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Karolina [17]
3 years ago
13

When the chemical name of a compound is being written, the subscripts will determine the symbols. elements. prefixes. suffixes.W

hen the chemical name of a compound is being written, the subscripts will determine the symbols. elements. prefixes. suffixes.
Physics
2 answers:
inn [45]3 years ago
6 0

Answer:

C is correcting taking test now

Explanation:

Xelga [282]3 years ago
5 0

Answer:

C. Prefixes

Explanation:

I took the quiz on edge

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A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this
Novay_Z [31]

Answer:

SEE EXPLANATION

Explanation:

p =  \frac{fd}{t}  \\ where \: \\p  = power \\  f = force \\ d = distance \\ and \: t = time \\  \\ p =  \frac{3800 \times 50}{10}  \\ p =  \frac{190000}{10}  \\ p = 19000w

7 0
3 years ago
What is free nitrogen
Alenkasestr [34]

Answer:

It is simply molecular nitrogen (N2). Nitrogen, in its molecular form, consists of two nitrogen atoms bound together with a tripple bond

Explanation:

4 0
3 years ago
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A coin mass 0.005 kg is dropped from a height of 3 m. How much mechanical energy does it have right before it hits the ground ?
lesya692 [45]
Mechanical energy equals the sum of potential and kinetic energy. During the process, all PE converts into KE, assuming air resistance is neglected. So, the mechanical energy does not change and is equal to the initial potential energy.
ME
=mgh
=0.005 x 9.81 x 3
=0.147J
6 0
3 years ago
We have an Atwood device, two blocks connect by a string strung over a pulley, but the twist this time is that both blocks are o
Zanzabum

The Acceleration of the system is 6.41 m/s².

Given,

α= 15°, m₁ = 7kg

β= 65°, m₂ = 11 kg

Let, a be the acceleration and T is the tensions at the end it's the cord.

Let, the mass m₂ be coming down along the inclined plane along the inclined surface towards downward m₂g sin β and the tension in the upward direction,

Resultant force, m₂a=m₂g sin β -T

11a=((11) ×g sin 65°) -T  ...(i)

Now, considering the motion of m₁ which moves downwards, the forces are m₁g sinα, and T both are acting downwards.

Resultant force m₁a = m₁g sin α+T

7a =7g sin 15°+T  ...(ii)

Solving both the equations by adding them,

18a=11gsin 65°+7g sin 15°-T+T

18a=11gsin 65°+7g sin 15°=115.45

a=115.45/18=6.41 m/s²

Hence, the Acceleration of the system is 6.41 m/s².

Learn more about the acceleration here:

brainly.com/question/22048837

#SPJ10

6 0
2 years ago
Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista
soldi70 [24.7K]

Answer:

v = 7793150 m/s

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

v = \frac{kQ}{r}

                 Where     k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }

and it is satisfy the superposition principle, thus

v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} +  \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}

v = 222.73v

The electrical potential at 10 cm from charge 1 is:

v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} +  \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}

v = 395.44 v

Since the work - energy theorem, we have:

q\Delta v = \frac{mv^{2} }{2}

                     where q is the electron's charge and m is the electron's mass

Therefore:

v = \sqrt{\frac{2q\Delta v}{m} }

v = 7793150 m/s

6 0
3 years ago
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