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Maslowich
3 years ago
7

Look at the equation. What detail is missing? 3 m/s2= (33 m/s - X)/30 S ​

Physics
1 answer:
Tems11 [23]3 years ago
6 0

Answer:

The starting velocity.

Explanation:

We must understand that this equation comes from the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity = 33 [m/s]

Vo = starting velocity [m/s]

a = acceleration = 3 [m/s²]

t = time = 30 [s]

So, these values can be assembly in the following way:

v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}

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A rifle with a mass of 0.9 kg fires a bullet with a mass of 6.0 g (0.006 kg). The bullet moves with a muzzle velocity of 750 m/s
SCORPION-xisa [38]

Answer:

a )  4.5 N.s

b) V =5 m/s    

Explanation:

given,

mass of rifle(M)  = 0.9 kg

mass of bullet(m)  = 6 g = 0.006 kg

velocity of the bullet(v)  = 750 m/s

a) momentum of bullet = m × v

                                  = 750 × 0.006

                                  = 4.5 N.s

b) recoil velocity                                                      

m × u + M × U = m × v + M × V

0  + 0  = 0.006 × 750 -  0.9 × V

V = \dfrac{4.5}{0.9}

V =5 m/s                    

 

5 0
3 years ago
A strong lightning bolt transfers an electric charge of about 31 C to Earth (or vice versa). How many electrons are transferred?
olchik [2.2K]

Answer:

m=5.78\times 10^{-3}\ g

n_e=1.935\times 10^{20} is the no. of electrons

Explanation:

Given:

  • quantity of charge transferred, Q=31\ C

<u>No. of electrons in the given amount of charge:</u>

As we have charge on one electron 1.602\times 10^{-19}\ C

so,

n_e=\frac{Q}{e}

n_e=\frac{31}{1.602\times 10^{-19}}

n_e=1.935\times 10^{20} is the no. of electrons

  • Now if each water molecules donates one electron:

Then we require n=1.935\times 10^{20} molecules.

<u>Now the no. of moles in this many molecules:</u>

n_m=\frac{n}{N_A}

where

N_A=6.022\times 10^{23} Avogadro No.

n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}

n_m=3.213\times 10^{-4}\ moles

  • We have molecular mass of water as M=18 g/mol.

<u>So, the mass of water in the obtained moles:</u>

n_m=\frac{m}{M}

where:

m = mass in gram

3.213\times 10^{-4}=\frac{m}{18}

m=5.78\times 10^{-3}\ g

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3 years ago
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Answer:

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Describe a process in which energy changes forms at least twice
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