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klemol [59]
3 years ago
7

The mass of a cannon ball is 10 kg. If the speed of the cannon ball is 50 m/s in

Physics
1 answer:
Maurinko [17]3 years ago
7 0

Answer:

Explanation:

Kinetic Energy = 0.5(Mass)(Velocity2)

Kinetic energy= 0.5 × 10kg × (50m/s)2

Kinetic Energy = 5kg × 2500m/s

Kinetic energy = 125000 J ( Ans)

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Why can light travel through outer space, but sound cannot?
WINSTONCH [101]
Usually, sound needs a medium to travel through, like a vacuum for example. 

Light does not need a medium to travel, and since air is considered a medium, light is not dependent on that. 

On the other hand, sound needs a medium to travel through, and outer space doesn't contain space, therefore, no sound. 
5 0
4 years ago
Read 2 more answers
Chris and Jamie are carrying Wayne on a horizontal stretcher. The uniform stretcher is 2.00 m long and weighs 100 N. Wayne weigh
BabaBlast [244]

Answer:

<h2>E. 650N</h2>

Explanation:

step one:

given

length of stretcher= 2m

weight of stretcher=100N

Wayne weighs =800N

distance of Wayne weighs from chris's end= 75cm= 0.75m

The force that Chris is exerting to support the stretcher, with Wayne on it, can be computed by taking moments of the weight of the stretcher and Wayne weighs  about Chris's end, the end result is the reaction at Chris's end

Taking moment about Chris's end

The moment of Wayne weight 75cm from Chris+ Half the weight of stretcher 1m from Chris

0.75*800+50*1=0

600+50=0

650N

6 0
3 years ago
When mapping the equipotentials on the plates with different electrode configurations you may find that some have significant ar
Olenka [21]

Answer:

v = 10 V and E = 2 10³ N/C

Explanation:

The electrical potentials and the electric field at one point are related by the expression

            ΔV = - ∫ E. dS

Where the bold indicates vector quantities, E is the electric field and S is the line of displacement of the load, in general displacement is perpendicular to the equipotential lines, which reduces the product scales to the ordinary product.

 If the potential difference is the most usual that is V = 10 V, the electric field is

   s = 0.5 cm = 0.5 10⁻² m

                E = ΔV / S

                E = 10/0.5 10⁻²

                 E = 2 10³ N / C

4 0
4 years ago
Consider a positive charge Q and a point B twice as far away from Q as point A. What is the ratio of the electric field strength
Vikentia [17]

Answer:

\frac{E_{A}}{E_{B}}=4

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

E=\frac{F}{q}.

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

F=\frac{kQq}{r^{2}}.

By substitution we get that

E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}

Now, letting E_{A} be the electric field at point A, letting E_{B} be the electric field at point B, and letting R be the distance from the charge to A:

E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}.

The ration of the electric fields is

\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4

This means that at half the distance, the electric field is four times stronger.

4 0
3 years ago
Khalid has been studying the gravitational attraction between three pairs of objects. The table shows the distance between each
SCORPION-xisa [38]

Answer:

Explanation:

Probably the most famous force of all is gravity. We humans on earth think of gravity as an apple hitting Isaac Newton on the head. Gravity means that stuff falls down. But this is only our experience of gravity. In truth, just as the earth pulls the apple towards it due to a gravitational force, the apple pulls the earth as well. The thing is, the earth is just so massive that it overwhelms all the gravity interactions of every other object on the planet. Every object with mass exerts a gravitational force on every other object. And there is a formula for calculating the strengths of these forces, as depicted in the diagram below:

Diagram of gravitational forces between two spheres

Diagram of gravitational forces between two spheres

Let’s examine this formula a bit more closely.

F refers to the gravitational force, the vector we ultimately want to compute and pass into our applyForce() function.

G is the universal gravitational constant, which in our world equals 6.67428 x 10^-11 meters cubed per kilogram per second squared. This is a pretty important number if your name is Isaac Newton or Albert Einstein. It’s not an important number if you are a ProcessingJS programmer. Again, it’s a constant that we can use to make the forces in our world weaker or stronger. Just making it equal to one and ignoring it isn’t such a terrible choice either.

m_1m  

1

​  

m, start subscript, 1, end subscript and m_2m  

2

​  

m, start subscript, 2, end subscript are the masses of objects 1 and 2. As we saw with Newton’s second law (\vec{F} = M\vec{A}  

F

=M  

A

F, with, vector, on top, equals, M, A, with, vector, on top), mass is also something we could choose to ignore. After all, shapes drawn on the screen don’t actually have a physical mass. However, if we keep these values, we can create more interesting simulations in which “bigger” objects exert a stronger gravitational force than smaller ones.

\hat{r}  

r

^

r, with, hat, on top refers to the unit vector pointing from object 1 to object 2. As we’ll see in a moment, we can compute this direction vector by subtracting the location of one object from the other.

r^2r  

2

r, squared refers to the distance between the two objects squared. Let’s take a moment to think about this a bit more. With everything on the top of the formula—G, m_1m  

1

​  

m, start subscript, 1, end subscript, m_2m  

2

​  

m, start subscript, 2, end subscript—the bigger its value, the stronger the force. Big mass, big force. Big G, big force. Now, when we divide by something, we have the opposite. The strength of the force is inversely proportional to the distance squared. The farther away an object is, the weaker the force; the closer, the stronger.

Hopefully by now the formula makes some sense to us. We’ve looked at a diagram and dissected the individual components of the formula. Now it’s time to figure out how we translate the math into ProcessingJS code. Let’s make the following assumptions.

We have two objects, and:

Each object has a PVector location: location1 and location2.

Each object has a numeric mass: mass1 and mass2.

There is a numeric variable G for the universal gravitational constant.

Given these assumptions, we want to compute a PVector force, the force of gravity. We’ll do it in two parts. First, we’ll compute the direction of the force \hat{r}  

r

^

r, with, hat, on top in the formula above. Second, we’ll calculate the strength of the force according to the masses and distance.

Remember when we figured out how to have an object accelerate towards the mouse? We're going to use the same logic.

4 0
3 years ago
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