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3 years ago

15 POINTS PLEASE HELP

Physics

1 answer:

SVETLANKA909090 [29]3 years ago

7 0

Answer:

Electron

Explanation:

See attached figure (it was missing in the question).

An atom consists of three types of particle:

- Proton: it is positively charged (charge $+e = +1.6\cdot 10^{-19} C$ ) and it is located in the nucleus. Its mass is approximately $1.67\cdot 10^{-27} kg$

- Neutron: it has no electric charge, and it is also located in the nucleus. Its mass is just slighly larger than that of the proton (but it is also approximated to $1.67\cdot 10^{-27} kg$ )

- Electron: it is negatively charged (charge $-e = -1.6\cdot 10^{-19} C$ ) and it orbits around the nucleus. Its mass is much smaller than that of the proton ( $m_e = 9.11 \cdot 10^{-31} kg$ )

From the figure, we see that the only particle orbiting around the nucleus is particle A: therefore, A must be an electron.

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The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by

Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, $\mu = 0.2$

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, $f_{y} = 40sin \theta$

$\sum f(y) = 0$

$N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N$

$\sum f(x) = 0$

$40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}$

$v^{2} = u^{2} + 2as\\u = 0 m/s\\v^{2} = 2 * 0.731 * 5\\v^{2} = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s$

Power, $P = Fvcos \theta$

$P = 40 *2.704 cos60\\P = 54.074 W$

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4 years ago

A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling

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m1=1500kg
m_2=3000kg
v_1=5m/s
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$\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3$

$\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3$

$\\ \sf\Rrightarrow 7500-21000=4500v_3$

$\\ \sf\Rrightarrow -13500=4500v_3$

$\\ \sf\Rrightarrow v_3=-3m/s$

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3 years ago

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Answer:

C. Blood

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An accepted value for the acceleration due to gravity is 9.801 m/s2. In an experiment with pendulums, you calculate that the val

Fed [463]

g Generally the accepted value of acceleration due to gravity is 9.801 $m/s^2$

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hence the error is - $\frac{0.381}{9.801} *100$

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4 years ago