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3 years ago

Calculate the heat energy needed to change the temperature of 2 kg of copper from 10C to 110C

Physics

1 answer:

Alina [70]3 years ago

7 0

Q = m c ΔT

Q = (2)(385)(110 - 10)

Q = 77000 Joules

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You run a race with your friend. At first you each have the same kinetic energy, but then you find that she is beating you. When

Blizzard [7]

Answer:

52.49 Kg

Explanation:

Let m1 and v1 denote your mass and velocity respectively

Let m2 and v2 denote your friends mass and velocity respectively

Kinetic energy is given by

$KE= 0.5mv^{2}$

Since your kinetic energies are the same hence

$0.5m1(v1)^{2}=0.5m2(v2)^{2}$

$m1(v1)^{2}=m2(v2)^{2}$ and making m2 the subject then

$m2=\frac { m1(v1)^{2}}{(v2)^{2}}$

Since v2 is v1+0.28v1=1.28v1

Substituting m1 for 86 Kg

$m2=\frac { 86 Kg(v1)^{2}}{(1.28v1)^{2}}= 52.49023\approx 52.49 Kg$

3 0

4 years ago

How many grams are in 6.53 moles of Mn?

Westkost [7]

Answer:

359 g Mn

General Formulas and Concepts:

Dimensional Analysis
Reading the Periodic Table of Elements

Explanation:

Step 1: Define

6.53 mol Mn

Step 2: Find conversion

1 mol Mn = 54.94 g Mn

Step 3: Dimensional Analysis

$6.53 \hspace{3} mol \hspace{3} Mn(\frac{54.94 \hspace{3} g \hspace{3} Mn}{1 \hspace{3} mol \hspace{3} Mn} )$ = 358.758 g Mn

Step 4: Simplify

We are given 3 sig figs.

358.758 g Mn ≈ 359 g Mn

3 0

3 years ago

3. What is the gravitational force between a 70 kg physics student and her 1 kg textbook, at a distance of 1 meter? (This number

Rina8888 [55]

ANSWER

$\begin{equation*} 4.67*10^{-9}\text{ }N \end{equation*}$

EXPLANATION

Parameters given:

Mass of the student, M = 70 kg

Mass of the textbook, m = 1 kg

Distance, r = 1 m

To find the gravitational force acting between the student and the textbook, apply the formula for gravitational force:

$F=\frac{GMm}{r^2}$

where G = gravitational constant

Therefore, the gravitational force acting between the student and the textbook is:

$\begin{gathered} F=\frac{6.67430*10^{-11}*70*1}{1^2} \\ \\ F=4.67*10^{-9}\text{ }N \end{gathered}$

That is the answer.

6 0

1 year ago

What is the kinetic energy of a 150 kg bear running at 3 m/s?

OverLord2011 [107]

Explanation:

KE = ½ mv²

KE = ½ (150 kg) (3 m/s)²

KE = 675 J

5 0

3 years ago

You have a pulley 10.4 cm in diameter and with a mass of 2.3 kg. You get to wondering whether the pulley is uniform. That is, is

madreJ [45]

Answer:

Explanation:

Given

Diameter of Pulley=10.4 cm

mass of Pulley(m)=2.3 kg

mass of book $(m_0)=1.7 kg$

height(h)=1 m

time taken=0.64 s

$h=ut+frac{at^2}{2}$

$1=0+\frac{a(0.64)^2}{2}$

$a=4.88 m/s^2and [tex]a=\alpha r$

where $\alpha$ is angular acceleration of pulley

$4.88=\alpha \times 5.2\times 10^{-2}$

$\alpha =93.84 rad/s^2$

And Tension in Rope

$T=m(g-a)$

$T=1.7\times (9.8-4.88)$

T=8.364 N

and Tension will provide Torque

$T\times r=I\cdot \alpha$

$8.364\times 5.2\times 10^{-2}=I\times 93.84$

$I=0.463\times 10^{-2} kg-m^2$

$I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2$

Thus mass is uniformly distributed or some more towards periphery of Pulley

4 0

3 years ago

Calculate the heat energy needed to change the temperature of 2 kg of copper from 10C to 110C ​

Calculate the heat energy needed to change the temperature of 2 kg of copper from 10C to 110C