Is N2 a triple bond

Calculate the average orbital speed of Ceres in

marin [14]

] Ceres is composed of rock and ice and is estimated to comprise approximately one third of the mass of the entire asteroid belt. Ceres is the only object in the asteroid belt known to be rounded by its own gravity (though detailed analysis was required to exclude 4 Vesta). From Earth, the apparent magnitude of Ceres ranges from 6.7 to 9.3, peaking once every 15 to 16 months,[21]hence even at its brightest it is too dim to be seen with the naked eye except under extremely dark skies.

8 0

3 years ago

A bird flies overhead from where you stand at an altitude of 270.0ĵ m and at a velocity horizontal to the ground of 14.0î m/is.

Varvara68 [4.7K]

Answer:

L = 8694 Kg.m²/s

Explanation:

r = 270 ĵ m

v = 14 î m/s

m = 2.3 kg

θ = 90º

L = ?

We can apply the equation

L = m*v*r*Sin θ

L = (2.3 kg)*(14 m/s)*(270 m)*Sin 90º = 8694 Kg.m²/s

8 0

3 years ago

To test the performance of its tires, a car

Rom4ik [11]

The coefficient of static friction is 0.222

Explanation:

In order for the car to remain in circular motion, the frictional force must be able to provide the necessary centripetal force. Therefore, the car will start skidding when the two forces are equal:

$\mu mg=m\frac{v^2}{r}$

where the term on the left is the frictional force, while the term on the right is the centripetal force, and where

$\mu$ is the coefficient of static friction

m is the mass of the car

g is the acceleration of gravity

v is the speed of the car

r is the radius of the track

In this problem, we have:

r = 564 m

v = 35 m/s

$g=9.8 m/s^2$

And re-arranging the equation for $\mu$ , we can find the coefficient of static friction:

$\mu = \frac{v^2}{gr}=\frac{35^2}{(9.8)(564)}=0.222$

Learn more about friction:

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#LearnwithBrainly

5 0

2 years ago

SHOW WORK

Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

$T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\$

For Point b:

$E=\frac{1}{2} m a^2 w^2\\\\$

$=\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J$

For Point C:

$V_{max}= a w$

$= (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}$

For point D:

$X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm$

5 0

2 years ago

Please help<br> types of stimuli the sensory organs receive.

Alborosie

Answer: Photoreceptor cell, Chemoreceptor, Thermoreceptor, Mechanoreceptor

Hope this helps(:

Explanation:

3 0

2 years ago