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2 years ago

Which is observed globally: "Earth Day" or "Arbor Day"?

Physics

2 answers:

sattari [20]2 years ago

8 0

Earth day is more globally observed. Arbor day is lowkey rare

neonofarm [45]2 years ago

5 0

Answer:

Earth Day is observed globally

Explanation:

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A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wal

asambeis [7]

Answer:

The distance is 3.1 m

Explanation:

The position vector of the fly relative to the corner of the wall is

r = (3.1, 0.5).

The distance of the fly from the corner will be calculated as the magnitude of the vector "r"

magnitude of vector $r = \sqrt{(3.1 m)^{2} + (0.5 m)^{2}} = 3.1 m$

Since the numbers to be added have only one decimal place 3.<u>1</u> and 0.<u>5</u>, the result of the sum will have to have one decimal place. The result of the square root will also have one decimal place.

5 0

2 years ago

At what angle two forces P + Q and (P - Q) act so that their resultant is :

stiv31 [10]

Use resultant formula

$\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}$

A be p+q and B be p-q

$\\ \rm\Rrightarrow R=\sqrt{3p^2+q^2}$

$\\ \rm\Rrightarrow \sqrt{(p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha}=\sqrt{3p^2+q^2}$

$\\ \rm\Rrightarrow 2p^2+2q^2+2(p^2-q^2)cos\alpha=3p^2+q^2$

$\\ \rm\Rrightarrow 2cos\alpha=1$

$\\ \rm\Rrightarrow cos\alpha=\dfrac{1}{2}$

$\\ \rm\Rrightarrow \alpha=\dfrac{\pi}{3}$

$\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)$

$\\ \rm\Rrightarrow 2cos\beta=0$

$\\ \rm\Rrightarrow cos\beta=0$

$\\ \rm\Rrightarrow \beta=\dfrac{\pi}{2}$

$\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2$

$\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)$

$\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}$

$\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)$

8 0

2 years ago

Read 2 more answers

At what time after being ejected is the boulder moving at a speed 20.7 m/s upward?

Svetlanka [38]

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

a = g = -9.8 $$$m / s^2$

downward (acceleration due to gravity).

By using Suvat equation:

v = u + at

where: v is the velocity at time t

u = 40.0 m/s is the initial velocity

a = g = -9.8 $$$m/s^2$ is the acceleration

To find the time t at which the velocity is v = 20.7 m/s

Therefore,

$$t=\frac{v-u}{a}=\frac{20.7-40}{-9.8}=2.0204 \mathrm{~s}$

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

The complete question is:

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?

To learn more about uniformly accelerated motion refer to:

brainly.com/question/14669575

#SPJ4

4 0

1 year ago

A system releases 255 cal of heat to the surroundings while delivering 428 cal of work. what is the change in internal energy of

vichka [17]

Answer:-683 cal

Explanation:

Given

Heat released by system Q=-255 cal

as heat released is taken as negative and vice-versa