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s344n2d4d5 [400]

2 years ago

14

A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And

oil are 0.5kj/kg.k and 3.5kj/kg.k calculate the entropy change in steel,oil and in the universe.

1 answer:

Naily [24]2 years ago

8 0

Answer:

700J

Explanation:

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A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it

vredina [299]

Answer:

The distance is 11 m.

Explanation:

Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

We need to calculate the acceleration

Using newton's second law

$F = ma$ ...(I)

Using formula of friction force

$F= \mu m g$ ....(II)

Put the value of F in the equation (II) from equation (I)

$ma=\mu mg$ ....(III)

$a = \mu g$

Put the value in the equation (III)

$a=0.24\times9.8$

$a=2.352\ m/s^2$

We need to calculate the distance,

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}2.352\times(3.0)^2$

$s=10.584\ m\ approx\ 11\ m$

Hence, The distance is 11 m.

3 0

2 years ago

Consider a force of 750 n (roughly the weight of an adult human). over what area (in cm2) would this force need to be applied in

polet [3.4K]

$p = \frac{f}{a}$
P=25x10^6 andF=750.So plug in everything to solve for A. which is 3x10^-5m^2 OR 0.3mm^2

7 0

2 years ago

An electric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an e

Oksi-84 [34.3K]

Answer:

0.69 ohm

Explanation:

Heat generated per second, H = 50 cal/s

Potential difference, V = 12 V

Let R is the resistance of coil.

The formula for the heat is given by

$H = \frac{V^{2}}{R}t$

$50\times 4.186 = \frac{12^{2}}{R}\times 1$

R = 0.69 ohm

3 0

2 years ago

A 5.0-kilogram box is sliding across a level floor. The box is acted upon by a force of 27 newtons east and a frictional force o

marshall27 [118]

Answer:

The magnitude of the acceleration of the box is 2 m/s².

Explanation:

Given:

Mass of the box, $m=5.0$ kg

Force acting towards east, $F=27$ N

Frictional force acting towards west, $f=17$ N

Let the acceleration be $a$ m/s².

Now, net force acting on the box towards east is given as:

$F_{net}=F-f=27-17=10\textrm{ N}$

From Newton's second law of motion,

$F_{net}=ma\\10=5.0\times a\\a=\frac{10}{5.0}=2\textrm{ }m/s^2$

Therefore, the magnitude of the acceleration of the box is 2 m/s².

6 0

2 years ago

Read 2 more answers

A power station burns 75 kilograms of coal per second. Each kg of coal contains 27 million joules of energy.

Kaylis [27]

Answer:

Explanation:

a )

one kg of coal gives energy of 27 x 10⁶ J

75 kg of coal gives energy of 27 x 10⁶ x 75 J

So rate which energy is coming out of coal per second

= 27 x 10⁶ x 75 J

= 2025 x 10⁶ J /s

2025 million watts .

b ) energy output = 800 million watts

efficiency = (800 / 2025) x 100

= 39.5 % .

3 0

2 years ago