collides head on with a second

Fynjy0 [20]

Answer:

it may be -6 m/s

Explanation:

3 0

4 years ago

A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.

Nataliya [291]

Answer:

Approximately $1.2 \times 10^{4}\; {\rm J}$ (assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The change in the gravitational potential energy ( ${\rm GPE}$ ) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass $m$ increased by $\Delta h$ , the ${\rm GPE}$ of that object would have increased by $m\, g\, \Delta h$ .

In this question, the height of this object increased by $\Delta h = 6.0\; {\rm m}$ . The mass of this object is $m = 200\; {\rm kg}$ . Thus, the ${\rm GPE}$ of this object would have increased by:

$\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}$ .

(Note that $1\; {\rm N \cdot m} = 1\; {\rm J}$ .)

3 0

2 years ago