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Hoochie [10]
3 years ago
15

A 1500-kilogram car is at rest on a level track. The driver gives the engine gas and the car begins to accelerate forward. The c

ar moves from 0 meters per second to 27 meters per second in 8 seconds. What is the acceleration of the car? What is the force of the car moving forward?
Physics
1 answer:
Nutka1998 [239]3 years ago
8 0

1. The acceleration of the car is 3.375 m/s².

2. The force moving the car is 5062.5 N

From the question given above, the following data were obtained:

Mass of car = 1500 Kg

Initial velocity (u) = 0 m/s

Final velocity (v) = 27 m/s

Time (t) = 8 s

<h3>Acceleration (a) =? </h3><h3>Force (F) =? </h3>

<h3>1. Determination of the acceleration </h3>

Initial velocity (u) = 0 m/s

Final velocity (v) = 27 m/s

Time (t) = 8 s

<h3>Acceleration (a) =?</h3>

a = \frac{v - u}{t} \\\\a = \frac{27 - 0}{8} \\\\a = \frac{27}{8}\\

<h3>a = 3.375 m/s²</h3>

Therefore, the acceleration of the car is 3.375 m/s²

<h3>2. Determination of the force moving the car.</h3>

Mass of car = 1500 Kg

Acceleration (a) = 3.375 m/s²

<h3>Force (F) =?</h3>

<h3>F = ma</h3>

F = 1500 × 3.375

<h3>F = 5062.5 N</h3>

Therefore, the force moving the car is 5062.5 N

Learn more: brainly.com/question/13001406

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jok3333 [9.3K]

Answer:

a)3000ohm

b)4.44mA

Explanation:

a) we were given a Nine tree lights connected inparallel across 120-V potential difference, since the resistor are in parallel we use the expresion below

1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉

But according to ohm'law which can be expressed below

V=IR

R=V/I

R(total)= 120/0.36

= 333.33ohm

1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉

R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉

1/R(total)=9/R

1/333.33= 9/R

R= 3000ohm

Therefore, the resistance is 3000ohm

b)the bulbs were connected in series here, then for series connection we use below expression

R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉

R(total)=9R

= 9*3000

=27000ohm

I=VR

I=V/R

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7 0
4 years ago
A particle passes through the point P=(−3,1,0)P=(−3,1,0) at time t=3t=3, moving with constant velocity v⃗ =⟨5,3,−2⟩v→=⟨5,3,−2⟩.
eimsori [14]

Answer:

The parametric equation for the position of the particle is (-18+5t,-8+3t,6-2t).

Explanation:

Given that,

The point is

P=(-3,1,0)

Time t = 3

Velocity v=(5,3,-2)

We need to calculate the parametric equation for the position of the particle

Using parametric equation for position

r(t)=r_{0}+v(t)t....(I)

at t = 3,

P=r(t)

Put the value into the formula

(-3,1,0)=r_{0}+(5,3,-2)\times3

(-3,1,0)=r_{0}+(15,9,-6)

r_{0}=(-18,-8,6)

Put the value of r₀ in equation (I)

r(t)=(-18,-8,6)+(5,3,-2)t

r(t)=(-18+5t,-8+3t,6-2t)

Hence, The parametric equation for the position of the particle is (-18+5t,-8+3t,6-2t).

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3 years ago
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Answer:

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b) F' = 0 N

c)F''=28195.5N

Explanation:

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F' = 0 N

For Third Drawing

F'' =\frac{ k q^2}{d^2} * \sqrt (2)

F'' = 9*10^9* (6.4*10^-6)^2 * \frac{\sqrt(2)}{(4.3*10^-3)^2}

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