All categories

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

You might be interested in

Whichiscorrect?(A)anopaqueobjectdoesnotallowlighttopassthroughit.(B)an

Anuta_ua [19.1K]

Answer:

option a is the correct

Explanation:

4 0

2 years ago

Read 2 more answers

A 70.0Kg cyclist Rides a 15.0kg

Inessa [10]

Answer:

$1275 kg*m/s$

Explanation:

We'll use the momentum equation:

$p=mv$

where:

p = momentum

m = mass

v = velocity

Since we're doing the magnitude of momentum of the system, we'll add the mass of the cyclist and the mountain bike together:

$70.0 kg + 15.0kg=85.0 kg$

Given that, we can now substitute our given values into the momentum equation:

$p=mv\\p=85.0kg*15.0m/s$

Our final answer is:

$p= 1275 kg*m/s$

4 0

1 year ago

Explain how you know that gasoline burning in a car engine is a chemical change.?

polet [3.4K]

Burning, in general, is a chemical change.
Why?
This is because we are breaking and/or forming new bonds. We form a totally different substance.

5 0

2 years ago

Read 2 more answers

the maximum displacement of an oscillatory motion is A=0.49m. determine the position x at which the kinetic energy of the partic

kipiarov [429]

Answer:0.4 m

Explanation:

Given

Maximum displacement A=0.49

The sum of kinetic and elastic potential energy is $\frac{1}{2}kA^2$

where k=spring constant

U+K.E.= $\frac{1}{2}kA^2$

when K.E.=U/2

K.E.=kinetic energy

U=Elastic potential Energy

$\rightarrow \ U+\frac{U}{2}=\frac{1}{2}KA^2\\\rightarrow \ \frac{3U}{2}=\frac{1}{2}KA^2\\\rightarrow \ U=\frac{1}{3}KA^2\\\rightarrow \ \frac{Kx^2}{2}=\frac{1}{3}KA^2\\\\x=\sqrt{\frac{2}{3}}A\\x=0.4\ m$

3 0

2 years ago

The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.

Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let $\rho_{m}$ be the density of metal and $\rho_{w}$ be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

$W =0.10400\ N$

$mg=0.10400$

$\rho V g=0.10400$

$Vg=\dfrac{0.10400}{\rho_{m}}$ .....(I)

The weight of metal in water is

Using buoyancy force

$F_{b}=0.10400-0.08400$

$F_{b}=0.02\ N$

We know that,

$F_{b}=\rho_{w} V g$ ....(I)

Put the value of $F_{b}$ in equation (I)

$\rho_{w} Vg=0.02$

Put the value of Vg in equation (II)

$\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02$

$1000\times\dfrac{0.10400}{0.02}=\rho_{m}$

$\rho_{m}=5200\ kg/m^3$

Hence, The density of the metal is 5200 kg/m³.

6 0

3 years ago