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3 years ago

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An object is launched from the top of a building which is 100 m tall (relative to the ground) at a speed of 22 m/s at an angle o

f 23 degrees BELOW the horizontal. How long, in seconds, does it take to reach the ground

1 answer:

klio [65]3 years ago

6 0

Answer:

$3.58\:\mathrm{s}$

Explanation:

We can use the kinematics equation $\Delta y=v_it+\frac{1}{2}at^2$ to solve this problem. To find the initial vertical velocity, find the vertical component of the object's initial velocity using basic trigonometry for right triangles:

$\sin28^{\circ}=\frac{y}{22},\\y=22\sin28^{\circ}=10.3283743813\:\mathrm{m/s}$

Now we can substitute values in our kinematics equation:

$\Delta y=-100$
$a=-9.8\:\mathrm{m/s^2}$ (acceleration due to gravity)
$v_i=-10.3283743813\:\mathrm{m/s}$
Solving for $t$

$-100=-10.3283743813t+\frac{1}{2}\cdot -9.8\cdot t^2,\\\\-4.9t^2-10.3283743813t+100=0,\\\\\boxed{t=3.5849312673637455}, t=-5.692762773751501\:\text{(Extraneous)}$

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3 years ago

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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3 years ago

Communicate the results

katen-ka-za [31]

What results do we communicate?

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4 years ago

A wheel in the shape of a flat, heavy, uniform, solid disk is initially at rest at the top of an inclined plane of height 2.00 m

olasank [31]

Answer:

Explanation:

If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.

mgh = ½mv²

v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s

However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½(½mR²)(v/R)²

2gh = v² + ½v²

2gh = 3v²/2

v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s

7 0

3 years ago