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loris [4]
2 years ago
14

How far does a car travel in 10 s if its initial velocity is 3 m/s and its acceleration

Physics
1 answer:
victus00 [196]2 years ago
4 0

To learn more about Acceleration please visit -
brainly.com/question/13563385
#SPJ4



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When does DNA replication occur?
Digiron [165]

Answer:

NA is replicated during the S phase (Synthesis phase) of Interphase.

Hope it helps!!!!!!!!!!

5 0
3 years ago
Read 2 more answers
A set of charged plates 0.00262 m apart has an electric field of 155 N/C between them. What is the potential difference between
mylen [45]

Answer: The potential difference between the plates = 0.4061V

Explanation:

Given that the

Electric field strength E = 155 N/C

Distance d = 0.00262 m

From the definition of electric field strength, is the ratio of potential difference V to the distance between the plates. That is

E = V/d

Substitute E and d into the above formula

155 = V/0.00262

Cross multiply

V = 155 × 0.00262

V = 0.4061 V

The potential difference between the plates is 0.4061 V

5 0
3 years ago
Write 200m and 2 ×10^4m in km<br><br><br>write 5000ms and 5×10^7ms in seconds
WINSTONCH [101]

Answer:

a) 20.2 kilometers

b)50 005 seconds

Explanation:

5 0
3 years ago
A point charge Q is located a distance d away from the center of a very long charged wire. The wire has length L &gt;&gt; d and
zzz [600]

Answer:

F = \frac{Qq}{2\pi \epsilon_0 L d}

Explanation:

As we know that if a charge q is distributed uniformly on the line then its linear charge density is given by

\lambda = \frac{q}{L}

now the electric field due to long line charge at a distance d from it is given as

E = \frac{2k\lambda}{d}

E = \frac{q}{2\pi \epsilon_0 d}

now the force on the other charge in this electric field is given as

F = QE

F = \frac{Qq}{2\pi \epsilon_0 L d}

5 0
4 years ago
Find the final temperature of 375 grams of tea (c = 4.184 J/g°C) if its initial temperature is 95°C just before it is placed in
Minchanka [31]

Answer:

the final temperature of the tea is 7.39⁰C.

Explanation:

Given;

mass of the tea, m = 375 g

specific heat capacity of the tea, C = 4.184 JJ/g°C

initial temperature of the tea, t₁ = 95°C

the final temperature of the tea, t₂ = ?

Energy lost by the refrigerator, Q = 137,460 J

The energy lost by the refrigerator is given by the following formula;

-Q = mc(t₂ - t₁)

-137,460 =375 x 4.184(t₂ - 95°C)

-137,460 = 1569(t₂ - 95°C)

t_2-95 = \frac{-137,460}{1569} \\\\t_2-95 = -87.61\\\\t_2 = -87.61 + 95\\\\t_2 = 7.39 \ ^0 C

Therefore, the final temperature of the tea is 7.39⁰C.

4 0
3 years ago
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