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zavuch27 [327]
3 years ago
11

A dog running in an open field has components of velocity vx = 3.1 m/s and vy = -1.3 m/s at time t1 = 11.3 s . For the time inte

rval from t1 = 11.3 s to t2 = 22.5 s , the average acceleration of the dog has magnitude 0.52 m/s2 and direction 27.5 ∘ measured from the +x−axis toward the +y−axis. At time t2 = 23.6s , what are the x-component of the dog's velocity?

Physics
1 answer:
ANTONII [103]3 years ago
3 0

Answer:

x -component of dog's velocity = 8.265m/s

y- component of dog's velocity = 1.389m/s

Explanation:

The detailed and step by step calculation is as shown in the attached file.

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Two glass rods are rubbed together. One of the glass rods are brought near (but
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The correct answer is - do nothing.

Explanation:

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Only two different objects are able to transfer electrons from one material to the other material if rubbed together, to produce the conditions for static electricity to be observed. Therefore there would be no charge on the glass rod so there would be no charge to balls to repel or attract.

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Simplify 45÷3+2×8-12+42​
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Answer:

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4 0
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A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The sprin
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Answer:

The velocity of block = 0.188 \frac{m}{s}

Explanation:

Mass m = 5.6 kg

k = 1040 \frac{N}{m}

\mu = 0.26

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x_{2} = 0.02 m

From work energy theorem

K_{1} + U_{1} + W_{other} = K_{2} + U_{2}  --------- (1)

Kinetic energy

K = \frac{1}{2} k x^{2}  ------- (1)

Potential energy

U = \frac{1}{2} k x^{2} ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

R_{N} = mg

R_{N}  = 5.6 × 9.81 = 54.9 N

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Now the work done by the friction

W_{f} = - 21.9 × 0.015

W_{f} = - 0.329 J

Now kinetic energy

At point 1

K_{1} = \frac{1}{2} m v^{2} _{1}

K_{1} = 0

U_{1} = \frac{1}{2} k x^{2}

U_{1} = \frac{1}{2}  (1040) 0.035^{2}

U_{1} = 0.637 J

At point 2

K_{2} = \frac{1}{2} (5.6) v^{2} _{2}

K_{2} = 2.8 v_{2} ^{2}

Potential energy

U_{2} = \frac{1}{2}  k x_2^{2}

U_{2} = \frac{1}{2}  (1040) 0.02^{2}

U_{2} = 0.208 J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8 v_{2} ^{2} + 0.208

2.8 v_{2} ^{2} = 0.1

v_{2} = 0.188 \frac{m}{s}

This is the velocity of block.

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