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3 years ago

Convert 0.000238 m to cm

Physics

1 answer:

enot [183]3 years ago

3 0

Answer:

0.0238 centimetre

Explanation:

multiply the length value by 100

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3 years ago

A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. a) What is the resultant velocity of the motorb

Nostrana [21]

Explanation:

It is given that,

Velocity in East, $v_1=4\ m/s$

Velocity in North, $v_2=3\ m/s$

(a) The resultant velocity is given by :

$v=\sqrt{4^2+3^2}=5\ m/s$

(b) The width of the river is, d = 80 m

Let t is the time taken by the boat to travel shore to shore. So,

$t=\dfrac{d}{v}$

$t=\dfrac{80\ m}{5\ m/s}$

t = 16 seconds

(c) Let x is the distance covered by the boat to reach the opposite shore. So,

$x=v_2\times t$

$x=3\ m/s\times 16\ s$

x = 48 meters

Hence, this is the required solution.

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3 years ago

Give two examples of chemical energy being transformed into electrical energy

julsineya [31]

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3 years ago

The submarine emits a pulse of sound to detect other objects in the sea. The sped of sound in sea water is 1500m/s. An echo is r

r-ruslan [8.4K]

Answer:

d = 375 m

Explanation:

The speed of sound is constant in any medium, therefore we can use the uniform motion relationships

v = x / t

x = v t

In this case it indicates that the time since the sound is emitted and received is t = 0.50 s, in this time the sound traveled a round trip distance

x = 2d

2d = v t

d = v t/2

let's calculate

d = 1500 0.5 / 2

d = 375 m

3 0

3 years ago

You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown

Kipish [7]

Answer:

The mass is $m_2 =21.75*10^{-27} \ kg$

The velocity is $v_2 = 3.0*10^{6} m/s$

Explanation:

From the question we are told that

The speed of the protons is $u_1 = 2.10*10^{7} m/s$

The mass of the protons is $m$

The speed of the rebounding protons are $v_1 = -1.80 * 10^{7} \ m/s$

The negative sign shows that it is moving in the opposite direction

Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as

$m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1$

Where $m_1$ is the mass of a single proton

So substituting values

$m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1$

$m_2 =13 m_1$

The mass of on proton is $m_1 = 1.673 * 10^{-27} \ kg$

So $m_2 =13 ( 1.673 * 10^{-27} )$

$m_2 =21.75*10^{-27} \ kg$

Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as

$m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2$