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3 years ago

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7) The maximum horizontal distance a boy can throw a ball is 50.0 m. Assume he can throw with the same initial speed at all angl

es. How high does he throw the ball when he throws it straight upward?

1 answer:

ivanzaharov [21]3 years ago

6 0

Given: v is constant at any value and x=50 m, then find the maximum hieght which is $y_{max}$ . We can also see that the motion of horizontal direction is given by $x=v_{x}t$ , and also, since $v_{x}=v_{i}$ it's $v_{i}t=50$ , also $t_{1}=\frac{t}{2}$ . So now just put this in the equation $y_{max}=v_{yi}+\frac{1}{2}gt^{2}_{1}$ you will get ymax being 12.6 along with using the equation $v_{yf}=v_{yi}-gt_{1}$ .

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)

- Cutiepatutie ☺❀❤

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At what point is the northern hemisphere pointed farthest away from the sun?

inna [77]

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3 years ago

A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of $1.055\times 10^{- 7}$ in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV = $250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, $m_{p} = 1.67\times 10^{- 27} kg$

The initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This energy of proton is $\simeq 250 GeV$

Thus the speed of the proton, v $\simeq c$

Now, the time taken to cover 1 km = 1000 m of the distance:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

Now, in accordance to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

$\Delta t_{o} = \Delta t$

where

$\Delta t$ = final width

3 0

3 years ago

A. I am a natural indicator. I turn my pink color to deep pink in an acid and green in a basic solution. I am …………………. B. I am a

olga nikolaevna [1]

Answer:

A. China rose

B. Methyl red

C. Phenolphthalein

Explanation:

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3 years ago

With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons

Shtirlitz [24]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

$m=\frac{Y-Y_{1}}{X-X_{1}}$ (1)

Where:

$m=5.7$ is the slope of the line

$Y_{1}=2390.7pounds$ is the airplane weight with 51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

$X_{1}=51gallons$ is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

$(X_{1},Y_{1})=(51,2390.7)$

Rewritting (1):

$Y=m(X-X_{1})+Y_{1}$ (2)

As Y is a function of X:

$Y=f_{(X)}=m(X-X_{1})+Y_{1}$ (3)

Substituting the known values:

$f_{(X)}=5.7(X-51)+2390.7$ (4)

$f_{(X)}=5.7X-290.7+2390.7$ (5)

$f_{(X)}=5.7X+2100$ (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

$f_{(81)}=5.7(81)+2100$ (7)

$f_{(81)}=2561.7$ (8) This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

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3 years ago

What do the elements in group 17 of the periodic table have in common

yulyashka [42]

They are toxic and all non metals

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3 years ago