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2 years ago

a force 4000 n accelerates a car of mass 800 kg from rest to 20 ms how far does the car travel while the force is acting?

Physics

1 answer:

Neko [114]2 years ago

4 0

Answer:

d = 40 m

Explanation:

Given that,

Force, F = 4000 N

Mass of a car, m = 800 kg

Initial velocity, u = 0 (at rest)

Final velocity, v = 20 m/s

We need to find the distance traveled. Force acting on a car is given by :

F = ma

Where

a is the acceleration of the car.

$a=\dfrac{F}{m}\\\\a=\dfrac{4000}{800}\\\\a=5\ m/s^2$

Let it has traveled d distance. Using third equation of motion.

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(20)^2-(0)^2}{2\times 5}\\\\d=40\ m$

So, the required distance is equal to 40 m.

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a car accelerates uniformly from rest to a speed of 65 km/h (18 m/s) in 12s. Find the distance the car travels during this time?

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The car's speed was zero at the beginning of the 12 seconds,
and 18 m/s at the end of it. Since the acceleration was 'uniform'
during that time, the car's average speed was (1/2)(0 + 18) = 9 m/s.

12 seconds at an average speed of 9 m/s ==> (12 x 9) = 108 meters .

==========================================

That's the way I like to brain it out. If you prefer to use the formula,
the first problem you run into is: You need to remember the formula !

The formula is D = 1/2 a T²

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   Acceleration = (change in speed) / (time for the change)
    = (18 m/s) / (12 sec)
    = 1.5 m/s² .

Distance = (1/2 x 1.5 m/s²) x (12 sec)²
      =     (0.75 m/s²) x (144 sec²) = 108 meters .

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3 years ago

A student practicing for a cross country meet runs 250 m in 30 s. What is her average speed?

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It's 8.3 m per second

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MamaMia's Pizza purchases its pizza delivery boxes from a printing supplier. MamaMia's delivers on-average 200 pizzas each month

Ira Lisetskai [31]

Answer:

a) 138 units

b) 17 units

c) 17 units

d) Total Cost = $353.35

Explanation:

Given:

Average pizzas delivered = 200

Charge of inventory holding = 30% of cost

Lead time = 7 days

Now,

a) Economic Order Quantity = $\sqrt\frac{2\times\textup{Annual Demand}\times\textup{Cost per Order}}{\textup{Carrying cost}}$

also,

Annual Demand = 200 × 12 = 2400

Cost per Order = Cost of Box + Processing Costs

= 30 cents + $10

= $10.30

and, Carrying Cost = $\frac{\textup{Total Inventory Cost}}{\textup{total annual demand}}$

= $\frac{\textup{Total Cost per order}\times\textup{Annual demand}\times\frac{25}{100}}{\textup{Annual demand}}$

= $\frac{\$10.30\times2400}\times\frac{25}{100}}{2400}$

= $2.575

Therefore,

Economic Order Quantity = $\sqrt\frac{2\times\textup{2400}\times\textup{10.30}}{\textup{2.575}}$

= 138.56 ≈ 138 units

b) Reorder Point

= (average daily unit sales × the lead time in days) + safety stock

= ( $\frac{200}{30}\times7$

= 46.67 ≈ 47 units

c) Number of orders per year = $\frac{\textup{Annual Demand}}{\textup{Economic order quantity}}$

= $\frac{\textup{2400}}{\textup{138}}$

= 17.39 ≈ 17 units

d) Total Annual Cost (Total Inventory Cost)

= Ordering Cost + Holding Cost

Now,

The ordering Cost = Cost per Order × Total Number of orders per year

= $10.30 × 17

= $175.1

and,

Holding Cost = Average Inventory Held × Carrying Cost per unit

Average Inventory Held = $\frac{\textup{0+138}}{\textup{2}}$ = 69

Carrying Cost per unit = $2.575

Holding Cost = 69 × $2.575 =

$177.675

Therefore,

Total Cost = Ordering Cost + carrying cost

= $175.1 + $177.675 = $353.35

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A system of ideal gas at 22°C undergoes an ischoric process with an internal energy decrease of 4.30 × 10 3 4.30×103 J to a fina

Komok [63]

Answer:

The approximate change in entropy is -14.72 J/K.

Explanation:

Given that,

Temperature = 22°C

Internal energy $U=4.30\times10^{3}\ J$

Final temperature = 16°C

We need to calculate the approximate change in entropy

Using formula of the entropy

$\Delta S=\dfrac{\Delta U}{T}$

Where, $\Delta U$ = internal energy

T = average temperature

Put the value in to the formula

$\Delta S=\dfrac{-4.30\times10^{3}}{\dfrac{22+273+16+273}{2}}$

$\Delta S=-14.72\ J/K$

Hence, The approximate change in entropy is -14.72 J/K.

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3 years ago

a force 4000 n accelerates a car of mass 800 kg from rest to 20 ms how far does the car travel while the force is acting?​

a force 4000 n accelerates a car of mass 800 kg from rest to 20 ms how far does the car travel while the force is acting?