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2 years ago

10

A softball is hit into the outfield during a game. The initial velocity is 33.52 m/s, and the ball is hit at an angle of 27°. Ho

w far away should an opposing player stand to catch the ball? As
players are the same height. Show your work.

1 answer:

Karo-lina-s [1.5K]2 years ago

7 0

Answer:

weeeeeeeeeebeeeeeeeeeeeeeee

Explanation:

weeeeeeeeeee

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Two waves are superposed. one wave has an amplitude of 5 cm, and other has an amplitude of 4 cm. What is the resultant amplitude

Vsevolod [243]

If both waves have the same wavelength, then the amplitude of
their sum could be anything between 1 cm and 9 cm, depending
on the phase angle between them.

If the waves have different wavelengths, then the resultant is a beat
with an amplitude of 9 cm.

4 0

3 years ago

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$

$7.05 + 30.5 = \frac{0.266}{r}$

$r = 7.08 \times 10^{-3} m$

so distance moved by the particle is given as

$d = r_1 - r_2$

$d = 0.0377 - 0.00708$

$d = 0.0306 m$

6 0

3 years ago

A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh

Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 $rad/sec^2$

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius $r=\frac{d}{2}=\frac{1.63}{2}=0.815m$

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

$\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec$

(b) We know that linear speed is given by

$v=r\omega =0.815\times 8.362=6.815m/sec$

(c) We have given final angular velocity $\omega _f=675rev/min$

And $\omega _i=79.9rev/min$

Time t = 63 sec

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2$

(d) Change in angle is given by

$\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev$

7 0

3 years ago

A skier of mass 82.9 kg starts from rest at the top of a frictionless incline of height 20 m. At the bottom of the incline, the

ehidna [41]

Answer: 170.67 N

Explanation:

Given

Mass of skier is $m=82.9\ kg$

Height of the inclination is $h=20\ m$

Here, the potential energy of the skier is converted into kinetic energy which is consumed by the friction force by applying a constant force that does work to stop the skier.

$\Rightarrow mgh=F\cdot x\quad \quad [\text{F=constant friction force}]\\\\\Rightarrow 82.9\times 9.8\times 20=F\cdot 95.2\\\\\Rightarrow F=\dfrac{16,248.4}{95.2}\\\\\Rightarrow F=170.67\ N$

Thus, the horizontal friction force is 170.67 N.

7 0

3 years ago

The space shuttle usually orbited Earth at altitudes of around 300.0 km. 1) Determine the time for one orbit of the shuttle abou

777dan777 [17]

Answer:

Explanation:

distance of shuttle from centre of the earth = radius of the orbit

= 6300 + 300 = 6600 km

= 6600 x 10³

Formula of time period of the satellite

T = 2π R /v₀ , v₀ is orbital velocity

v₀ = √gR , ( if height is small with respect to radius )

T = 2π R /√gR

= 2π√ R /√g

= 2 x 3.14 x √ 6600 x 10³ / √9.8

= 2 x 3.14 x 256.9 x 10 / 3.13

= 5154.41 s

= 5154.41 / 60 minutes

= 85.91 m

85.9 minutes.

2 ) No of sunrise per day = no of rotation per day

= 24 x 60 / 85.9

= 16.76

or 17 sunrises.

3 0

3 years ago