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3 years ago

What is heat of combustion?

1 answer:

4 0

The heat of combustion is the total energy released as heat when a substance undergoes complete combustion with oxygen under standard conditions. The chemical reaction is typically a hydrocarbon or other organic molecule reacting with oxygen to form carbon dioxide and water and release heat.

According to Wikipedia :)

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A toy cart at the end of a string 0.70 m long moves in a circle on a table. The cart has a mass of 2.0 kg and the string has a b

luda_lava [24]

Given that the mass of the toy cart is 2.0 kg and and the acceleration is unknown, the normal formula would be a=f/m where a is acceleration, f is force and m is mass but the string's breaking strength is 40n so I think the formula in this case will be f is greater than m*a
40 is greater than 2a
40 is greater than 2a
40/2 is greater than 2a/2
20m/s² is greater than a
Therefore the maximum speed the toy cart should have should be less than 20m/s²

8 0

4 years ago

If a an object is traveling at a rate of 5 meters per second squared with a force of 10 Newtons, what is the mass of the object?

Svetllana [295]

Answer:

2 kg

Explanation:

Acceleration = 5 m/s^2

Force = 10 N

Force = mass * acceleration

mass = force / acceleration

mass = 10 / 5

mass = 2 kg

8 0

3 years ago

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$

and re-arranging the equation, we find

$v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s$

7 0

3 years ago

How much work is done on a 75 newton bowling ball when you carry it horizontally across a 10 meter room

svlad2 [7]

F = force applied to hold the weight of the bowling ball = weight of the bowling ball = 75 N

d = distance through which the bowling ball is moved horizontally = 10 meter

θ = angle between the force in vertically upward direction and displacement in horizontal direction = 90

W = work done on the bowling ball

work done on the bowling ball is given as

W = F d Cosθ

inserting the values

W = (75) (10) Cos90

W = (75) (10) (0)

W = 0 J

6 0

3 years ago

Question 3 of 10

Mazyrski [523]

Answer:

B . energy cannot be created or destroyed

3 0

3 years ago