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2 years ago

What happens to the change in the value of the speed as you increased the amount of force applied on your

Physics

1 answer:

nataly862011 [7]2 years ago

3 0

The change of speed is proportional to the Force. If you double the Force, you double the change of speed.

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I NEED a LOT of help PLEASE!

Minchanka [31]

Infrared, visible light, then ultraviolet. Infrared is light that the human eye can not see and visible light is clearly light we can see then ultraviolet is has such a high frequency we can't see it either.

3 0

2 years ago

Read 2 more answers

Four charges are on the four corners of a square. Q1 = +5μC, Q2 = -10μC, Q3 = +5μC, Q4 = -10μC. The side length of the square is

Marat540 [252]

Answer:

Explanation:

Electric field due to a point charge Q at a point at distance d is given by the relation

E = $\frac{K\times Q}{d^2}$

Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.

The charges are situated on the corners of a square in such a way that

equal charges of Q1 and Q3 are situated on the diametrically opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two charges will be zero.

On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero

Overall, net field due to all the four charges will be zero

3 0

3 years ago

group of students is making model cars that will be propelled by model rocket engines. These engines provide a nearly constant t

iris [78.8K]

Answer:

increase.

Explanation:

According to the newton’s second law of motion force is expressed as product of mass and acceleration.

F = m a

If the force acting is constant, then.

m∝ $\frac{1}{a}$

That is if the mass of object increases the acceleration decreases and vice versa. The above equation is used when the force acting on the body is constant.

As the thrust force from the rocket engine is constant throughout there will be a variation in the mass or acceleration.

Thus, it won't stay the same.

As the weight of the car is maximum at the start because of the fuel present in the rocket engine and minimum at the end as the fuel burns throughout the journey of the car. Weight will be minimum at the end and hence acceleration is maximum at the end.

Thus, it won't decrease.

As the acceleration is going from minimum at the start to maximum at the end, therefore it is continuously increases throughout its journey.

Thus, it will increase.

7 0

2 years ago

Create the following matrix H:

sladkih [1.3K]

Answer:

a) {[1.25 1.5 1.75 2.5 2.75]

[35 30 25 20 15] }

b) {[1.5 2 40]

[1.75 3 35]

[2.25 2 25]

[2.75 4 15]}

Explanation:

Matrix H: {[1.25 1.5 1.75 2 2.25 2.5 2.75]

[1 2 3 1 2 3 4]

[45 40 35 30 25 20 15]}

Its always important to get the dimensions of your matrix right. "Roman Columns" is the mental heuristic I use since a matrix is defined by its rows first and then its column such that a 2 X 5 matrix has 2 rows and 5 columns.

Next, it helps in the beginning to think of a matrix as a grid, labeling your rows with letters (A, B, C, ...) and your columns with numbers (1, 2, 3, ...).

For question a, we just want to take the elements A1, A2, A3, A6 and A7 from matrix H and make that the first row of matrix G. And then we will take the elements B3, B4, B5, B6 and B7 from matrix H as our second row in matrix G.

For question b, we will be taking columns from matrix H and making them rows in our matrix K. The second column of H looks like this:

{[1.5]

[2]

[40]}

Transposing this column will make our first row of K look like this:

{[1.5 2 40]}

Repeating for columns 3, 5 and 7 will give us the final matrix K as seen above.

4 0

2 years ago

The period of an oscillating particle is 32 s, and its amplitude is 15 cm. at t = 0, it is at its equilibrium position. find the

morpeh [17]

At t=0, the particle was at its equilibrium position. The time period is 32 seconds, so in 8 seconds, it will reach the extreme location once, and hence, in 8 seconds, it will cover a distance equivalent to its amplitude.

5 0

2 years ago