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3 years ago

11

when a man throws a stone in the direction of maximum range the stone rises to a height of 20m . what is the range and how long

is the stone in air

1 answer:

jeyben [28]3 years ago

3 0

Answer:

50m

Explanation:im just smart thank me later

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How many calories are needed to vaporize 1g of boiling water at 100?

vfiekz [6]

Using the equation: Q = mL
Q = 1(100) Q = 100cal
I hope this is right and helps :)

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3 years ago

A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal

Sati [7]

(a) The ball has a final velocity vector

$\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j$

with horizontal and vertical components, respectively,

$v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}$

$v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}$

The horizontal component of the ball's velocity is constant throughout its trajectory, so $v_{x,i}=v_{x,f}$ , and the horizontal distance <em>x</em> that it covers after time <em>t</em> is

$x=v_{x,i}t=v_{x,f}t$

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

$103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s$

The vertical component of the ball's velocity at time <em>t</em> is

$v_{y,f}=v_{y,i}-gt$

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

$-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}$

So, the initial velocity vector is

$\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which carries an initial speed of

$\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}$

and direction <em>θ</em> such that

$\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}$

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is

$y=v_{y,i}t-\dfrac12gt^2$

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of

$y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}$

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4 years ago

Ted lifts a 10N weight at a height of 1.5 m in 1

masha68 [24]

Answer:

D. Ted expanded more power.

Explanation:

Given the following data;

For Ted.

Force = 10N

Height = 1.5m

Time = 1 seconds

To find Ted's power;

Power = workdone/time

But workdone = force * distance

Workdone = 10 * 1.5

Workdone = 15 Nm

Power = 15/1

Power = 15 Watts.

For Johnny.

Force = 10N

Height = 1.5m

Time = 2 seconds

To find Ted's power;

Power = workdone/time

But workdone = force * distance

Workdone = 10 * 1.5

Workdone = 15 Nm

Power = 15/2

Power = 7.5 Watts

Therefore, from the calculations we can deduce and conclude that Ted expanded more power.

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3 years ago

The MRI (Magnetic Resonance Imaging) body scanners used in hospitals operate at a frequency of 400 MHz (M = 106 ). Calculate the

horrorfan [7]

Answer:

Energy, $E=2.65\times 10^{-25}\ J$

Explanation:

It is given that,

The MRI (Magnetic Resonance Imaging) body scanners used in hospitals operate at a frequency of 400 MHz,

$\nu=400\ MHz=400\times 10^6\ Hz=4\times 10^8\ Hz$

We need to find the energy for a photon having this frequency. The energy of a photon is given by :

$E=h\nu$

$E=6.63\times 10^{-34}\ J-s\times 4\times 10^8\ Hz$

$E=2.65\times 10^{-25}\ J$

So, the energy of the photon is $2.65\times 10^{-25}\ J$ . Hence, this is the required solution.

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3 years ago

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mario62 [17]

The absolute awser is b

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