Encontrar el peso de un elefante cuya masa es de 4500kg. Recordar que la gravedad es de 9.8m/s2

A frog is at the bottom of a 17-foot well. Each time the frog leaps, it moves up 3 feet. If the frog has not reached the top of

docker41 [41]

Answer:

The frog takes 8 jumps to reach top of well

Explanation:

Given data

Frog at bottom=17 foot

Each time frog leaps 3 feet

Frog has not reached the top of the well, then the frog slides back 1 foot

To Find

Total number of leaps the frog needed to escape from well

Solution

in 1 jump distance jumped=3+(-1)

=2 feet

=2×1 feet

The "-1" is because the frog goes back

Now After 2 jumps the distance jumped as:

Distance Jumped=2+2

Distance Jumped=2*2

=4 feet

Similarly after 7 jumps

Distance Jumped=2+2+......+2

Distance Jumped=2*7

=14 feet

Now after 8th jump the frog climbs but doesnot slide back as it is reached to the top of well.

So

Distance Jumped=(Distance Jumped after 7 jumps)+3

=14+3

=17 feet

The frog takes 8 jumps to reach top of well

7 0

3 years ago

Explain how sociology will contribute to the understanding of daily life and what is happening within it?

Helen [10]

Answer:

Explain how sociology will contribute to the understanding of daily life and what is happening within it?

Sociology is the study of society of human being, how they develop and those things put in place for such development

Explanation:

Sociology helps to develop culture, how people interact with one another and this boost the relationship and development as far as human society is concerned.

8 0

3 years ago

Many television sets show 25 images, called 'frames, each second. What is the time interval between one

Firlakuza [10]

Answer:

Given,

Frame rate = 25 frames per second

To find,

Time interval between one frame and the next.

Solution,

We can simply solve this numerical problem by using the following process.

Now,

Number of frames = 25

Total time taken to display the given number of frames (ie. 25 frames) = 1 second

To calculate the time interval between one frame and next, we need to divide the time taken to display total number of frames by total number of frames.

So,

Time interval between one frame and next :

= Time taken to display total number of frames / Total frames

= 1/25

= 0.04 second

Hence, time interval between one frame and next is 0.04 second.

8 0

2 years ago

a 2,000-kilogram railroad car moving at 8m/s to the right collides with a 6,000-kilogram railroad car moving at 3m/s to the west

astra-53 [7]

A freight car of mass 20,000 kg moves along a frictionless level railroad track ... After the push the skateboarder II moves with a velocity of 2 m/s to ... After the collision the cars stick to each other and ... diver jumps with a velocity of 3 m/s in opposite ... A 10 kg object moves at a constant velocity 2 m/s to the right and collides

3 0

3 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago