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lyudmila [28]
2 years ago
7

If a car change the velocity from 36M/S to 28M/S with an acceleration of -2.0 M/S then how much time does it take for the vehicl

e to slow down? please help I NEDD it fasstttt
Physics
1 answer:
leonid [27]2 years ago
6 0

The time it will take if a car change the velocity from 36m/s to 28m/s with an acceleration of -2.0m/s² is 4s.

<h3>How to calculate deceleration?</h3>

Deceleration is the amount by which a speed or velocity decreases (and so a scalar quantity or a vector quantity).

The deceleration of a body can be calculated using the following formula:

-a = (v - u)/t

Where;

  • -a = deceleration
  • v = final velocity
  • u = initial velocity
  • t = time in seconds

-2 = (28 - 36)/t

-2t = -8

t = 4s

Therefore, the time it will take if a car change the velocity from 36m/s to 28m/s with an acceleration of -2.0m/s² is 4s.

Learn more about deceleration at: brainly.com/question/4403243

#SPJ1

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Answer:

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V = (14 ft)(15 ft)(8 ft)(30.48 cm/1 ft)³ = 0.0593 cm³

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What is the relationship between the internal energy of a substance and its state of matter? A) As a gas loses internal energy i
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4 0
3 years ago
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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

7 0
3 years ago
When water freezes, its volume increases by 9.05% (that is, equation). What force per unit area is water capable of exerting on
PIT_PIT [208]

Answer:

1.991 × 10^(8) N/m²

Explanation:

We are told that its volume increases by 9.05%.

Thus; (ΔV/V_o) = 9.05% = 0.0905

To find the force per unit area which is also pressure, we will use bulk modulus formula;

B = Δp(V_o/ΔV)

Making Δp the subject gives;

Δp = B(ΔV/V_o)

Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²

Thus;

Δp = 2.2 × 10^(9)[0.0905]

Δp = 1.991 × 10^(8) N/m²

3 0
2 years ago
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