A car accelerate from a stop to 27 m/sec in 5 seconds. What was the car's acceleration?

You can see your image in a shiny, flat surface because the light waves are bouncing off the surface back at you. This is an exa

Alexxx [7]

I think it would be Reflection because the light in the ray reflects off of any flat or shiny object.... Hope this helps ^-^....

4 0

2 years ago

Read 2 more answers

Ball A is allowed to fall and strike ball B. Assume that all of ball A's energy is transferred to ball B at point I, and that th

Alika [10]

Answer:

Explanation:

If no Energy is lost ( either as Sound energy or any other energy) the total energy of ball A equals the total energy of ball B

Therefore the Kinetic Energy of Ball B equals the Total Energy of Ball A minus the Potential Energy of Ball A

K. E = T. E - P.E

5 0

3 years ago

Read 2 more answers

Which element has the fewest valence electrons?

xxTIMURxx [149]

Answer

the answer is c because i did that before

Explanation:

8 0

2 years ago

What could be used as another word for electrical potential

tekilochka [14]

Answer:

hey mate

answer is probably voltage as per me

as

Explanation:

Voltage, electric potential difference, electric pressure or electric tension is the difference in electric potential between two points, which is defined as the work needed per unit of charge to move a test charge between the two points

5 0

2 years ago

Read 2 more answers

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

2 years ago