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AleksAgata [21]
3 years ago
14

1.80 kJ of heat is added to a slug of gold and a separate 1.80 kJ of heat is added to a slug of manganese. The heat capacity of

the gold slug is 149J/°C while the heat capacity of the manganese slug is 505 J/°C. If the slugs are each originally at 25.00°C, what is the final temperature of each slug?
Physics
1 answer:
Svetlanka [38]3 years ago
7 0

Answer:

final temperature of  slug of gold is 37°C

final temperature of  slug of manganese is 28.56°C

Explanation:

Hello! To solve this problem we must take into account the concept of heat capacity, this is defined as the ratio between energy and temperature rise.

In other words it is the amount of energy that is required to increase a temperature degree.

Taking into account the above we infer the following equation

where  

C=heat capacity

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Water equal to iron is greater than cooper
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A shopper does 157 J of work pushing a cart with 10.9 N force
Tanzania [10]

The cart travelled a distance of 14.4 m

Explanation:

The work done by a force when pushing an object is given by:

W=Fd cos \theta

where:

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this problem we have:

W = 157 J is the work done on the cart

F = 10.9 N is the magnitude of the force

\theta=0^{\circ}, assuming the force is applied parallel to the motion of the cart

Therefore we can solve for d to find the distance travelled by the cart:

d=\frac{W}{F cos \theta}=\frac{157}{(10.9)(cos 0)}=14.4 m

Learn more about work:

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#LearnwithBrainly

4 0
3 years ago
While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.95 6.95 m/s. The st
qwelly [4]

Answer:

18.1347 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-6.95^2}{2\times -9.81}\\\Rightarrow s=2.4619\ m

Total height the ball falls is 2.4619+14.3 = 16.7619 m

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 16.7619+0^2}\\\Rightarrow v=18.1347\ m/s

The speed at which the stone reaches the ground is 18.1347 m/s

8 0
3 years ago
You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f
Verdich [7]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,

u = 0m/s

Also, it is given that the acorn takes 0.201s to pass the length of the meter stick.

s = ut+\frac{1}{2} at^2

Replacing,

1 = u(0.141)+ \frac{1}{2} (9.8)(0.141)^2

u =6.4013m/s

The height of the acorn above the meter stick can be calculated as,

v^2 = u^2 +2gh

h = \frac{v^2-u^2}{2g}

h = \frac{6.4013^2-0^2}{2(9.8)}

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Also the top of the meter stick is 1.87m above the ground hence the height of the acorn above the ground is

h = 2.0906+1.87

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When a puddle dries up what are the particles really doing
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The particles are either being absorbed or evaporating
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