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Darina [25.2K]
2 years ago
10

Sebanyak 80g emas bersuhu 30°C diberi kalor sebesar 1512 J.Jika kalor jenis emas 126 J/kg K,suhu akhir emas setelah diberi kalor

sebesar
Physics
1 answer:
Leno4ka [110]2 years ago
8 0

Answer:

180 ° C

Explanation:

Dari pertanyaan yang diberikan di atas, diperoleh data sebagai berikut:

Massa (M) emas = 80 g

Suhu awal (T1) = 30 ° C

Kalor (Q) = 1512 J.

Kapasitas kalor jenis (C) = 126 J / kgK Suhu akhir (T2) =?

Selanjutnya, kita akan mengubah 80g menjadi kilogram (Kg). Ini dapat diperoleh sebagai berikut:

1000 g = 1 Kg

Karena itu,

80 g = 80 g × 1 Kg / 1000 g

80 g = 0,08 Kg

Jadi, 80 g setara dengan 0,08 Kg

Selanjutnya, kita akan mengubah suhu 30 ° C ke Kelvin. Ini dapat diperoleh sebagai berikut:

T (K) = T (° C) + 273

Suhu awal (T1) = 30 ° C

Suhu awal (T1) = 30 ° C + 273 = 303 K.

Selanjutnya, kami akan menentukan suhu akhir emas. Ini dapat diperoleh sebagai berikut:

Massa (M) emas = 0,08 Kg

Suhu awal (T1) = 303 K

Kalor (Q) = 1512 J.

Kapasitas kalor jenis (C) = 126 J / kgK Suhu akhir (T2) =?

Q = MC (T2 - R1)

1512 = 0,08 × 126 (T2 - 303)

1512 = 10,08 × (T2 - 303)

Bagilah kedua sisinya dengan 10,08 1512 / 10,08 = (T2 - 303)

150 = T2 - 303

Kumpulkan istilah serupa

150 + 303 = T2

T2 = 453 K.

Akhirnya, kita akan mengubah suhu 453 K menjadi celsius. Ini dapat diperoleh sebagai berikut:

T (° C) = T (K) - 273

T2 = 453 K.

T (° C) = 453 - 273

T (° C) = 180 ° C

Oleh karena itu, temperatur akhir emas setelah dipanaskan adalah 180 ° C

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B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 - 14.7t - 4.9t^2

4.9t^2 + 14.7t - 19.6 = 0

t^2 + 3t - 4 = 0

(t + 4)(t - 1) = 0

(t - 1) = 0

\boxed {t = 1 ~ second}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 + 14.7t - 4.9t^2

4.9t^2 - 14.7t - 19.6 = 0

t^2 - 3t - 4 = 0

(t - 4)(t + 1) = 0

(t - 4) = 0

\boxed {t = 4 ~ seconds}

The difference in the two ball's time in the air is:

\Delta t = 4 ~ seconds - 1 ~ second

\large {\boxed {\Delta t = 3 ~ seconds} }

<h2>Question B:</h2><h3>First Ball</h3>

v^2 = u^2 - 2gH

v^2 = (-14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

<h3>Second Ball</h3>

v^2 = u^2 - 2gH

v^2 = (14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 11.025 ~ m}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 25.725 ~ m}

The difference in the two ball's height after 0.500 s is:

\Delta h = 25.725 ~ m - 11.025 ~ m

\large {\boxed {\Delta h = 14.7 ~ m} }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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