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3 years ago

In a elastic lab what did you change and why?

2 answers:

7 0

I changed my undershorts. The elastic on the old ones I put on that day was deteriorated, and it completely failed when I dripped lab coffee on it, causing falldown.

Setler79 [48]3 years ago

4 0

Answer:

mass to see how it affected stretch length of a rubber band

Explanation:

took the test

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Armand hands the rock to Pierre, who is standing next to the trampoline. Explain how Armand moves, in terms of the forces acting

Nikolay [14]

Answer:

Armando's weight ,restored force created by the trampoline

a harmonic movement within the trampoline

Explanation:

In a trampoline we have two forces that actuate Armando's weight and the restored force created by the trampoline that depends on the deformation distance of the elastic canvas.

Amando's weight is vertical and directed towards the center of the Earth and has a constant value, this weight is balanced with the elastic force the springboard exerts on Armando in a vertical direction.

In general, when entering the trampoline, a small jump is made, this creates a speed that deforms the canvas until the speed is reduced to zero, at this point the elastic force is greater than the weight and the boy begins to climb, After the boy leaves the canvas he meets only the force of gravity and his speed decreases to zero and begins his fall.

In Summary Armando is in a harmonic movement within the trampoline

6 0

3 years ago

Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s

ozzi

Answer:

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe ..... (1)

T Cosθ = mg .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

$tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}$

$tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}$

$tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}$

$tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}$

As θ is very small, so tanθ and Sinθ is equal to θ.

$\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}$

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

7 0

2 years ago

A resistor of 5 Ω is placed in a circuit. The voltage drop across the resistor is 12 V. What is the current through the resistor

Kaylis [27]

Data:

R = 5 Ω
U = 12 V
i = ?

Formula:

 $U = R*i$ → $i = \frac{U}{R}$ 

Solving:

 $i = \frac{U}{R}$
$i = \frac{12}{5}$
$\boxed{i = 2.4A}$

Answer:

The current through the resistor is 2.4 Amperes

5 0

2 years ago

What is the kinetic energy of a an 80kg football player running at 8 m/s?

Ugo [173]

In the question it is already given that the football player is 80 kg.
Then the mass of the football player = 80 kg
Velocity at which the football player is running = 8 m/s
Kinetic Energy = 0.5 • mass • square of velocity
Now we have to put the known data in this equation to find the actual velocity of the footballer.
 So
Kinetic Energy of the footballer = 0.5 * 80 * (8 * 8)
= 0.5 * 80 * 64
= 2560
So the Kinetic energy of the footballer is 2560 joules

4 0

3 years ago

True or false—If a rock is thrown into the air, the increase in the height would increase the rock’s kinetic energy, and then th

nlexa [21]

I think that this is false but I am not sure

5 0

2 years ago