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1 year ago

A 600 kg car is at rest, and then accelerates to 5 m/s.

Physics

1 answer:

Crank1 year ago

3 0

Answer:

7500J

Explanation:

Given parameters:

Mass of car = 600kg

Velocity = 5m/s

Unknown:

Original kinetic energy = ?

Final kinetic energy = ?

Work used = ?

Solution:

The kinetic energy of a body is the energy due to the motion of a body.

It can be solve mathematically using expression below;

K.E = $\frac{1}{2}$ m v²

where m is mass

v is velocity

original kinetic energy;

The car started at rest and v = 0, therefore K.E = 0

Final kinetic energy;

K.E = $\frac{1}{2}$ x 600 x 5² = 7500J

Work done:

Work done = Final K.E - Initial K.E = 7500 - 0 = 7500J

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Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68

den301095 [7]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, $y_0=1500\ m$

Initial horizontal velocity is, $u_x=688\ m/s$

Initial vertical velocity is, $u_y=0(\textrm{Horizontal velocity only initially)}$

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is $S=y-y_0=0-1500=-1500\ m$

Acceleration in the vertical direction is due to gravity, $g=-9.8\ m/s^2$

Therefore, the displacement of the bomb is given as:

$S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s$

So, the bomb will remain in air for 17.5 s before hitting the ground.

6 0

2 years ago

A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later

UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity. If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

(64.2 m/s) - [ (9.8 m/s² ) x (1.5 sec) ]

= (64.2 m/s) - [ 14.7 m/s ]

= 49.5 m/s . (upward)

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2 years ago

The food calorie, equal to 4186J , is a measure of how much energy is released when food is metabolized by the body. A certain b

vovangra [49]

<h2>The hiker will go up to 850 m on the hill</h2>

Explanation:

The total energy gained by the hiker = 140 x 4186 J

This energy is consumed in the potential energy acquired , while climbing up the hill.

The potential energy P.E = mass of hiker x acceleration due to gravity x height

Thus

140 x 4186 = 69 x 10 x h

or h = $\frac{4186x140}{69x10}$ = 850 m

If the 20% of the total energy is used

the height h₀ = $\frac{0.2x4186x140}{69x10}$ = 170 m

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2 years ago

You've just solved a problem and the answer is the mass of an electron, me=9.11×10−31kilograms. How would you enter this number

mamaluj [8]

The mass of an electron, me, 9.11 × 10⁻³¹ kilograms shows there are 3 important numbers in them, namely numbers 9, 1, and 1

<h3><em>Further explanation</em></h3>

Significant numbers are obtained from measurement results of exact numbers and the last number estimated

This is called significant numbers

the scientific notation can be described as:

a, ... x 10ⁿ

a, ... called a significant number

10ⁿ is called a big order

Rules for significant numbers in general:

1. All non-zero numbers are significant numbers

2. a zero which is located between two non-zero numbers including a significant number

3. all zeros are located in the final row written behind the decimal point of the include significant number

4. zero decimal point is the not significant number

From the numbers known in the question amounting to 9.11 × 10⁻³¹ kilograms shows there are 3 important numbers in them,(symbol a, before a big order ,called a significant number) namely numbers 9, 1, and 1

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Keywords : significant number, the scientific notation, decimal point, a big order