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2 years ago

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A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal

( s = 0.900 J/g·°C), find the final temperature of the block and the water

1 answer:

Nataly_w [17]2 years ago

6 0

Answer:

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature $T_i$ = 93.2°C

We know

$Q = m c \Delta T$ ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in $J/g^{\circ}C$

$\Delta T$ = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal = $125\times 0.9\times (93.2-T_f)$

Heat gained by the water = $100\times 4.184\times(T_f -18.3)$

thus, we have

$125\times 0.9\times (93.2-T_f)$ = $100\times 4.184\times(T_f -18.3)$

$10485-112.5T_f = 418.4T_f - 7656.72$

⇒ $T_f = 34.17^oC$

Therefore, the final temperature will be = 34.17°C

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A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C.

Sedaia [141]

Answer:

335°C

Explanation:

Heat gained or lost is:

q = m C ΔT

where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

Heat gained by the water = heat lost by the copper

mw Cw ΔTw = mc Cc ΔTc

The water and copper reach the same final temperature, so:

mw Cw (T - Tw) = mc Cc (Tc - T)

Given:

mw = 390 g

Cw = 4.186 J/g/°C

Tw = 22.6°C

mc = 248 g

Cc = 0.386 J/g/°C

T = 39.9°C

Find: Tc

(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)

Tc = 335

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You hear the engine roaring on a race car at the starting line. Predict the changes in the sound as the race starts and the car

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B. The sound of the engine will get louder and the pitch higher.

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For each property listed, identify the type of element it describes. Very good electrical conductivity: Amphoteric, able to form

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The elements that is very good in electrical conductivity are gold and copper: elements that is amphoteric are copper, zinc, tin, lead, aluminum and beryllium: elements that is gaseous at room temperature are hydrogen, nitrogen, oxygen, fluorine and chlorine: elements that is solid at room temperature are all metal except mercury and perhaps some unseen radioactive elements. Lastly, elements that is brittle are hydrogen, carbon, nitrogen, oxygen, phosphorus, sulfur and selenium

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A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

yanalaym [24]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial

Being said that, we can calculate the initial velocity of the ball

a) First we analyze its horizontal motion

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

That would be our first equation

Now, we need to analyze its vertical motion

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Knowing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Solving for t

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

So, the magnitude of the velocity is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The <u><em>direction of the ball</em></u> is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

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2 years ago

A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity

Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2 = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches = 0.0508 m

n = 1750 rpm

$H_{nom}=2hp=1491.4W$

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

$w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m$

$V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s$

$F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N$

$(F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N$

$T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm$

$F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N$

$F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N$

b)

$H_a=1491*1.25=1863.75W$

$n_f_s=\frac{H_a}{H_{nom}K_S }=1$

dip = $\frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm$

7 0

2 years ago