1.5 x 10^3 standard notation

A Navy vessel is traveling due north during wartime. A torpedo has been launched by an enemy directly toward the stern (rear) of

Anika [276]

Answer:

The correct option is;

B) No, the Navy vessel is slower

Explanation:

The speed of some torpedoes can be as high as 370 km/h. The average speed of a fast Navy vessel is approximately 110 km/h

Therefore, the torpedoes travel approximately 3 times as fast as the (slower) Navy vessel, such that the torpedo covers three times the distance of the Navy vessel in the same time and therefore, if the Navy vessel and the torpedo continue in a straight line (in the same direction) due north the vessel can not outrun the torpedo

Therefore, no the Navy vessel travels slower than a torpedo.

8 0

2 years ago

Two obiect accumulated a charge of

tamaranim1 [39]

Answer:

A. 181.24 N

Explanation:

The magnitude of hte electrostatic force between two charged objects is given by the equation

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

In this problem, we have:

$q_1=4.5\mu C=4.5\cdot 10^{-6}C$ is the magnitude of the 1st charge

$q_2=2.8\mu C=2.8\cdot 10^{-6}C$ is the magnitude of the 2nd charge

r = 2.5 cm = 0.025 m is the separation between the charges

Therefore, the magnitude of the electric force is:

$F=\frac{(9\cdot 10^9)(4.5\cdot 10^{-6})(2.8\cdot 10^{-6})}{(0.025)^2}=181.44 N$

So, the closest answer is

A) 181.24 N

3 0

3 years ago

What force is acting on the rainwater in the model?

Vikki [24]

The force is gravitational because when something is falling is call gravitational

8 0

3 years ago

Read 2 more answers

An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec

lana66690 [7]

Answer:

91.87 m/s

Explanation:

Given:

x = initial distance of the electron from the proton = 6 cm = 0.06 m

y = initial distance of the electron from the proton = 3 cm = 0.03 m
u = initial velocity of the electron = 0 m/s

Assume:

m = mass of an electron = $9.1\times 10^{-31}\ kg$
v = final velocity of the electron
e = magnitude of charge on an electron = $1.6\times 10^{-19}\ C$

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

$\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\$

$\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\$

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0

3 years ago

2 ways weak nuclear forces and gravity are alike PLEASE HELP ):

shtirl [24]

They are attractive
They don’t depend on charge

7 0

2 years ago