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3 years ago

9

Find the energy turned into heat.

1 answer:

SOVA2 [1]3 years ago

3 0

To calculate the amount of heat released in a chemical reaction, use the equation Q = mc ΔT, where Q is the heat energy transferred (in joules), m is the mass of the liquid being heated (in kilograms), c is the specific heat capacity of the liquid (joule per kilogram degrees Celsius), and ΔT is the change in ...

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The formula of force is newton and

Arlecino [84]

Answer:

The formula for calculating force is F= Mass × Acceleration

Explanation:

Newton is what force is measured in

8 0

3 years ago

A horizontal 2.00\ m2.00 m long, 5.00\ kg5.00 kg uniform beam that lies along the east-west direction is acted on by two forces.

Sunny_sXe [5.5K]

Answer: $240\ rad/s^2$

Explanation:

Given

Length of beam $l=2\ m$

mass of beam $m=5\ kg$

Two forces of equal intensity acted in the opposite direction, therefore, they create a torque of magnitude

$\tau =F\times l=200\times 2=400\ N.m$

Also, the beam starts rotating about its center

So, the moment of inertia of the beam is

$I=\dfrac{ml^2}{12}=\dfrac{5\times 2^2}{12}\\\\I=\dfrac{5}{3}\ kg.m^2$

Torque is the product of moment of inertia and angular acceleration

$\Rightarrow \tau=I\alpha\\\\\Rightarrow 400=\dfrac{5}{3}\times \alpha\\\\\Rightarrow \alpha =240\ rad/s^2$

7 0

3 years ago

Which of the following processes charges an object by friction

Olenka [21]

An example is when u rub your pen on your hair hard that is friction

7 0

4 years ago

Read 2 more answers

A car moves 20 km towards the North and then 35 km at an angle of 60o towards west of North. Its magnitude of displacement from

Law Incorporation [45]

Answer:

15

Explanation:

displacement = initial position - final position

8 0

3 years ago

An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Dete

dezoksy [38]

Answer:43.34 m

Explanation:

Given

acceleration(a) $=2 m/s^2$

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

$v=0+2\times 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+\frac{at^2}{2}$

$s=0+\frac{2\times 6^2}{2}$

s=36 m

After fuel run out distance traveled in upward direction is

$v^2-u^2=2as_0$

here v=0

u=12 m/s

$a=9.8 m/s^2$

$0-12^2=2(-9.8)(s)$

$s_0=\frac{144}{2\times 9.8}=7.34 m$

$s+s_0=36+7.34=43.34 m$

7 0

3 years ago