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3 years ago

A velocity time graph shows how____ changes over time

Physics

1 answer:

irga5000 [103]3 years ago

5 0

Answer: A velocity-time graph shows how velocity changes over time

Explanation:

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Give an example of Newton's 1st law

Vesna [10]

Answer:

A ball moving until gravity pulls it back down to the ground

Explanation:

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3 years ago

Select the correct answer.

lutik1710 [3]

I’m sorry i haven’t found the answer to this

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3 years ago

An infinitely long line of charge has a linear charge density of 7.50×10^−12 C/m . A proton is at distance 14.5 cm from the line

Nata [24]

Answer:

10.22 cm

Explanation:

linear charge density, λ = 7.5 x 10^-12 C/m

distance from line, r = 14.5 cm = 0.145 m

initial speed, u = 3000 m/s

final speed, v = 0 m/s

charge on proton, q = 1.6 x 10^-19 C

mass of proton, m = 1.67 x 10^-27 kg

Let the closest distance of proton is r'.

The potential due t a line charge at a distance r' is given by

$V=-2K\lambda ln\left (\frac{r'}{r} \right )$

where, K = 9 x 10^9 Nm^2/C^2

W = q V

By use of work energy theorem

Work = change in kinetic energy

$qV = 0.5m(u^{2}-v^{2})$

By substituting the values, we get

$V=\frac{mu^{2}}{2q}$

$-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}$

$- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }$

$- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }$

$- ln\left ( \frac{r'}{r} \right )=0.35$

$\frac{r'}{r} =e^{-0.35}$

$\frac{r'}{r} =0.7047$

r' = 14.5 x 0.7047 = 10.22 cm

5 0

3 years ago

The magnitude R, measured on the Richter scale, of an earthquake of intensity I is defined as Requalslog StartFraction Upper I

lapo4ka [179]

Answer:

R = 6.8

Explanation:

Given data:

Richter scale $R = log(\frac{I}{I_o})$

where R - magnitude of earthquake of Richter scale

I - quake's intensity $= 10^{6.8} \times I_o$

$I_o$ - minimum intensity earthquake

Plugging all information in the equation to get Richter's scale

$R = log(\frac{10^{6.8} \times I_o}{I_o})$

$R = log(10^{6.8})$

R = 6.8

6 0

3 years ago

a car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achiev

atroni [7]

T=Vf-Vi/s
25m/s -15m/s/ 125m
10m/s /125m
=0.08s
I hope it’s correct !

6 0

3 years ago