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Schach [20]
3 years ago
13

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

Physics
2 answers:
sattari [20]3 years ago
8 0
Earth day is more globally observed. Arbor day is lowkey rare
neonofarm [45]3 years ago
5 0

Answer:

Earth Day is observed globally

Explanation:

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What will happen to the fringe spacing if the wavelength of the light is decreased?.
REY [17]

Answer:

The formula is the form -

m λ = d sin Θ

As the wavelength λ is decreased sin Θ will also decrease.

One can see from the derivation that as the wavelength, being considered,

is decreased the dispersion will also decrease.

3 0
3 years ago
A pendulum is swinging. It swings 85 complete swings and this takes 102 seconfs. What is the frequency? A. 0.83 Hz B.17Hz C.187
Serga [27]

Answer:

A: 0.83 Hz

Explanation:

Frequency can be calculated in a multitude of ways. The one way that is going to help you solve this problem is (# of times/seconds)

so you would divide 85 swings/102 seconds

= 5/6 or 0.8333 Hz

so your answer is A

3 0
3 years ago
. A car starts to move from rest. If its velocity becomes 90km/hr after 3s, calculate its acceleration​
Likurg_2 [28]

Answer:

32.5/s squared

Explanation:

6 0
3 years ago
A particle of mass m= 2.5 kg has velocity of v = 2 i m/s, when it is at the origin (0,0). Determine the z- component of the angu
melomori [17]

Answer:

please read the answer below

Explanation:

The angular momentum is given by

|\vec{L}|=|\vec{r}\ X \ \vec{p}|=m(rvsin\theta)

By taking into account the angles between the vectors r and v in each case we obtain:

a)

v=(2,0)

r=(0,1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

b)

r=(0,-1)

angle = 90°

L=(2.5kg)(1)(2\frac{m}{s})sin90\°=5.0kg\frac{m}{s}

c)

r=(1,0)

angle = 0°

r and v are parallel

L = 0kgm/s

d)

r=(-1,0)

angle = 180°

r and v are parallel

L = 0kgm/s

e)

r=(1,1)

angle = 45°

L = (2.5kg)(2\frac{m}{s})(\sqrt{2})sin45\°=5kg\frac{m}{s}

f)

r=(-1,1)

angle = 45°

the same as e):

L = 5kgm/s

g)

r=(-1,-1)

angle = 135°

L=(2.5kg)(2\frac{m}{s})(\sqrt{2})sin135\°=5kg\frac{m}{s}

h)

r=(1,-1)

angle = 135°

the same as g):

L = 5kgm/s

hope this helps!!

4 0
4 years ago
A motorboat accelerates uniformly from a velocity of 6.5 m/s west to a velocity of 1.5 m/s west. If its acceleration was 2.7 m/s
expeople1 [14]

Answer:

<em>The motorboat ends up 7.41 meters to the west of the initial position </em>

Explanation:

<u>Accelerated Motion </u>

The accelerated motion describes a situation where an object changes its velocity over time. If the acceleration is constant, then these formulas apply:

\vec v_f=\vec v_o+\vec a.t

\displaystyle \vec r=\vec v_o.t+\frac{\vec a.t^2}{2}

The problem provides the conditions of the motorboat's motion. The initial velocity is 6.5 m/s west. The final velocity is 1.5 m/s west, and the acceleration is 2.7 m/s^2 to the east. Since all the movement takes place in one dimension, we can ignore the vectorial notation and work with the signs of the variables, according to a defined positive direction. We'll follow the rule that all the directional magnitudes are positive to the east and negative to the west. Rewriting the formulas:

v_f=v_o+a.t

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Solving the first one for t

\displaystyle t=\frac{v_f-v_o}{a}

We have

v_o=-6.5,\ v_f=-1.5,\ a=2.7

Using these values

\displaystyle t=\frac{-1.5+6.5}{2.7}=1.852\ s

We now compute x

\displaystyle x=(-6.5)(1.852)+\frac{(2.7)(1.852)^2}{2}

x=-7,41\ m

The motorboat ends up 7.41 meters to the west of the initial position

5 0
3 years ago
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