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hammer [34]
1 year ago
9

Calculate the force needed to give a car of mass 800 kg an acceleration of 2.0 ms−2. plss quick

Physics
1 answer:
erik [133]1 year ago
4 0

The force needed to give a car of mass 800 kg an acceleration of 2.0 ms-² is 1600N.

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

The force needed to push an object can be calculated by multiplying the mass of the object by its acceleration as follows:

Force = mass × acceleration

According to this question, a car of mass 800 kg has an acceleration of 2.0 ms−². The force is calculated as follows:

Force = 800kg × 2m/s²

Force = 1600N

Therefore, the force needed to give a car of mass 800 kg an acceleration of 2.0 ms-² is 1600N.

Learn more about force at: brainly.com/question/13191643

#SPJ1

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Answer:

it is 2.2 m

Explanation:

because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2

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The fragment of an asteroid or any interplanetary material is known as a
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Some element can be either solid or liquid. At the melting point, the liquid has 8 × 10-22 J more enthalpy per atom than the sol
OlgaM077 [116]

Answer:

481.76 J/mol

133.33 K

Explanation:

N_A = Avogadro's number = 6.022\times 10^{23}

Change in enthalpy is given by

\Delta H=8\times 10^{-22}\times 6.022\times 10^{23}\\\Rightarrow \Delta H=481.76\ J/mol

Entropy is given by

\Delta S=6\times 10^{-24}\times 6.022\times 10^{23}\\\Rightarrow \Delta S=3.6132\ J/mol K

Latent heat of fusion is given by

L_f=\Delta H\\\Rightarrow L_f=481.76\ J/mol

The latent heat of fusion is 481.76 J/mol

Melting point is given by

T_m=\dfrac{L_f}{\Delta S}\\\Rightarrow T_m=\dfrac{8\times 10^{-22}\times 6.022\times 10^{23}}{6\times 10^{-24}\times 6.022\times 10^{23}}\\\Rightarrow T_f=133.33\ K

Melting occurs at 133.33 K

3 0
3 years ago
The waste products of a nuclear fission powerplant can best be described as
marishachu [46]
The waste products of a nuclear fission power plant can best be described as radioactive waste. 
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At what speed does a typical electron pass by any given point in the wire?
attashe74 [19]
A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.

a) How many electrons pass through the light bulb each second?

b) What is the current density in the wire? (answer in A/m^2)

<span>c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)

</span>a) 5.0 A = 5.0 C/s 
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s

b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²

c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³

(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed

(q/m³)(A)(v) = i

v = i.[(q/m³)A]ˉ¹

<span>v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s</span>
7 0
3 years ago
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