If you wanna make it easier you can number them going down 1,2,3, or 4 & just tell me the number

What is the Kinetic Energy of a tennis ball after 40.0cm of freefall?

OlgaM077 [116]

The kinetic energy is the same as the potential energy of raising it 40cm (0.4m). That's mgh where m is mass of ball. Its then 3.924*m, whatever m is equal to in kg.

3 0

3 years ago

Put the following events in the order they occurred to lead to the formation of the solar system:

AfilCa [17]

Answer:

Third → First → Second → Fourth

Explanation:

Initially helium and Hydrogen in abundance started to form atoms

Following the atoms formation by hydrogen and helium bombardment of new matter into the Earth took place.

The bombardment of matter resulted in separation of Earth into respective layers based on the weight of the matter.

A primitive ocean was created as a result of off gassing of the volcanoes.

8 0

2 years ago

Cual es el deporte que le da fortaleza y flexibilidad al cuerpo

AlexFokin [52]

Answer:

Aesthetic sports

Explanation:

Aesthetic sports are the one's that need well-developed physical qualities such as strength, agility, stamina, flexibility, and technical knowledge and artistry, in addition to technical ability and artistry. Elite athletes in these sports generally have a low abdominal fat , and the ranking is subjective.

In aesthetic sports like gymnastics, swimming, and figure skaters, dynamic and proactive flexibility is required.

4 0

2 years ago

A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i

Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

$y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]$ (1)

n = 4

Now,

We know the standard eqn is given by:

$y = AsinKxcos\omega t$ (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = $14.4 m^{- 1}$

$\omega = 166\ rad/s$

where

A = Amplitude

K = Propagation constant

$\omega$ = angular velocity

Now, to calculate the string's wavelength,

(a) $K = \frac{2\pi}{\lambda}$

where

K = propagation vector

$\lambda = \frac{2\pi}{K}$

$\lambda = \frac{2\pi}{14.4} = 0.436\ m$

(b) The length of the string is given by:

$l = \frac{n\lambda}{2}$

$l = \frac{4\times 0.436}{2} = 0.872\ m$

(c) Now, we first find the frequency of the wave:

$\omega = 2\pi f$

$f = \frac{\omega}{2\pi}$

$f = \frac{2\pi}{166} = 26.42\ Hz$

Now,

Speed of the wave is given by:

$v = f\lambda$

$v = 26.419\times 0.436 = 11.518\ m/s$

4 0

2 years ago

1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc

Colt1911 [192]

Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of $3.0 m/s^2.$

So, the acceleration of the express train, $a=-3 m/s^2$ .

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, $a=-3 m/s^2$

So, 0=36+(-3)t

$\Rightarrow t= 36/3=12$ seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

$v^2=u^2+2as$

$\Rightarrow 0^2=36^2+2(-3)s$

$\Rightarrow s=\frac{36\times36}{6}$

$\Rightarrow s=216$ meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

6 0

2 years ago