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3 years ago

1. A car accelerates uniformly from 12 m/s to 39 m/s in 12 seconds. What is the car's average acceleration

Physics

1 answer:

Alinara [238K]3 years ago

5 0

Answer:

(1) 2.25m/s^2

(2) 45.6m

Explanation:

(1) A car accelerates uniformly from 12m/s to 39m/s in 12 seconds

Therefore the average acceleration can be calculated as follows

a = 39-12/12

a = 27/12

a= 2.25m/s^2

(2) A butterfly is flying at 4m/s , it accelerates uniformly at 1.2 m/s for 6 seconds

u= 4

a= 1.2

t= 6

Therefore the distance can be calculated as follows

S= ut + 1/2at^2

= 4×6 + 1/2 × 1.2 × 6^2

= 24 + 1/2 × 1.2 × 36

= 24 + 1/2 × 43.2

= 24 + 21.6

S = 45.6m

Hence the butterfly travels at 45.6m

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Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

The temperature of water in a beaker is 45°C. What does this measurement represent?

julia-pushkina [17]

Your answer would be C, with a quick lookup of this question online lol

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3 years ago

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The weather map shows some conditions in the atmosphere at noon on a

earnstyle [38]

The geographical region where I would expect the warmest weather is: C. in the southeast.

<h3>What is a weather map?</h3>

A weather map can be defined as a type of chart that is typically used to provide information about the average atmospheric condition of a particular geographical region over a specific period of time.

Based on the weather map shown in the image attached below, we can infer and logically deduce that the geographical region where the warmest weather is expected is in the southeast due to its very high atmospheric pressure.

Read more on weather here: brainly.com/question/24730207

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2 years ago

What is the current flowing through a 10 resistor in a parallel circuit with a 120 V potential difference across it?

Elena L [17]

Answer:

let current flow be x

Explanation:

than we know potential difference = energy supply / charge then 120 = 10 /x then x=12 therefore pd = 12 v

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3 years ago

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In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera

Talja [164]

Answer:

η = 40 %

Explanation:

Given that

Qa ,Heat addition= 1000 J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

$\eta=\dfrac{W(net)}{Q(heat\ addition)}$

Now by putting the values in the above equation ,then we get

$\eta=\dfrac{400}{1000}$

η = 0.4

The efficiency in percentage is given as

η = 0.4 x 100 %

η = 40 %

Therefore the answer will be 40%.

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4 years ago