Ask question

Ask question

All categories

2 years ago

13

The specific heat of water is 4.184 J g—1 K—1. A piece of iron (Fe) weighing 40 g is heated to 80EC and dropped into 100 g of wa

ter at 25EC. (Specific heat of Fe is 0.446 J g—1 K—1) What is the temperature when thermal equilibrium has been reached

1 answer:

CaHeK987 [17]2 years ago

5 0

Explanation:

Below is an attachment containing the solution.

You might be interested in

What is the acceleration of an object with a mass of

aev [14]

Heya!!

For calculate aceleration, lets applicate second law of Newton:

$\boxed{F=ma}$

<u>Δ Being Δ</u>

F = Force = 183 N

m = Mass = 367 kg

a = Aceleration = ?

⇒ Let's replace according the formula and clear "a":

$\boxed{a=183 \ N / 367\ kg}$

⇒ Resolving

$\boxed{ a = 0.49 \ m/s^{2}}$

Result:

The aceleration is <u>0,49 meters per second squared (m/s²)</u>

Good Luck!!

3 0

2 years ago

Please help me.

vesna_86 [32]

A) the tension in the string.

4 0

3 years ago

Read 2 more answers

The dipole moment of HI is 0.42D. What is the dipole moment of HI in C⋅m?

sleet_krkn [62]

So in your question where ask to find the dipole moment of HI in C.M. So in my calculation by converting the given data from Debyes to Coulomb Meteres is that 1 Dyebes is equals to 3.33X10^(-30) C.M and the answer would be 1.40X10(-30)C.M. I hope you are satisfied with my answer and feel free to ask for more

5 0

2 years ago

What is not a factor affecting impact in a collision; vehicle weight, speed, kinetic energy, centrifugal force

navik [9.2K]

I believe the answer is vehicle weight

6 0

2 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

3 years ago