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frozen [14]
3 years ago
5

A 50.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the

cord is 13.0 m. The jumper reaches reaches the bottom of her motion 40.0 m below the bridge before bouncing back. We wish to find the time interval between her leaving the bridge and her arriving at the bottom of her motion. Her overall motion can be separated into an 13.0-m free-fall and a 27.0-m section of simple harmonic oscillation.] (a) For the free-fall part, what is the appropriate analysis model to describe her motion.
particle under constant acceleration particle in simple harmonic motion particle under constant angular acceleration
Physics
1 answer:
lina2011 [118]3 years ago
8 0

Answer:

The the analysis for the free fall part should be done under the constant acceleration.

Explanation:

In the given problem, the jumper is falling under the free fall. Since, no external force is acting on the body therefore, the fall will be under the action gravity only. also, the acceleration due to gravity is always constant.

Therefore, the the analysis for the free fall part should be done under the constant acceleration.

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Light does not pass through some materials. What do you think happens
Hatshy [7]

Answer:

When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster, (it has just been "hit" by a rapidly moving photon)

A photon is a quantum of EM radiation. Its energy is given by E = hf and is related to the frequency f and wavelength λ of the radiation by. E=hf=hcλ(energy of a photon) E = h f = h c λ (energy of a photon) , where E is the energy of a single photon and c is the speed of light.

7 0
2 years ago
A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

6 0
3 years ago
A box weighs 100N and its base area of 2 m2. What pressure does it exert on the ground?
kotegsom [21]

Answer:

P = F/S = 100/2 =50 (N/m2)

5 0
3 years ago
24. An elevator is moving vertically up with an acceleration a. The force exerted on the floor by a passenger of mass m is
Wewaii [24]

<u>Given </u><u>:</u><u>-</u>

  • An elevator is moving vertically up with an acceleration a.

<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>

  • The force exerted on the floor by a passenger of mass m .

<u>Solution</u><u> </u><u>:</u><u>-</u>

As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .

  • The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .

For the FBD refer to the attachment . From that ,

\implies Weight_{apparent}= mg + ma

<u>Hence</u><u> </u><u>option</u><u> </u><u>d </u><u>is </u><u>correct</u><u> </u><u>choice </u><u>.</u>

<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>

3 0
2 years ago
Find the force between 2C and -1C separated by a distance 1m. Indicate if it is an attractive force (negative) or a repulsive fo
telo118 [61]

Answer:

Explanation:

Its definitely an Attractive force since the two charges are Unlike.

From Coulombs Law

F=kq1q2/R²

Given

K=9x10^9

R=1m

q1=2C

q2=-1C

F=(9x10^9 x 2 x -1)/1²

F= - 1.8x10^10N. (Attractive).

6 0
3 years ago
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