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3 years ago

Which of the following are advantages of

Physics

1 answer:

jolli1 [7]3 years ago

4 0

Answer:

A. They can provide an accurate picture of someone’s personality.

C. They can be used to target employee training.

E. They may lead to lower turnover rates.

Explanation:

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2. What is the speed of a car that travels 73.4 kilometers (km) in 5 hours?

maksim [4K]

Speed =dist./time
=73.4/5
=14.68 km/hr

7 0

3 years ago

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You are growing a little taller each year. What units could you use to measure how fast

vlada-n [284]

Answer:

inches and feet (or even centameters)

Explanation:

these are all common units used to measure height

6 0

2 years ago

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A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.30 m/s2. At 20.0 s after bla

lesya [120]

Answer:

Explanation:

v = u +at

u = 0

a = 2.3 m /s²

t = 20 s

v = 2.3 x 20

= 46 m /s

Distance covered under acceleration of 2.3 m/s²

s = ut + 1/2 at²

= 0 + .5 x 2.3 x 20²

= 460 m

After that it moves under free fall ie g acts on it downwards .

v² = u² - 2gh , h is height moved by it under free fall

0 = 46² - 2 x 9.8 h

h = 107.96 m

Total height attained

= 460 + 107.96

= 567.96 m

b ) At its highest point ,it stops so its velocity = 0

c ) rocket's acceleration at its highest point = g = 9.8 downwards .

At highest point , it is undergoing free fall so its acceleration = g

6 0

3 years ago

Which chemical equation is balanced?

nika2105 [10]

Answer:

Explanation:

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6 0

2 years ago

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In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53×10−10m

Ksenya-84 [330]

Answer

given,

radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m

mass of electron, M = 9.11 x 10⁻³¹ Kg

charge of electron, q₁ = 1.6 x 10⁻¹⁹ C

q₂ = 1.6 x 10⁻¹⁹ C

we know, force between two charges

$F = \dfrac{kq_1q_2}{r^2}$

$F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.53\times 10^{-10})^2}$

F = 8.20 x 10⁻⁸ N

b) using newton's second law

F = m a

m a = 8.20 x 10⁻⁸

$a =\dfrac{8.20\times 10^{-8}}{9.11\times 10^{-31}}$

a = 9 x 10²² m/s²

c) speed of the electron

$a =\dfrac{v^2}{r}$

$9\times 10^{22} =\dfrac{v^2}{0.53\times 10^{-10}}$

v² = 4.77 x 10¹²

v = 2.18 x 10⁶ m/s

d) the period of the circular motion.

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi\times 0.53\times 10^{-10}}{2.18\times 10^6}$

T = 1.53 x 10⁻¹⁶ s

8 0

3 years ago

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