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3 years ago

9

Explain why a moving object cannot come to a stop instantly in zero seconds

2 answers:

sineoko [7]3 years ago

7 0

Well because it has momentum and it needs to be slowed down.

Harman [31]3 years ago

5 0

<span>The scientific term is <span>Inertia.
</span>
</span><span>"An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force."
</span>
There isn't an unbalanced force that can stop something instantly.
The object must be decelerated, due to the momentum a moving object is carrying with it.

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Write the adverbs use in sentences no.1-5 ty

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Answer:

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3 years ago

PLEASE ANSWER THIS ASAP!!!

Burka [1]

Answer:

I guess B

Explanation:

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2 years ago

A car is moving initially at 30 m/s comes gradually at stop at 900 m . What is acceleration of the car ?

ziro4ka [17]

The acceleration of the car is 0.5 meters per seconds square.

Given the following data:

Initial velocity, U = 30 m/s

Distance, S = 900 meters

Note: The final velocity (V) of the car would be zero (0) m/s when it comes to a stop.

To find the acceleration of the car, we would use the third equation of motion;

Mathematically, the third equation of motion is given by the formula;

$V^2 = U^2 - 2aS\\\\0 = 30^2 - 2a(900)\\\\0 = 900 - 1800a\\\\1800a = 900\\\\a = \frac{900}{1800}$

<em>Acceleration, a </em><em>=</em><em> 0.5 </em> $m/s^2$ <em></em>

Therefore, the acceleration of the car is 0.5 meters per seconds square.

Read more: brainly.com/question/8898885

4 0

2 years ago

Which describes how chemical changes are different from physical changes?

lions [1.4K]

Answer:

B. Chemical changes involve the formation of a new substance.

Explanation:

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4 years ago

2.65 In a standard tensile test a steel rod of 22-mm diameter is subjected to a tension force of 75 kN. Knowing that n 5 0.30 an

Schach [20]

To solve this problem it is necessary to apply the concepts related to uniaxial deflection for which the training variable is applied, determined as

$\delta = \frac{Pl}{AE}$

Where,

P = Tensile Force

L= Length

A = Cross sectional Area

E = Young's modulus

PART A) The elongation of the bar in a length of 200 mm caliber, could be determined through the previous equation, then

$\delta = \frac{Pl}{AE}$

$\delta = \frac{(75*10^3)(200)}{(\pi/4*22^2)(200*10^3)}$

$\delta = 0.1973mm$

Therefore the elongaton of the rod in a 200mm gage length is $0.1973mm$

PART B) To know the change in the diameter, we apply the similar ratio of the change in length for which,

$\upsilon = \frac{l_{a}}{l_{s}}$

Where,

$\upsilon =$ Poission's ratio

$l_{a}$ = Lateral strain

$l_{s}$ = Linear strain

$\upsilon = \frac{\delta/d}{\delta/l}$

$0.3 = \frac{\delta/22}{0.1973/200}$

$0.3(\frac{0.1973}{200}) = \frac{\delta}{22}$

$\delta = 6.5109*10^{-3}mm$

Therefore the change in diameter of the rod is $6.5109*10^{-3}mm$

5 0

3 years ago