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1 year ago

A baseball is batted from a height of 1.09 m with a speed of

1 answer:

5 0

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

$H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m$

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

$V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s$

<h3>Resultant speed of the ball and crosswind</h3>

$V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s$

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

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3 years ago

Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made o

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Answer:

There is a loss of fluid in the container of 0.475L

Explanation:

To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.

The formula that describes this thermal expansion process is given by:

$\Delta V = \beta V_0 \Delta T$

Where,

$\Delta V =$ Change in volume

$V_0 =$ Initial Volume

$\Delta T =$ Change in temperature

$\beta =$ coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

$\Delta V_T = \Delta V_l - \Delta V_c$

Where,

$\Delta V_l$ = Change in the volume of liquid

$\Delta V_c$ = Change in the volume of copper

Then replacing with the previous equation we have:

$\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T$

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

Our values are given as,

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Replacing we have that,

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

$\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)$

$\Delta V = 0.475L$

Therefore there is a loss of fluid in the container of 0.475L

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3 years ago

An engine draws energy from a hot reservoir with a temperature of 1250 K and exhausts energy into a cold reservoir with a temper

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Answer:

The power output of this engine is $P = 17.5 W$

The the maximum (Carnot) efficiency is $\eta_c = 0.7424$

The actual efficiency of this engine is $\eta _a = 0.46$

Explanation:

From the question we are told that

The temperature of the hot reservoir is $T_h = 1250 \ K$

The temperature of the cold reservoir is $T_c = 322 \ K$

The energy absorbed from the hot reservoir is $E_h = 1.37 *10^{5} \ J$

The energy exhausts into cold reservoir is $E_c = 7.4 *10^{4} J$

The power output is mathematically represented as

$P = \frac{W}{t}$

Where t is the time taken which we will assume to be 1 hour = 3600 s

W is the workdone which is mathematically represented as

$W = E_h -E_c$

substituting values

$W = 63000 J$

$P = \frac{63000}{3600}$

$P = 17.5 W$

The Carnot efficiency is mathematically represented as

$\eta_c = 1 - \frac{T_c}{T_h}$

$\eta_c = 1 - \frac{322}{1250}$

$\eta_c = 0.7424$

The actual efficiency is mathematically represented as

$\eta _a = \frac{W}{E_h}$

substituting values

$\eta _a = \frac{63000}{1.37*10^{5}}$

$\eta _a = 0.46$

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3 years ago

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Answer:

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2 years ago

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Answer:

22145.27733 ft

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Explanation:

The equation of pressure is

$P=P_0e^{-kh}$

where,

$P_0$ =Atmospheric pressure = 800 mbar

k = Constant

h = Altitude = 35000 ft

$P=\dfrac{1}{3}P_0$

$\dfrac{1}{3}P_0=P_0e^{-k35000}\\\Rightarrow \dfrac{1}{3}=e^{-k35000}\\\Rightarrow 3=e^{k35000}\\\Rightarrow ln3=k35000\\\Rightarrow k=\dfrac{ln3}{35000}\\\Rightarrow k=3.13\times 10^{-5}$

Now

$P=\dfrac{1}{2}P_0$

$ln2=kh\\\Rightarrow h=\dfrac{ln2}{k}\\\Rightarrow h=\dfrac{ln2}{3.13\times 10^{-5}}\\\Rightarrow h=22145.27733\ ft$

The altitude will be 22145.27733 ft

$P=0.02P_0$

$0.02P_0=P_0e^{-kh}\\\Rightarrow 0.02=e^{-3.13\times 10^{-5}h}\\\Rightarrow ln0.02=-3.13\times 10^{-5}h\\\Rightarrow h=\dfrac{ln0.02}{-3.13\times 10^{-5}}\\\Rightarrow h=124984.76055\ ft$

The elevation is 124984.76055 ft

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3 years ago