Someone please help??

Which type of energy will definitely not be used in the lighting of a match?

In-s [12.5K]

I say that the answere would be B

4 0

3 years ago

Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4

dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

6 0

3 years ago

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

wel

Answer:

Normal force = 8.75 N

Explanation:

given,

frictional force between the steel spatula and the Teflon frying pan=0.350 N

coefficient of friction between material =0.04

normal force = ?

using formula,

Frictional force = coefficient of friction × normal force

$normal\ force = \dfrac{Frictional\ force}{coefficient\ of\ friction}$

$normal\ force = \dfrac{0.350}{0.04}$

Normal force = 8.75 N

8 0

3 years ago

What part of an atom bonds with other atoms? ? A the innermost electrons B valence electronsany C the outermost protons D any el

kaheart [24]

Ionic bonds occur when electrons are donated from one atom to another. Each type of atom forms a characteristic number of covalentbonds with other atoms. An example of that is a hydrogen atom with one electron in its outer shell forms only one bond, its out most orbital becomes filled with two electrons.
the outermost protons

8 0

3 years ago

Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs

Temka [501]

The motion of the ball on the vertical axis is an accelerated motion, with acceleration
$a=g=-9.81 m/s^2$
The following relationship holds for an uniformly accelerated motion:
$2aS=v_f^2 - v_i^2$
where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
$v_f =0$
So we can rewrite the equation as
$2(-9.81 m/s^2) h=-v_i^2$
from which we can isolate h
$h= \frac{v_i^2}{19.62}$ (1)

Now let's assume that $v_i$ is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: $2v_i$ . So the maximum height of the second ball is
$h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62}$ (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.

8 0

3 years ago