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2 years ago

A baseball player hit a ball with an upward velocity of 46

Physics

1 answer:

masya89 [10]2 years ago

3 0

Its height h is feet after t seconds is given by the function h= 16t^2 + 46t + 6.
I hope this helped.

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A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

3 0

2 years ago

How do we work out the world time zone differences

Virty [35]

To find the time zone in hours of a particular location, you can take the longitude -- in degrees -- and divide it by 15. So, for example, 75° E would be 75/15 which equals 5. That translates to the time zone being 5 hours ahead of UTC or GMT time, which can also be labeled as UTC+5. i hope this helped if not then sorry

4 0

2 years ago

Please help me this is important!

Mariulka [41]

145 Grams!
It asks for the “Total Mass” basically asking to add, If you add 20 to 125, you get 145! Correct me if im wrong

4 0

2 years ago

Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s

Ne4ueva [31]

The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
$F=qvB \sin \theta$
where $\theta$ is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is $q=2.7 \mu C=2.7 \cdot 10^{-6}C$ . In this case, v and B are perpendicular, so $\theta=90^{\circ}$ , therefore we have:
$B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T$

2) In this second case, the angle between v and B is $\theta=55^{\circ}$ . The charge is now $q=42.0 \mu C=42.0 \cdot 10^{-6}C$ , and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
$F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N$

5 0

2 years ago

Read 2 more answers

Waves are observed passing under a dock. Wave crests are 8.0 meters apart. The time for a complete wave to pass by is 4.0 second

ki77a [65]

To answer that question, we don't care what the highest and lowest
levels of the wave are, or how far apart they are. We only need to be
able to identify the highest point on the wave, and keep track of how
often those pass by us.

You said it takes 4 seconds for a complete wave to pass by.
Through the sheer power of intellect, I'm able to take that information
and calculate that 1/4 of the wave passes by in 1 second.

There's your frequency . . . 1/4 per second, or 0.25 Hz.

6 0

2 years ago