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3 years ago

What kind of energy does a burning log have? Can't figure this out for homework. It just needs to be a short answer.

1 answer:

lina2011 [118]3 years ago

6 0

I would assume thermal energy, since it's concerning heat and all. Although judging off of the fact that this is for physics, I don't think that would be the right answer in this case.

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A tennis ball of mass 44.0 g is held just above a basketball of mass 594 g. With their centers vertically aligned, both are rele

ZanzabumX [31]

Answer:

u = 4.6 m/s

h = 8.01 m

Explanation:

Given:

Mass of the tennis ball, m = 44.0 g

Mass of the basket ball, M = 594 g

Height of fall, h = 1.08m

Now,

we have

$u^2-u'^2 = 2as$

where, s = distance = h

a = acceleration

u = final speed before the collision

u' = initial speed

since it is free fall case

thus,

a = g = acceleration due to gravity

u' = 0

thus we have

$u^2-0^2 = 2\times9.8\tiimes1.08$

$u = \sqrt{21.168}$

u = 4.6 m/s

b) Now after the bounce, the ball moves with the same velocity

thus, v = v₂

thus,

final speed ( $v_f$ ) = v = 4.6 m/s

Then conservation of energy says

$\frac{1}{2}mu_1^2+\frac{1}{2}Mu_2^2 = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2$

also

applying the concept of conservation of momentum

we have

mu₁ + Mu₂ = mv₁ + Mv₂

u₁ =velocity of the tennis ball before collision = -4.6 m/s

u₂ = velocity of the basketball before collision= 4.6 m/s

v₁ = velocity of the tennis ball after collision

v₂ = velocity of the basketball after collision

substituting the values in the equation, we get

Now,

solving both the equations simultaneously we get

$v = (\frac{2M}{m+M})u_1+(\frac{m-M}{m+M})u_2$

substituting the values in the above equation we get

$v = (\frac{2\times594}{44+594})(-4.6)+(\frac{44-594}{44+594})4.6$

$v = -8.565-3.965$

$v = -12.53m/s$

here negative sign depicts the motion of the ball in the upward direction

now the kinetic energy of the tennis ball

$K.E = \frac{1}{2}mv^2$

$K.E = \frac{1}{2}44\times 10^{-3}kg\times 12.53^2$

K.E = 3.45 J

also at the height the K.E will be the potential energy of the tennis ball

thus,

3.45 J = mgh

3.45 = 44 × 10⁻³ × 9.8 × h

h = 8.01 m

5 0

3 years ago

A JFET has a drain current of 5mA. If IDSS = 10mA and VGS ( off )= -6 v. find The Value Of

levacccp [35]

$\underline {\huge \boxed{ \sf \color{skyblue}Answer : }}$

Given :

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{D} = 5mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{DSS} = 10mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS(off)} = -6V$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS} = {?}$

$\: \: \:$

Let's Slove :

$\tt \large I_{D} = I_{(DSS)} (1 - \frac {V_{GS}}{V_{GS(off)}} )^{2}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times V_{GS(off)}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times { - 6}$

$\: \:$

$\underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}$

3 0

2 years ago

A 70.0 kg person climbs stairs, gaining 3.90 meters in height. Find the work done (in J) to accomplish this task.

Veronika [31]

Answer:

Work done, W = 2675.4 J

Given:

mass, m = 70.0 kg

height, H = 3.90 m

Solution:

According to the question, as the person jumps the stairs up, there is an increase in the potential energy of the person which is provided by the work done in climbing the stairs and is given by:

Work done, W = mgH

where

g = acceleration due to gravity = $9.8 m/s^{2}[tex][tex]W = 70.0\times 9.8\times 3.90 = 2675.4 J$