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7nadin3 [17]
2 years ago
14

a closed systems internal energy changes by 178 j as a result of being heated with 658 j of energy. the energy used to do work b

y the system is j
Physics
1 answer:
Akimi4 [234]2 years ago
7 0
<h3><u>Answer;</u></h3>

= 480 Joules

<h3><u>Explanation;</u></h3>

We use the formula, Q - W = ΔU

Where, Q = Heat transferred to the system

W = Work done by the system

ΔU = Change of internal energy.

As per the question, Q = 658 J

ΔU = 178 J

Thus, W = Q - ΔU = (658 - 178) J = 480 J.

The energy used to do work by the system is 480 J.

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A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.
Pie

#1

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

m = 100 * 11.34 = 1134 g

m = 1.134 kg

so its weight in air will be given as

W = mg = 1.134* 9.8 = 11.11 N

now the buoyant force on the lead is given by

F_B = W - F_{net}

F_B = 11.11 - 11 = 0.11 N

now as we know that

F_B = \rho V g

0.11 = 1000* V * 9.8

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V = 11.22 cm^3

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(iii) Buoyant force = 0.11 N

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#2

buoyant force on the lead block is balancing the weight of it

F_B = W

\rho V g = W

13* 10^3 * V * 9.8 = 11.11

V = 87.2 cm^3

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight =  11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury


#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0
2 years ago
A 2 ft x 2 ft x 2 ft box weighs 100 pounds, and the weight is evenly distributed. What is the magnitude of the minimum horizonta
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Answer:

Explanation:

Let the force required be F . It is applied at the top of the box . The box is likely to turn about a corner . Torque of this force about this corner

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This torque of weight

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F x 2  = 100

F = 50 pounds .

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2 years ago
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If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45. 
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