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3 years ago

13

Plz help, its an emergency, giving lots of points for it. and srry if its in the wrong subject, not really sure? I just don't un

derstand it at all, it's about atoms and all that stuff.

2 answers:

Nonamiya [84]3 years ago

8 0

What is the question being asked?

exis [7]3 years ago

4 0

Anything I can help with

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What is the connection between the x- and y-motions of a projectile?

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3 years ago

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NEED HELP ASAP.. ANSWER AS MANY AS YOU CAN

Yuri [45]

Answer:q1=a q2=c

Explanation:

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3 years ago

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A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to h

timofeeve [1]

To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

PART A) Normal Force.

$F_n = \frac{mv^2}{r}$

$N+mgcos\theta = \frac{mv^2}{r}$

Here,

Normal reaction of the ring is N and velocity of the ring is v

$N+mgcos\theta = \frac{mv^2}{r}$

$N+Wcos\theta = \frac{W}{g} (\frac{v^2}{r})$

$N+2cos30\° = \frac{2}{32.2}*\frac{10^2}{2}$

$N = 1.374lb$

PART B) Acceleration

$F_t = ma_t$

$-mgsin\theta = ma_t$

$-W sin\theta = \frac{W}{g} a_t$

$-2Sin30\° = (\frac{2}{32.2})a_t$

$a_T = -16.10ft/s^2$

Negative symbol indicates deceleration.

<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>

8 0

3 years ago

Si un ciclista se mueve a una velocidad de 5 m/s y acelera 1 m/s2, a los 10 segundos su velocidad será

MrRa [10]

Answer:

A los 10 segundos su velocidad será 15 $\frac{m}{s}$

Explanation:

La aceleración de un objeto es una magnitud que indica cómo cambia la velocidad del objeto en una unidad de tiempo.

En otras palabras, la aceleración relaciona los cambios de la velocidad con el tiempo en el que se producen, es decir que mide cómo de rápidos son los cambios de velocidad:

Una aceleración grande significa que la velocidad cambia rápidamente.
Una aceleración pequeña significa que la velocidad cambia lentamente.
Una aceleración cero significa que la velocidad no cambia.

La aceleración "a" puede ser calculada mediante la expresión:

$a=\frac{vfinal - vinicial}{tiempo}$

En este caso:

a= 1 $\frac{m}{s^{2} }$

vfinal= ?
vinicial= 5 $\frac{m}{s}$
tiempo= 10 s

Reemplazando:

$1\frac{m}{s^{2} }=\frac{vfinal - 5\frac{m}{s} }{10 s}$

Resolviendo se obtiene:

1 $\frac{m}{s^{2} }$ *10 s= vfinal - 5 $\frac{m}{s}$

10 $\frac{m}{s}$ = vfinal - 5 $\frac{m}{s}$

10 $\frac{m}{s}$ + 5 $\frac{m}{s}$ = vfinal

15 $\frac{m}{s}$ = vfinal

<u><em>A los 10 segundos su velocidad será 15 </em></u> $\frac{m}{s}$ <u><em></em></u>

5 0

2 years ago