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4 years ago

If a car can go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50

mi/hr?

Physics

2 answers:

Harman [31]4 years ago

6 0

$a= \frac{V}{t} \ \ \ \Rightarrow\ \ a= \frac{60\ [mi/hr]}{8\ [s]} \\ \\the\ final\ speed\ after\ 5.0\ seconds\\ \\V=50\ [mi/hr]+5\ [s]\ \cdot \frac{60\ [mi/hr]}{8\ [s]}=50\ [mi/hr] + \frac{5}{8} \cdot 60\ [mi/hr]=\\ \\=50\ [mi/hr]+37.5\ [mi/hr]=87.5\ [mi/hr]$

Lady_Fox [76]4 years ago

5 0

$v_1=60\ \frac{mi}{h}\\\\t_1=8.0\ s=\frac{8}{3600}\ h=\frac{1}{450}\ h\\--------------\\a=\frac{v}{t}\\\\a=\frac{60}{\frac{1}{450}}\ \frac{mi}{h^2}=27000\ \frac{mi}{h^2}\\--------------\\v_o=50\ \frac{mi}{h}\\\\t=5.0\ s=\frac{5}{3600}\ h=\frac{1}{720}\ h\\--------------$

$v=v_o+at\\\\\\v=50\ \frac{mi}{h}+27000\ \frac{mi}{h^2}\cdot\frac{1}{720}\ h=50\ \frac{mi}{h}+37.5\ \frac{mi}{h}=87.5\ \frac{mi}{h}\leftarrow Answer$

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Calculate amount of current flowing in wire if number of charge is 8 and e = 1.69 x10^-19 , t= 8millsec

irina1246 [14]

The amount of current flowing in the wire is 1.69 × 10⁻¹⁶ A

From the question,

We are to calculate, the amount of current flowing in the wire.

From the formula,

$I =\frac{nQ}{t}$

Where I is the current

n is the number of charge

Q is the quantity of charge

t is the time

From the given information

n = 8

Q = e = 1.69 × 10⁻¹⁹ C

t = 8 msecs = 8 × 10⁻³ sec

Putting the parameters into the formula, we get

$I = \frac{8 \times 1.69 \times 10^{-19} }{8\times 10^{-3} }$

$I = 1.69 \times 10^{-16} \ A$

Hence, the amount of current flowing in the wire is 1.69 × 10⁻¹⁶ A

Learn more here: brainly.com/question/22135459

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2 years ago

As his socks tumbled in the dryer, they became charged. If a small piece of lint with a charge of +1.25 E -19 C is attracted to

qwelly [4]

Answer:

$2.4\cdot 10^{10}N$

Explanation:

When an electric charge is immersed in an electric field, it experiences a force given by the equation

$F=qE$

where

q is the charge

E is the magnitude of the electric field

In this problem, we have:

$q=+1.25\cdot 10^{-19}C$ is the charge on the small piece of lint

$F=3.0\cdot 10^{-9}N$ is the force experienced by the charge

Therefore, we can find the magnitude of the electric field by re-arranging the equation and solving for E:

$E=\frac{F}{q}=\frac{3.0\cdot 10^{-9}}{1.25\cdot 10^{-19}}=2.4\cdot 10^{10}N$

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Answer:

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Explanation:

given,

F = 14.1 i + 0 j + 5.1 k

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Assuming block is moving in x- direction

we know,

dW = F dx

$\int dW = F\int dx$

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$W = F[x]_0^6$

$W = 14.1 \times 6$

$W =84.6\ Nm$

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