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2 years ago

A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume calorimeter. Consequently,

1 answer:

3 0

Mass of methanol = 1.922g; Change in temperature = 5.14° C; Heat capacity of the bomb calorimeter + water = 8.69kJ/°C. Number of moles.

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9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the

mojhsa [17]

Answer:

Moment about SHOULDER ∑ τ = 3.17 N / m,

Moment respect to ELBOW Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

∑ τ = I α

The moment is requested on the elbow and shoulder at the initial instant, just when the movement begins.

They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

∑ τ = I α

I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

I_mass = m L²

we substitute

∑ τ = (⅓ m L² + M L²) α

let's calculate

∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

Στ = I α

I = I_ forearm + I_mass

I = ⅓ m (L-0.03)² + M (L-0.03)²

let's calculate

i = ⅓ 2.3 0.47² + 0.5 0.47²

I = 0.2798 Kg m²

Στ = 0.2798 10

Στ= 2.80 N m

3 0

3 years ago

half life worksheet answer key 3. What percent of a sample As-81 remains un-decayed after 43.2 seconds

Ivahew [28]

The final mass after decay can be obtained by using under given relation:

half life period of As-81 = 33 seconds

mf = mi x (1/2^n)

= 100 x ( 1/2^(43.2/33))

= 40.4 %

3 0

3 years ago

Obi Wan hears the destruction of a planet and all of its people through 'the force'. These sounds are only in his head and are n

trapecia [35]

Answer:

medium

Explanation:

A sound medium is defined as channel through which sound can travel or be transmitted.

Sound medium could be in the form gases, liquids, solids or plasmas. Space is made up of vacuum and therefore, has no medium within it. Hence, space cannot transmit sound in any form or allows sound to travel through it.

6 0

3 years ago

Two 0.2304 cm x 0.2304 cm square aluminum electrodes, spaced 0.5974 mm apart are connected to a 61 V battery. What is the charge

Arisa [49]

Answer:

The charge on each plate is 0.0048 nC

Explanation:

for the distance between the plates d and given the area of plates, A, and ε = 8.85×10^-12 C^2/N.m^2, the capacitance of the plates is given by:

C = (A×ε)/d

=[(0.2304×10^-2)(0.2304×10^-2)×(8.85×10^-12))/(0.5974×10^-3)

= 7.86×10^-14 F

then if the plates are connected to a battery of voltage V = 61 V, the charge on the plates is given by:

q = C×V

= (7.86×10^-14)×(61)

= 4.80×10^-14 C

≈ 0.0048 nC

Therefore, the charge on each plate is 0.0048 nC.

3 0

3 years ago

A plank of length L=2.200 m and mass M=4.00 kg is suspended horizontally by a thin cable at one end and to a pivot on a wall at

baherus [9]

Hello!

Let's begin by doing a summation of torques, placing the pivot point at the attachment point of the rod to the wall.

$\Sigma \tau = 0$

We have two torques acting on the rod:
- Force of gravity at the center of mass (d = 0.700 m)

- VERTICAL component of the tension at a distance of 'L' (L = 2.200 m)

Both of these act in opposite directions. Let's use the equation for torque:
$\tau = r \times F$

Doing the summation using their respective lever arms:

$0 = L Tsin\theta - dF_g$

$dF_g = LTsin\theta$

Our unknown is 'theta' - the angle the string forms with the rod. Let's use right triangle trig to solve:

$tan\theta = \frac{H}{L}\\\\tan^{-1}(\frac{H}{L}) = \theta\\\\tan^{-1}(\frac{1.70}{2.200}) = 37.69^o$

Now, let's solve for 'T'.

$T = \frac{dMg}{Lsin\theta}$

Plugging in the values:
$T = \frac{(0.700)(4.00)(9.8)}{(2.200)sin(37.69)} = \boxed{20.399 N}$

3 0

1 year ago