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WITCHER [35]
3 years ago
13

the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg​

Physics
1 answer:
lukranit [14]3 years ago
3 0

Answer:

Mass of the steel cube =  7800 kg

Volume of the steel  = 1.025 cubic centimetre

Explanation:

Given:

The density of the steel   =  7.8

Side of the cube = 12 cm

<u>(1)The mass of steel cube :</u>

We know that,

Density  = \frac{Mass}{Volume}

We are given with density and sides of the cube

then volume of the cube

= (side)^3

= 10^3

= 1000 cubic centimetre

Now

7.8 = \frac{mass}{1000}

mass = 7.8 \times 1000

mass =  7800 kg

<u>(2)volume of steel:</u>

Given the mass  = 8 kg

Density  = \frac{Mass}{Volume}

Substituting the values

7.8 = \frac{8}{volume}

volume = \frac{8}{7.8}

volume = 1.025 cubic centimetre

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A 440-g cylinder of brass is heated to 97.0 degree Celsius and placed in a 2 points
nydimaria [60]

Answer:

Specific heat of brass is 0.40 J g⁻¹ °C⁻¹ .

Explanation:

Given :

Mass of brass, m₁ = 440 g

Temperature of brass, T₁ = 97° C

Mass of water, m₂ = 350 g

Temperature of water, T₂ = 23° C

Specific heat of water, C₂ = 4.18 J g⁻¹ °C⁻¹

Equilibrium temperature, T = 31° C

Let C₁ be the specific heat of brass.

Heat loss by brass = Heat gain by water

m₁ x C₁ x ( T₁ -T ) = m₂ x C₂ x ( T - T₁ )

Substitute the suitable values in above equation.

440 x C₁ x (97 - 31) = 350 x 4.18 x (31 - 23)

C₁ = \frac{11704}{29040}

C₁ = 0.40 J g⁻¹ °C⁻¹

4 0
3 years ago
A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Tems11 [23]

Answer:

a) V_{wf} = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

m_{b}V{b_{i} }  + V_{wi}  = m_{w} V_{wf} + m_{b}V_{bf}

where m_{b},V{b_{i} are the mass and the initial velocity of the bullet, m_{w} and V_{wi} are the mass and the initial velocity of the wooden block, and V_{wf} and V_{bf} are the final velocities of the wooden block and the bullet

The wooden block is initial at rest (V_{wi} = 0) this yields

m_{b}V{b_{i} }  = m_{w} V_{wf} + m_{b}V_{bf}

By solving for V_{wf} adn substitute the givens

V_{wf} = \frac{m_{b}V_{bi} - m_{b} V_{bf}    }{m_{W} }

= \frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}

V_{wf} = 4.67m/s

b) The center of mass speed is defined as

V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}

substituting:

V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)

V = 8.29 m/s

7 0
3 years ago
2. What is one reason why the author includes information about religious beliefs during the time of Johannes Fabricius' discove
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A 0.500 kg rock is whirled in a vertical circle of a radius 0.60 m . the velocity of the rock at the bottom of the swing is 4.0
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Explanation:

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a = (4.0 m/s)² / 0.60 m

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