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Harrizon [31]
3 years ago
11

the gravitational force exerted on a baseball is 2.21 N down, a pitcher throws the ball horizontally with velocity 18.0 m/s by u

niformly accelerating it along a straight horizontal line for a time interval of 170 ms. the ball starts from rest. Through what distance does it move before its release? what are the magnitude and direction of the force of the pitcher exerts on the ball?

Physics
2 answers:
Ilia_Sergeevich [38]3 years ago
8 0
To have a weight of 2.21N., the ball's mass is (2.21/9.8) = .226kg. 
<span>a) d = 1/2 (vt), = 1/2 (18 x .17), = 1.53m. </span>
<span>b) Acceleration of the ball = (v/t), = 18/.17, = 105.88m/sec^2. </span>
<span>f = (ma), = .226 x 105.88, = 23.92N. </span>
erik [133]3 years ago
7 0

Hello!

The gravitational force exerted on a baseball is 2.21 N down, a pitcher throws the ball horizontally with velocity 18.0 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 170 ms. the ball starts from rest. Through what distance does it move before its release? what are the magnitude and direction of the force of the pitcher exerts on the ball ?

We have the following data:

P (force-weight or gravitacional force) = 2.21 N

m (mass) = ? (in kg)

g (acceleration of gravity) ≈ 9.8 m/s²

Vf (final velocity) = 18 m/s

Vi (initial velocity) = 0 m/s (in rest)

Δt (time interval) = 170 ms = 0.17 s

a (acceleration) = ? (in m/s²)

F (force or magnitude) = ? (in N or kg.m/s²)

[First Step]  Let's find the mass of the ball, let's see:

P = m*g

2.21 = m * 9.8

2.21 = 9.8\:m

9.8\:m = 2.21

m = \dfrac{2.21}{9.8}

\boxed{m \approx 0.226\:kg}

[Second Step] Let's find the acceleration, let'see:

a = \dfrac{\Delta{V}}{\Delta{T}}

a = \dfrac{V_f - V_i}{\Delta{T}}

a = \dfrac{18.0\:m/s - 0\:m/s}{0.17\:s}

a = \dfrac{18\:m/s}{0.17\:s}

\boxed{a \approx 105.88\:m/s^2}

[Third Step] Through what distance does it move before its release?

By the formula of the space of the Uniformly Varied Movement, it is:

d = V_i*t + \dfrac{a*t^2}{2}

d = 0*0.17 + \dfrac{105.88*0.17^2}{2}

d = 0 + \dfrac{105.88*0.0289}{2}

d = \dfrac{3.059932}{2}

\boxed{\boxed{d \approx 1.53\:m}}\:\:\:\:\:\:\bf\blue{\checkmark}

*** [Fourth Step] What are the magnitude....

F =m*a

F = 0.226\:kg*105.88\:m/s^2

F = 23.92888\:kg*m/s^2

\boxed{\boxed{F \approx 24\:N}}\:\:\:\:\:\:\bf\green{\checkmark}

***.... and direction of the force of the pitcher exerts on the ball ?

tan\:C= \dfrac{F_1\downarrow}{F_2\rightarrow}

tan\:C= \dfrac{2.21}{24}

tan\:C \approx 0.093 \to \boxed{\boxed{C = arctan\:0.093 \approx 5.3\:\º}}\Longrightarrow(above\:the\:horizontal)\:\:\:\:\:\:\bf\green{\checkmark}

________________________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}

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0.699 L of the fluid will overflow

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Substituting these values into the equation, we have

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3 years ago
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Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

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c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

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You use the equation (3) with n=5:

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