Nodes and antinodes.are part of an _____ wave A active B standing

A car of mass 1,000 kilograms is moving initially at the speed of 22 meters/second. When the brakes are applied, it takes the ca

Mariulka [41]

i believe it is C....tell me if im right please<3

4 0

3 years ago

Read 3 more answers

si se deja caer un carrito desde el punto mas alto de ua psta de coches cuya altura es de 1.4m cual es la velocidad maxima que p

forsale [732]

Answer:

v = 5.24[m/s]

Explanation:

Este problema se puede resolver por medio del principio de la conservación de la energía, donde la energía potencial es igual a la energía cinética. Es decir a medida que el carrito desciende su energía potencial disminuye, pero su energía cinética aumenta.

$E_{kin}=E_{pot}$

Donde:

$E_{kin}=\frac{1}{2} *m*v^{2} \\\\E_{pot}=m*g*h$

Ahora reemplazando:

$\frac{1}{2} *m*v^{2}=m*g*h\\\\0.5*v^{2}=9.81*1.4\\v=\sqrt{\frac{9.81*1.4}{0.5} } \\\\v=5.24[m/s]$

6 0

3 years ago

What factor determines an object's momentum?

vlada-n [284]

Momentum = mass x velocity

So both mass and velocity affect an object's momentum.

3 0

3 years ago

How does the Earth's rotation affect our view of stars?

rodikova [14]

you only see the stars once every twenty for hours so you can have daylight so because of Earth's rotation you only see the stars for a certain amount of hours

Explanation:

the Earth makes a full rotation so that's why my answer is what it is

5 0

3 years ago

The force between a pair of charges is 900 newtons. The distance between the charges is 0.01 meters. If one of the charges is 2e

Maksim231197 [3]

Answer:

$\fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}$

Explanation:

Given:

Force between pair of charges= 900 newtons

The distance between the charges = 0.01 meters

Strength of Charge first q1 = 2e-10 Coulomb

To find:

Strength of Charge second q2 = ____ Coulomb?

Solution:

We know that,

Force between two charges separate by distance r is given by the equation,

$|F| = K_e \frac{q1 \cdot \: q2}{ {r}^{2} } \\ 900 =K_e \frac{(2e - 10)\cdot \: q2}{ {0.01}^{2} } \\ 900 \times {10}^{ - 4} = K_e {(2e - 10)\cdot \: q2} \\ q2 = \frac{9 \times {10}^{ - 2} }{(2e - 10) K_e} \\ \\ \fbox{We \: know \: that \: e = 2.71 } \\ substituting \: the \: value \: \\ q2 = \frac{9 \times {10}^{ - 2} }{(2 \times 2.71 - 10)K_e} \\ q2 = \frac{0.09}{ - 4.58 K_e} \\ q2 = \frac{-0.0196}{K_e}\: coulomb$

$\fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}$

Thanks for joining brainly community!

3 0

2 years ago