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GalinKa [24]
3 years ago
13

What is the difference between malleability and ductile ??

Physics
1 answer:
Vitek1552 [10]3 years ago
3 0

Answer:

A malleable material is one in which a thin sheet can be easily formed by hammering or rolling. ... In contrast, ductility is the ability of a solid material to deform under tensile stress.

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two skateboarders of mass 50 kg and 60 kg push each other with force 70N.what is the acceleration of each skaters.step by step.
Keith_Richards [23]

Answer:

f=ma

f=70N

m1=50kg

m2= 60kg

a1= f/m

a1= 70/50

a1= 1.4N/kg

a2= 70/60

a2=1.17N/kg

6 0
3 years ago
In a nuclear reactor 1 g of mass is converted into energy. How much energy in Joules is produced?
KengaRu [80]

Answer:

3 x 10^5  J

Explanation:

mass of substance, m = 1 g = 0.001 kg

Velocity of light, c = 3 x 10^8 m/s

According to the Einstein mass energy equivalence, the energy associated with the mass is given by

E = m c^2

E = 0.001 x 3 x 10^8

E = 3 x 10^5  J

5 0
4 years ago
Pls help me its due today ​
poizon [28]

Answer:

pls give me brainliest

Explanation:

3 0
3 years ago
What is the net force on a car if the force of friction is 15 N and the forward force due to the engine is 20 N?
anygoal [31]
D. 35n forwards....................
8 0
3 years ago
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
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