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2 years ago

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What is displacement???

2 answers:

AnnZ [28]2 years ago

8 0

Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

I hope it'll help you...

il63 [147K]2 years ago

4 0

Answer:

What is displacement???

<em>In geometry and mechanics, a displacement is a vector whose length is the shortest distance from the initial to the final position of a point P undergoing motion.[1] It quantifies both the distance and direction of the net or total motion along a straight line from the initial position to the final position of the point trajectory. A displacement may be identified with the translation that maps the initial position to the final position.</em>

Explanation:

<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> will</em><em> help</em><em> you</em>

<em> $kai6418$ </em>

<em>#</em><em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em>o</em><em>n</em><em>l</em><em>e</em><em>r</em><em>a</em><em>n</em><em>i</em><em>n</em><em>g</em>

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Which of the following is the best reason that scientific models are used? (1 point)

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A 2,000 kg car is moving at 15m/s when it collides with a 1200kg car sitting still. Briefly compare the impulse imparted on the

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What rock type would most likely be found in the location marked by the black rectangle in the diagram below?

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How fast does a 2 MeV fission neutron travel through a reactor core?

Artemon [7]

Answer:

The answer is $1.956 \times 10^7\ m/s$

Explanation:

The amount of energy is not enough to apply the relativistic formula of energy $E = mc^2$ , so the definition of energy in this case is

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From the last equation,

$v= \sqrt{2E/m}$

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and the mass of the neutron is

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3 years ago

A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately

Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car $V_{c}$ = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

$V_{f}$ = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

$V_{f}$ = $V_{i}$ + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁ = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁ will be

$d_{c}$ = $V_{c}$ × t₁

we substitute

$d_{c}$ = 22.352 × 7

$d_{c}$ = 156.46 m

now distance between polive car and speeding car

Δd = $d_{c}$ - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( $V_{f}$ - $V_{c}$ )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = $V_{f}$ × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b) how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0

3 years ago