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aniked [119]
3 years ago
14

What is the equation for finding the acceleration of an object moving in a straight line?

Physics
1 answer:
kiruha [24]3 years ago
4 0

Acceleration can be found using one of the following suvat equations:

v=u+at\\s=ut+\frac{1}{2}at^2\\s=vt-\frac{1}{2}at^2\\v^2-u^2=2as

Explanation:

The acceleration of an object is defined as the rate of change of velocity of the object:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time taken for the velocity to change from u to t

There are several equations used to find the acceleration of an object moving in a straight line (they are usually called suvat equation). Depending on which quantities are given in the problem, you can use one of the following equations:

v=u+at\\s=ut+\frac{1}{2}at^2\\s=vt-\frac{1}{2}at^2\\v^2-u^2=2as

where

a is the acceleration

s is the distance covered

t is the time

u is the initial velocity

v is the final velocity

Learn more about acceleration and suvat equations:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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What effect does a catalyst have on a system in equilibrium?
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ANSWER:

What effect does a catalyst have on a system in equilibrium?

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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}

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.

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