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3 years ago

During a baseball game, a batter hits a high

Physics

1 answer:

My name is Ann [436]3 years ago

3 0

Answer: 43.01 m

Explanation:

Given that :

Time ball uses in the air = 5.93, time it takes to reach maximum height and return = time of flight(T) = 5.93 s

TIME of flight (T) = 2 * time taken(t)

Where g = a = acceleration due to gravity = 9.8m/s²

S = 0.5at²

S = maximum height

Tjme taken (t) = time of flight / 2 = 5.93/2 = 2.965 s

Hence,

S = 0.5at²

S = 2 × 0.5 × 9.8 × 2.965²

S = 43.0770025

S = 43.01 m

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Looking at the wave diagram which best describes the wave

Ad libitum [116K]

Answer:

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2 years ago

13) The condenser pressure is 417.4 psig for r-410A and the condenser outlet temperature is 108f. how much subcooling is there i

Tanzania [10]

Answer:

12°F

Explanation:

Calculation for how much subcooling is there in the condenser

Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

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Therefore the amount of subcooling that is there in the condenser will be 12°F

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3 years ago

Se aplican dos fuerzas concurrentes a un objeto de 4N a la derecha y 5N a la izquierda. ¿Hacia donde se movió y con cuanta fuerz

Maksim231197 [3]

The forces move strongly towards the left by 1N

Given the following

Force towards the right = 4N

Force towards the left = 5N

Note that the force acting towards the left is negative, hence the force acting towards the left is -5N

Take the sum of force

Resultant force = -5N + 4N

Resultant force = -1N

This shows that the forces move strongly towards the left by 1N

Learn more here: brainly.com/question/24629099

3 0

3 years ago

How fast is the sixth cosmic velocity?

denpristay [2]

<span>25,000 miles per hour

hope that this helps</span>

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3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

1 year ago

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