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3 years ago

I NEED HELP SOLVING THIS!!!!!!!!!!!!

Physics

1 answer:

Neko [114]3 years ago

7 0

Answer:

Yes

Explanation:

The given parameters are;

The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)

The angle in which the fastball is hit, θ = 22°

The distance of the field = 96 m (315 ft)

The range of the projectile motion of the fastball is given by the following formula

$Range = \dfrac{u^2 \times sin(2\cdot \theta)}{g}$

Where;

g = The acceleration due to gravity = 9.81 m/s², we have;

$Range = \dfrac{49.1^2 \times sin(2\times22^{\circ})}{9.81} \approx 170.71 \ m$

Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.

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How high would the level be in a gasoline barometer at normal atmospheric pressure?

Sergio [31]

Answer:

h = 13.06 m

Explanation:

Given:

- Specific gravity of gasoline S.G = 0.739

- Density of water p_w = 997 kg/m^3

- The atmosphere pressure P_o = 101.325 KPa

- The change in height of the liquid is h m

Find:

How high would the level be in a gasoline barometer at normal atmospheric pressure?

Solution:

- When we consider a barometer setup. We dip the open mouth of an inverted test tube into a pool of fluid. Due to the pressure acting on the free surface of the pool, the fluid starts to rise into the test-tube to a height h.

- The relation with the pressure acting on the free surface and the height to which the fluid travels depends on the density of the fluid and gravitational acceleration as follows:

P = S.G*p_w*g*h

Where, h = P / S.G*p_w*g

- Input the values given:

h = 101.325 KPa / 0.739*9.81*997

h = 13.06 m

- Hence, the gasoline will rise up to the height of 13.06 m under normal atmospheric conditions at sea level.

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2 years ago

A particle starts from rest and has an acceleration function 5 − 10t m/s2 . (a) What is the velocity function? (b) What is the p

frez [133]

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

$a(t)=(5-10t)\ m/s^2$

(a) Since, $a=\dfrac{dv}{dt}$

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$dv=a.dt$

$v=\int(a.dt)$

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t = 1 seconds

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