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3 years ago

The objective of baseball is to OUTSCORE your opponent. True or false Help ?

Physics

2 answers:

Yuki888 [10]3 years ago

5 0

Answer:

True

Explanation:

The team with the most points at the end of the game is the winner.

Evgesh-ka [11]3 years ago

3 0

Yes it is true - hope it help

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Zina [86]

I believe you are right. pH is the concentration of H+. Therefore the more acidic the more H+.

3 0

4 years ago

Two mirrors are touching so they have an angle of 35.4 degrees with one another. A light ray is incident on the first at an angl

alexandr1967 [171]

Answer:

54.6°

Explanation:

From law of reflection i=r.

So, construct the reflected ray at 55.7°degrees from the normal and let it fall on the other mirror.

Now draw the second normal at the point of incidence and again measure the angle of incidence, and draw the angle of reflection.

If you consider triangle AOB, one angle is ∠AOB=90°

and ∠OAB is 54.6°

From angle sum property third angle ie ∠ABO=180°-90°-54.6°=35.4°

So, the second incident angle will be 54.6°

Hence, the second reflected angle will be 54.6 degrees.

8 0

4 years ago

A brick is thrown vertically upward with an initial speed of 3.00 m/s from the roof of a building. If the building is 78.4 m tal

masya89 [10]

Answer:

0.918 sec

Explanation:

time of flight:

t=u²/g

=3²/9.8=0.918sec

6 0

3 years ago

Read 2 more answers

A planet of mass m 6.75 x 1024 kg is orbiting in a circular path a star of mass M 2.75 x 1029 kg. The radius of the orbit is R 8

Umnica [9.8K]

Answer:

The orbital period of the planet is 387.62 days.

Explanation:

Given that,

Mass of planet $m =6.75\times10^{24}\ kg$

Mass of star $m'=2.75\times10^{29}\ kg$

Radius of the orbit $r =8.05\times10^{7}\ km$

Using centripetal and gravitational force

The centripetal force is given by

$F = \dfrac{mv^2}{r}$

$F=m\omega^2r$

We know that,

$\omega=\dfrac{2\pi}{T}$

$F=m(\dfrac{2\pi}{T})^2r$ ....(I)

The gravitational force is given by

$F = \dfrac{mm'G}{r^2}$ ....(II)

From equation (I) and (II)

$m(\dfrac{2\pi}{T})^2r=\dfrac{mm'G}{r^2}$

Where, m = mass of planet

m' = mass of star

G = gravitational constant

r = radius of the orbit

T = time period

Put the value into the formula

$T^2=\dfrac{4\pi^2R^3}{m'G}$

$T^2=\dfrac{4\times(3.14)^2\times(8.05\times10^{7})^3}{2.75\times10^{29}\times6.67\times10^{-11}}$

$T=2\times3.14\times\sqrt{\dfrac{(8.05\times10^{10})^3}{2.75\times10^{29}\times6.67\times10^{-11}}}$

$T =3.34\times10^{7}\ s$

$T= 387.62\days$

Hence, The orbital period of the planet is 387.62 days.

4 0

3 years ago

Read 2 more answers

A tennis ball is hit into the air with a racket. When is the ball’s kinetic energy the greatest? Ignore air resistance.

Lady bird [3.3K]

Answer B is the correct answer

We know that kinetic energy $E = \frac{1}{2} mv^2$ , where m is the mass of object and v is the velocity of object.

In this case only velocity is the variable, mass remains constant.

So point having higher velocity has higher kinetic energy.

When it leaves the racket, the ball will be having a certain height, but just before it reaches the ground it will not having any height. So maximum velocity of ball is at that time when it reaches just above the ground.

So option B is the correct answer.

4 0

3 years ago