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3 years ago

Which set of shapes contains members that are always similar to one another?

1 answer:

Colt1911 [192]3 years ago

3 0

<span>The correct answer is option C. i.e.Equilateral triangles. All equilateral triangle have equal sides. Due to this these set of shapes will always be similar. Similarity in the sense that only the size of the triangles can change to large or smaller but the shape will be be similar always.</span>

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If a proton and an electron are released when they are 4.00 x 10^-10 m apart, find the initial acceleration of each of them.

SVEN [57.7K]

Answer:

For proton

$a=8.8\times 10^{17}\ m/s^2$

For electron

$a=1.5\times 10^{21}\ m/s^2$

Explanation:

We know that

The mass of electron

$m=9.1\times 10^{-31}\ kg$

The mass of proton

$m=1.67\times 10^{-27}\ kg$

Charge on electron and proton

q₁=q₂=q

$q=1.6\times 10^{-19}\ C$

Electrostatics force

$F=K\dfrac{q_1q_2}{r^2}$

Now by putting the values

$F=9\times 10^9\times \dfrac{1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(4\times 10^{-10})^2}$

$F=1.44\times 10^{-9}\ N$

For proton

F = m a

a =F/m

$a=\dfrac{1.4\times 10^{-9}}{1.67\times 10^{-27}}\ m/s^2$

$a=8.8\times 10^{17}\ m/s^2$

For electron

$a=\dfrac{1.4\times 10^{-9}}{9.1\times 10^{-31}}\ m/s^2$

$a=1.5\times 10^{21}\ m/s^2$

5 0

2 years ago

Read 2 more answers

What force is necessary to accelerate 32kg object by 2 m/s

Oxana [17]

Answer:

i8nnndnsjdjcicicixidocin where dosoldmxbeus sosojsnnd

Explanation:

nzjwiemddsisjxnxn,jisiskxnxjxjxiisnz MN nsjs. snxjsisisis s. xjsisisk

6 0

2 years ago

A pendulum consisting of a 0.5 kg mass tied to a 0.3 m string is set into oscillation at the same moment that a stone is dropped

lara31 [8.8K]

Answer:

2.72 cycles

Explanation:

First of all, let's find the time that the stone takes to reaches the ground. The stone moves by uniform accelerated motion with constant acceleration g=9.8 m/s^2, and it covers a distance of S=44.1 m, so the time taken is

$S=\frac{1}{2}at^2\\t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(44.1m)}{9.8 m/s^2}}=3 s$

The period of the pendulum instead is given by:

$T=2 \pi \sqrt{\frac{L}{g}}=2 \pi \sqrt{\frac{0.3 m}{9.8 m/s^2}}=1.10 s$

Therefore, the number of oscillations that the pendulum goes through before the stone hits the ground is given by the time the stone takes to hit the ground divided by the period of the pendulum:

$N=\frac{t}{T}=\frac{3 s}{1.10 s}=2.72$