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2 years ago

A block of density 8.9g/cm³ measures 5cm by 2cm, Given that the force of gravity is 10N/kg. Determine the maximum pressure

Physics

1 answer:

UNO [17]2 years ago

7 0

Answer:

Maximum pressure = 4450 N/m²

Explanation:

We'll begin by calculating the volume of the block. This can be obtained as follow:

Volume (V) = 5 cm × 3 cm × 2 cm

V = 30 cm³

Next, we shall determine the mass of the block. This can be obtained as follow:

Density = 8.9 g/cm³

Volume = 30 cm³

Mass =?

Density = mass /volume

8.9 = mass / 30

Cross multiply

Mass = 8.9 × 30

Mass = 267 g

Next, we shall covert 267 g to Kg.

1000 g = 1 Kg

Therefore,

267 g = 267 g × 1 Kg / 1000

267 g = 0.267 Kg

Thus, the mass of the block is 0.267 Kg

Next, we shall determine the force.

Force of gravity (g) = 10 N/Kg

Mass (m) = 0.267 Kg

Force (F) =?

F = mg

F = 0.267 × 20

F = 2.67 N

Next, we shall determine the minimum area since we are trying to obtain the maximum pressure. This can be obtained as follow:

Minimum area = 3 cm × 2 cm

Covert each measurement to metre (m) by dividing each measurement by 100

Minimum area = 0.03 m × 0.02 m

Minimum area = 0.0006 m²

Finally, we shall determine the maximum pressure. This can be obtained as follow:

Force (F) = 2.67 N

Minimum area = 0.0006 m²

Maximum pressure =?

Maximum pressure = Force /minimum area

Maximum pressure = 2.67 / 0.0006

Maximum pressure = 4450 N/m²

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3 years ago

An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Alexeev081 [22]

Answer:

a) $c=1822.3214\ J.kg^{-1}.K^{-1}$

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

mass of aluminium, $m_a=0.1\ kg$

mass of water, $m_w=0.25\ kg$

initial temperature of the system, $T_i=10^{\circ}C$

mass of copper block, $m_c=0.1\ kg$

temperature of copper block, $T_c=50^{\circ}C$

mass of the other block, $m=0.07\ kg$

temperature of the other block, $T=100^{\circ}C$

final equilibrium temperature, $T_f=20^{\circ}C$

We have,

specific heat of aluminium, $c_a=910\ J.kg^{-1}.K^{-1}$

specific heat of copper, $c_c=390\ J.kg^{-1}.K^{-1}$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

$Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)$

$Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)$

$Q_{in}=11375\ J$

The heat released by the blocks when dipped into water:

$Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)$

where

$c=$ specific heat of the unknown material

For the conservation of energy : $Q_{in}=Q_{out}$

so,

$11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)$

$c=1822.3214\ J.kg^{-1}.K^{-1}$

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

The material is peat, possibly.

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

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3 years ago

Read 2 more answers

Help please! Will give brainliest to first correct answer!

Olegator [25]

Answer:

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