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Vlad1618 [11]
3 years ago
8

A 5 kg object slows from +6 m/s to +2 m/s. What is the impulse?

Physics
1 answer:
Liula [17]3 years ago
5 0

Impulse on the object: -20 kg m/s

Explanation:

The impulse exerted on an object is a vector quantity equal to the product between the force exerted on the object and the time interval during which the force is applied.

The impulse exerted on an object is also equal to its change in momentum. Mathematically,

I=\Delta p = m(v-u)

where:

m is the mass of the object

u is the initial velocity

v is the final velocity

For the object in this problem, we have

m = 5 kg

u = +6 m/s

v = +2 m/s

Therefore, its impulse is:

I=(5)(+2-(+6))=-20 kg m/s

The negative sign means the direction of the impulse is opposite to the direction of motion of the object.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

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Answer:3W

If it takes an amount of work W to move two q point charges from infinity to a distance d apart from each other, then how much work should it take to move three q point charges from infinity to a distance d apart from each other?

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B) 3W

C) 4W

D) 6W

Explanation: calculating work done,W, in moving two positive q point charges from infinity to a valued distance d from each other  is

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work done in moving 3 equal positive charges from infinity to a finite distance is given by

W₂=W₄=W₆=k(+q)(+q)/ d

Total work done, W' =k(+q)(+q)/ d + k(+q)(+q)/ d + k(+q)(+q)/ d

= W + W + W = 3W

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A particle (m = 4.3 × 10^-28 kg) starting from rest, experiences an acceleration of 2.4 × 10^7 m/s^2 for 5.0 s. What is its de B
Novay_Z [31]

Answer:

Wavelength, \lambda=1.28\times 10^{-14}\ m

Explanation:

Given that,

Mass of the particle, m=4.3\times 10^{-28}\ kg

Acceleration of the particle, a=2.4\times 10^7\ m/s^2

Time, t = 5 s

It starts from rest, u = 0

The De Broglie wavelength is given by :

\lambda=\dfrac{h}{mv}

v = a × t

\lambda=\dfrac{h}{mat}

\lambda=\dfrac{6.67\times 10^{-34}}{4.3\times 10^{-28}\times 2.4\times 10^7\times 5}

\lambda=1.28\times 10^{-14}\ m

Hence, this is the required solution.

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3 years ago
Charge of uniform linear density (6.7 nC/m) is distributed along the entire x axis. Determine the magnitude of the electric fiel
ad-work [718]

Thw question is not complete. The complete question is;

Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC

Answer:

Option C: E = 75 N/C

Explanation:

We are given;

Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m

Distance on the y-axis; d = 1.6 m

Now, the formula for electric field with uniform linear density is given as;

E = λ/(2•π•r•ε_o)

Where;

E is electric field

λ is uniform linear density = 6.7 × 10^(-9) C/m

r is distance = 1.6m

ε_o is a constant = 8.85 × 10^(-12) C²/N.m²

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E = 75.31 N/C ≈ 75 N/C

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3 years ago
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