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2 years ago

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What will a spring scale read for the weight of a 57.0-kg woman in an elevator that moves upward with constant speed of 5.0 m/s

?

1 answer:

Tanzania [10]2 years ago

4 0

The spring scale will read 559 Newton's or 125.7 pounds.

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HELP PLEASE<br> 60 POINTS <br> HAVE A GREAT REST OF YOUR DAY PEOPLE :>

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2 years ago

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A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed verti

Korvikt [17]

Answer:

The strength of the electric field is $1.35\times10^{4}\ N/C$ .

Explanation:

Given that,

Speed $v= 5.05\times10^{5}\ m/s$

Time $t= 3.90\times10^{-7}\ s$

Angle = 45°

We need to calculate the acceleration

Using equation of motion

$v = u+at$

$5.05\times10^{5}=0+a\times3.90\times10^{-7}$

$a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}$

$a=1.29\times10^{12}\ m/s^2$

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

$F= ma=qE$

$ma = qE$

$E=\dfrac{ma}{q}$

Put the value into the formula

$E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}$

$E=1.35\times10^{4}\ N/C$

Hence, The strength of the electric field is $1.35\times10^{4}\ N/C$ .

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2 years ago

Can an object be moving downward and have an upward acceleration

dusya [7]

Answer: no

Explanation:

because the object is moving downwards so it will be called decceleration . which is the opposite of acceleration .

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2 years ago

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A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

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