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3 years ago

The number 3 in hydrogen in NH33 is the

Physics

1 answer:

frosja888 [35]3 years ago

4 0

Answer:

no of atoms

Explanation:

for each amonia molecule one nitrogen atom bind with 3 hydrogen atoms

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A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

$g=9.8 m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

3 0

3 years ago

The lower the value of the coefficient of friction,the_the resistance to sliding

malfutka [58]

Answer:

lower

Explanation:

The lower the value of the coefficient of friction, the lower the resistance to sliding.

The coefficient of friction is the ratio of the frictional force and the normal force pressing two surfaces in contact together.

U = $\frac{F}{N}$

U is the coefficient of friction

F is the frictional force

N is the normal force

We see that coefficient of friction is directly proportional to frictional force.

7 0

2 years ago

Star #1 is approaching the Earth with speed v. Star #2 is receding from the Earth with the same speed v. Measurements of the sam

leva [86]

Answer:

b. Both stars will have the same shift.

Explanation:

It's a very simple problem to solve. Star 1 is approaching toward Earth with a speed v, so let's assume that the change in Doppler Shift is +F and Star 2 is moving away so the change in Doppler shift is -F. But it's time to notice the speed of both stars and that is same but only directions are different. speed is the main factor here. The magnitude of both shifts is F as we can see and + and - are showing there direction of motion. So, because of same amount of speed, both stars will have same shift magnitude. (Just the directions are different)

4 0

3 years ago

Why are collisions between galaxies more likely than collisions between stars within a galaxy?

KatRina [158]

Answer:

stars share a gravitational force with the galaxy while nearby galaxies do not share a gravitational field.

Explanation:

stars will not collide because they are bound by a gravitational orbit around the galaxy

8 0

3 years ago

If viewed at the correct time and in the right conditions, asteroids, comets, and meteors are all visible to the nked eye.

UkoKoshka [18]

Answer:

True

Explanation:

If it is at right conditions and correct time (night time most likely) they will all be visible

8 0

1 year ago