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LUCKY_DIMON [66]
3 years ago
12

A 567-g empty iron kettle is put in a hot stove the kettle absorbs 18,100 j of heat to raise its temperature from 15.0c to a fin

al temperature what is the final temperature? The specific heat capacity of iron is 470j/kg•k
Physics
1 answer:
ipn [44]3 years ago
7 0

Answer:

T2 = 355.92 Kelvin or 82.92°C

Explanation:

Given the following data;

Mass = 567g to kilograms = 567/1000 = 0.567 kg

Quantity of heat = 18,100J

Initial temperature = 15°C to Kelvin = 15 + 273 = 288K

Specific heat capacity of iron = 470j/kg•k

To find the final temperature;

Heat capacity is given by the formula;

Q = mcdt

Where;

  • Q represents the heat capacity or quantity of heat.
  • m represents the mass of an object.
  • c represents the specific heat capacity of water.
  • dt represents the change in temperature.

Making dt the subject of formula, we have;

dt = \frac {Q}{mc}

Substituting into the equation, we have;

dt = \frac {18100}{0.567*470}

dt = \frac {18100}{266.49}

dt = 67.92K

Now, the final temperature T2 is;

But, dt = T2 - T1

T2 = dt + T1

T2 = 67.92 + 288

T2 = 355.92 Kelvin or 82.92°C

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1.
lyudmila [28]

Answer:

100m \div  \frac{9 \: m}{1sec}  = 11.11 \: sec \:

I guess you can round it to 11 seconds.

Explanation:

Going with a speed 9m/s means you are going 9 meters in each second.

If you are going 9 meters in second how many seconds will it take to 100 meters?

Visually;

9 meters - - - 1 second

100 meters - - - ?seconds.

When you write like this 9 times ?seconds equal to 100 meters time 1 second. (you probably know this but just in case)

So to find ?second you multiply 100meters by 1 and divide it by 9 whixh will give you 11.1111 seconds whixh again I believe you can round it to 11.

(Kind of a) Proof;

If 9m * ?sec = 100 m * 1 sec

you send 9 meters to other side.

?sec = (100 m * 1 sec) ÷ 9m

Hope it was clear and it helps! Please let me know if you have any questions.

7 0
3 years ago
Convert 15 litre into cubic metre​
atroni [7]

Answer:

0.015m^3

Explanation:

1 m^3 = 1000 liters

x m^3 = 15 liters

Cross multiply

xm^3 x 1000 l = 15 l

Divide both sides by 1000

xm^3 x1000/1000 = 15/1000

xm^3 = 0.015m^3

Therefore 15 liter = 0.015m^3

5 0
3 years ago
A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×
HACTEHA [7]

Answer:

f_r = 150.47 N

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta

∑ F_y = 0

0 = N cos \theta - f_r sin \theta - mg

N = \dfrac{f_rsin \theta + mg}{cos \theta}

\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta

              = f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta

        f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}

         f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}

f_r = 150.47 N

8 0
4 years ago
Can somebody re-answer the true and false questions!!!
stiv31 [10]

Answer:

I think no 3 is false

and 4 is true

and the the ones you did are correct

if wrong correct me pls

7 0
3 years ago
Read 2 more answers
a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet
ANTONII [103]

Answer:

Explanation:

Given

radius of circular path r=4\ in.

Position is given by

\theta =\cos 2t---1

Differentiate 1  to angular velocity we get

\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2

Differentiate 2 to get angular acceleration

\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3

Net acceleration is the vector summation of tangential and centripetal force

a_t=\alpha \times r

a_t=-4\cos 2t\times 4=-16\cos 2t

a_r=\omega ^2\cdot r

a_r=(-2\sin 2t)^2\cdot 4

a_r=16\sin^2(2t)

a_{net}=\sqrt{a_r^2+a_t^2}

a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}

a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}                                                    

6 0
3 years ago
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