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4 years ago

Which three characteristics of an object are represented by a motion map

Physics

1 answer:

nadya68 [22]4 years ago

8 0

There are three things that can be represented on a motion map

These three things are:

1)Motion

2)Acceleration

3)Velocity

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NEEDD SOME HELP ASAP HELP MEHHH<br> 50 pOINTS

Vadim26 [7]

An electron shell can hold 2(n^2) electrons (technically) where n is the shell number, i.e. shell 1 can hold 2, shell 2 can hold 8, 3 holds 18 and so on.
The atomic number of Nitrogen is 7, i.e. it has 7 electrons (to match its 7 protons, assuming it isn't an ion).

With the atomic number, you simply start from shell 1 and work out. So we put 2 electrons in shell 1, leaving us with 5 left. Shell 2 can hold 6 so we can fit all 5 in.

In other words, you should have 2 electron shells on the atom, shell 1 with 2 e- and shell 2 with 5 e-.

3 0

3 years ago

Read 2 more answers

PLEASE HELP!!!!! which statements correctly conpare the masses of protons,neutrons,and electrons

Anastasy [175]

Protons and neutrons have similar mass
Electrons are smaller then a proton or a neutron

8 0

3 years ago

In 1995 a research group led by Eric Cornell and Carl Wiemann at the University of Colorado successfully cooled Rubidium atoms t

saveliy_v [14]

Answer:

0.00493 m/s

Explanation:

T = Temperature of the isotope = 85 nK

R = Gas constant = 8.341 J/mol K

M = Molar mass of isotope = 86.91 g/mol

Root Mean Square speed is given by

$v_r=\sqrt{\dfrac{3RT}{M}}\\\Rightarrow v_r=\sqrt{\dfrac{3\times 8.314\times 85\times 10^{-9}}{86.91\times 10^{-3}}}\\\Rightarrow v_r=0.00493\ m/s$

The Root Mean Square speed is 0.00493 m/s

6 0

4 years ago

An accelerating voltage of 2.47 x 10^3 V is applied to an electron gun, producing a beam of electrons originally traveling horiz

Dmitry [639]

Answer:

6.3445×10⁻¹⁶ m

Explanation:

E = Accelerating voltage = 2.47×10³ V

m = Mass of electron

Distance electron travels = 33.5 cm = 0.335 cm

$E=\frac{mv^2}{2}\\\Rightarrow v=\sqrt{\frac{2E}{m}}\\\Rightarrow v=\sqrt{\frac{2\times 2470\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=29455356.08671\ m/s$

Deflection by Earth's Gravity

$\Delta =\frac {gt^2}{2}$

Now, Time = Distance/Velocity

$\Delta =\frac {g\frac{s^2}{v^2}}{2}\\\Rightarrow \Delta =\frac{9.81\frac{0.335^2}{29455356.08671^2}}{2}\\\Rightarrow \Delta =6.3445\times 10^{-16}\ m$

∴ Magnitude of the deflection on the screen caused by the Earth's gravitational field is 6.3445×10⁻¹⁶ m

3 0

3 years ago

A 5 kg block is resting on a ramp inclined at 35 degrees above the horizontal. What is the magnitude of the normal force acting

Mamont248 [21]

I'm pretty sure the answer is b 28n hope helps :)

4 0

3 years ago