When was the copernican treatise published

Grant sprints 50m to the right with an average velocity of 3.0 m/s

zlopas [31]

I need help with my math

6 0

3 years ago

In a typical rear-end collision, the victim's ________ is/are accelerated faster and harder than the torso

Georgia [21]

The victim's head is accelerated faster and harder than the torso when the victom is involved in a typical rear-end collision.

The traffic accident where a vehicle crashes into another vehicle that is directly in front of it is called a rear-end collision.

One of the most common accident in the United States is the rear-end collision, and in a lot of cases, rear-end collisions are prompted by drivers who are inattentive, unfavorable conditions of the road, and poor following distance.

<span>An enough room in front of your car so you can stop when the car in front of you stops suddenly is one basic driving rule. The person isn’t driving safely if he / she is behind you and couldn’t stop.</span>

8 0

3 years ago

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La tensión en newtons necesaria para que una onda transversal cuya longitud de onda es 3.33 cm vibre a razón de 625 ciclos por s

NemiM [27]

Answer:

9.34 N

Explanation:

First of all, we can calculate the speed of the wave in the string. This is given by the wave equation:

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

For the waves in this string we have:

$f=625 Hz$ , since it completes 625 cycles per second

$\lambda=3.33 cm = 0.033 m$ is the wavelength

So the speed of the wave is

$v=(625)(0.0333)=20.6 m/s$

The speed of the waves in a string is related to the tension in the string by

$v=\sqrt{\frac{T}{\mu}}$ (1)

where

T is the tension in the string

$\mu=\frac{m}{L}$ is the linear density

In this problem:

$m=16.5 g = 16.5\cdot 10^{-3} kg$ is the mass of the string

L = 0.75 m is the its length

Solving the equation (1) for T, we find the tension:

$T=\mu v^2 = \frac{m}{L} v^2 = \frac{16.5\cdot 10^{-3}}{0.75}(20.6)^2=9.34 N$

8 0

3 years ago

In a Broadway performance, an 84.5-kg actor swings from a R = 4.30-m-long cable that is horizontal when he starts. At the bottom

Arada [10]

Answer:

1.57772 m

Explanation:

M = Mass of actor = 84.5 kg

m = Mass of costar = 55 kg

v = Velocity of costar

V = Velocity of actor

$h_i$ = Intial height of actor = 4.3 m

g = Acceleration due to gravity = 9.81 m/s²

As the energy of the system is conserved

$\frac{1}{2}MV^2=Mgh_i\\\Rightarrow V=\sqrt{2gh_i}\\\Rightarrow V=\sqrt{2\times 9.81\times 4.3}\\\Rightarrow V=9.18509\ m/s$

As the linear momentum is conserved

$MV=(m+M)v\\\Rightarrow v=\frac{MV}{m+M}\\\Rightarrow V=\frac{84.5\times 9.18509}{84.5+55}\\\Rightarrow v=5.56372\ m/s$

Applying conservation of energy again

$\frac{1}{2}(m+M)v^2=(m+M)gh_f\\\Rightarrow h_f=\frac{v^2}{2g}\\\Rightarrow h_f=\frac{5.56372^2}{2\times 9.81}\\\Rightarrow h_f=1.57772\ m$

The maximum height they reach is 1.57772 m

3 0

3 years ago

A cyclist going downhill is accelerating at 1.2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is t

Blababa [14]

Answer:

Initial Velocity is 4 m/s

Explanation:

What is acceleration?

It is the change in velocity with respect to time, or the rate of change of velocity.

We can write this as:

$a=\frac{\Delta v}{t}$

Where

a is the acceleration

v is velocity

t is time

$\Delta$ is "change in"

For this problem , we are given

a = 1.2

t = 10

Putting into formula, we get:

$a=\frac{\Delta v}{t}\\1.2=\frac{\Delta v}{10}\\\Delta v = 1.2*10\\\Delta v = 12$

So, the change in velocity is 12 m/s

The change in velocity can also be written as:

$\Delta v = Final \ Velocity - Initial \ Velocity$

It is given Final Velocity = 16, so we put it into formula and find Initial Velocity. Shown Below:

$\Delta v = Final \ Velocity - Initial \ Velocity\\12=16-Initial \ Velocity\\Initial \ Velocity = 16 - 12 = 4$

hence,

Initial Velocity is 4 m/s

3 0

4 years ago

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