A car with a mass of 833 kg rounds an unbanked curve in the road at a speed of 28.0 m/s. If the

The electric field strength is 5.50×10^4 N/C inside a parallel-plate capacitor with a 2.50 mm spacing. A proton is released from

valentina_108 [34]

Answer:

$v=1.6\times10^{5}m/s$

Explanation:

The equation that relates electric field strength and force that a charge experiments by it is F=qE.

Newton's 2nd Law states F=ma, which for our case will mean:

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

We know from accelerated motion that $v^2=v_0^2+2ad$

In our case the proton is released from rest, so $v_0=0m/s$ and we get $v=\sqrt{2ad}$

Substituting, we get our final velocity equation:

$v=\sqrt{\frac{2qEd}{m}}$

For the <em>proton </em>we know that $q=1.6\times10^{-19}C$ and $m=1.67\times10^{-27}kg$ . Writing in S.I. d=0.0025m, we obtain:

$v=\sqrt{\frac{2(1.6\times10^{-19}C)(5.5\times10^{4}N/C)(0.0025m)}{(1.67\times10^{-27}kg)}}=162318.5m/s=1.6\times10^{5}m/s$

6 0

3 years ago

The conclusion to an experiment often includes new questions to be tested. This is why science is considered to be _____. A. dif

Veronika [31]

I would say D. Testable

6 0

3 years ago

A student is doing a lab experiment in which she uses diffraction to measure the width of a shaft of her hair. She shines a 530

professor190 [17]

Answer:

.09 mm

Explanation:

In case of diffraction by single slit with width a the width of central maxima is given below

width = 2 λD/a

where λ is wavelength of light , D is distance of screen , a is width o slit

substituting the given values

14 x 10⁻³ = $\frac{2\times530\times10^{-9}\times1.2}{a}$

a = .09 mm

b) If a is greater , width of central maxima will be less wide.

7 0

4 years ago

A car is traveling at a constant speed of 12m/s , when the driver accelerates the car reaches a speed of 26 m/s in 6s what is th

DaniilM [7]

Acceleration = (change in speed) / (time for the change)

Change in speed = (end speed) - (start speed)

Change in speed = (26 m/s) - (12 m/s) = 14 m/s

Time for the change = 6 s

Acceleration = (14 m/s) / (6 s)

Acceleration = (14/6) (m/s²)

<em>Acceleration = 2.33 m/s²</em>

7 0

4 years ago

une lampe soumise à une tension continue de 12V , consomme unepuissance de 1.8 W. L'intensité du courant traversant la lampe est

Eduardwww [97]

Answer:

I = 0.15 A

Explanation:

The question is "a lamp subjected to a continuous voltage of 12V , consumes a power of 1.8 W. The intensity of the current passing through the lamp is what"

Given that,

Power of the lamp, P = 1.8 W

Voltage, V = 12 V

We need to find the current through the lamp. The power of the lamp is given by :

$P=V\times I\\\\I=\dfrac{P}{V}\\\\I=\dfrac{1.8}{12}\\\\I=0.15\ A$

So, the current through the lamp is 0.15 A.

6 0

3 years ago