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3 years ago

What energy transformation takes place when you stretch a bungee cord?

Physics

1 answer:

Rzqust [24]3 years ago

4 0

Answer:

<h3>D.</h3><h2>Potential energy transforms into elastic energy.</h2>

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Light from two lasers is incident on an opaque barrier with a single slit of width 4.0 x 10^-4 m. One laser emits light of wavel

Sergio [31]

Answer:

a) y = 2.4 x 10⁻³ m = 0.24 cm

b) y = 3.2 x 10⁻³ m = 0.32 cm

Explanation:

The formula of Young's Double Slit experiment will be used here:

$y = \frac{\lambda L}{d}\\\\$

where,

y = distance between dark spots = ?

λ = wavelength

L = distance of screen = 2 m

d = slit width = 4 x 10⁻⁴ m

a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:

$y = \frac{(4.8\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}$

y = 2.4 x 10⁻³ m = 0.24 cm

a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:

$y = \frac{(6.4\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}$

y = 3.2 x 10⁻³ m = 0.32 cm

7 0

3 years ago

Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram

Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

Given the following data:

Force 1 = 500 Newton

Distance 1 = 1.2 meter

Force 2 = 500 Newton

Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

$Work\;done = Force \times distance$

For student 1:

$Work\;done = 500 \times 1.2$

Work done = 600 Joules

For student 2:

$Work\;done = 500 \times 5$

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0

3 years ago

Read 2 more answers

Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric

Goshia [24]

Complete Question:

Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:

A. 2F

B. F/4

C. F/2

D. F

E. 4F

Answer:

Explanation:

If both spheres can be treated as point charges, they must obey the Coulomb's law, that can be written as follows (in magnitude):

$F =\frac{kQ*2Q}{r^{2} }$

As it can be seen, this force is proportional to the product of the charges, so it must be the same for both charges.

As this force obeys also the Newton's 3rd Law, we conclude that the magnitude of the electric force on sphere A due to sphere B, must be equal to the the magnitude of the force on the sphere B due to the sphere A, i.e., just F.

3 0

3 years ago

What fraction of all the electrons in a 25 mg water

mihalych1998 [28]

Answer:

$9.11\times 10^{-15}.$

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of -40 nC is removed from the water droplet.

The charge on one electron, $\rm e=-1.6\times 10^{-19}\ C.$

Let the N number of electrons have charge -40 nC, such that,

$\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.$

Now, mass of one electron = $\rm 9.11\times 10^{-31}\ kg.$

Therefore, mass of N electrons = $\rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.$

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is $25\ \rm mg = 25\times 10^{-6}\ kg.$

Then,

$\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.$

It is the required fraction of mass of the droplet.

3 0

3 years ago

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land

Marina CMI [18]

Answer:

(a) $\vec{J}=176.4\frac{kg*m}{s} \hat{j}$

(b) $F=2125.30N$

Explanation:

(a) According to the law of conservation of energy, the potential energy of the person at 0.40 m is equal to its kinetic energy before the colision with the floor:

$\Delta U=\Delta K\\mgh=\frac{mv^2}{2}\\v=\sqrt{2gh}\\v=\sqrt{2(9.8\frac{m}{s^2})(0.40m)}\\v=2.8\frac{m}{s}$

This is the initial velocity in the negative y-direction. Impulse is given by:

$\vec{J}=\Delta \vec{p}\\\vec{J}=m\vec{v_f}-m\vec{v_i}\\\vec{J}=63kg(0 \hat{j})-63kg(-2.8\frac{m}{s} \hat{j})\\\vec{J}=176.4\frac{kg*m}{s} \hat{j}$

(b) The average force is:

$F=\frac{J}{\Delta t}\\F=\frac{176.4\frac{kg*m}{s}}{0.083s}\\F=2125.30N$

6 0

3 years ago