Match the organs to the functions they perform during digestion.<br> passes food to esophagus

Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a

Gnom [1K]

Answer:

a = 640 m/s²

Explanation:

From work-kinetic energy principles,

The net force acting on the second object is the gravitational force and the electric force due to the first object.

So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²

Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²

The net Force F = ma

So, -F₁ + F₂ = F (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)

-Gm²/r² + kq²/r² = ma

ma = -Gm²/r² + kq²/r²

a = (-Gm²/r² + kq²/r²)/m

a = (-G + kq²/m²)m/r²

Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.

Substituting the variables into the equation, we have

a = (m/r²)(-G + k(q/m)²)]

a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg² ]

a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)

a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²

a = 0.064 × 10⁴N/kg

a = 64 × 10 N/kg)

a = 640 m/s²

8 0

2 years ago

A busy chipmunk runs back and forth along a straight line of acorns that has been set out between his burrow and a nearby tree.

saveliy_v [14]

Answer: a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

Explanation:

Acceleration is the rate of change in the velocity per time

a = change in velocity/time

a = ∆v/t

average acceleration a = (v2 -v1)/t. ....1

Given;

Final velocity v2 = 1.63m/s

Initial velocity v1 = -1.15ms

time taken t = 2.11s

Substituting into eqn 1

a = [1.63 - (-1.15)]/2.11

a = (1.63+1.15)/2.11

a = 2.78/2.11

a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

6 0

3 years ago

Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in

postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is $\rho = \frac{m}{V}$ , and the volume of a pyramid is (confusingly written on the proposal) $V=\frac{1}{3} Ah$ , so we can write:

$m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h$

Where s is the side of the base, being $s^2$ the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that $\frac{1m}{100cm}=1$ , and we can use this factor to convert anything written in cm to anything written in m. Example:

$500cm=500cm\frac{1m}{100cm}=5m$

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

$2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3$

Where we have used the fact that $1^3=1$ (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

$m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg$

And now we will convert to short tons using two conversion factors at the same time:

$m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons$

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0

2 years ago

If a snowboarder’s initial speed is 4 m/s and comes to rest when making it to the upper level. With a slightly greater initial s

Brrunno [24]

(a) At a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).

(b) If the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.

<h3>Conservation of mechanical energy</h3>

The effect of height and gravity on speed on the given planet Epislon is determined by applying the principle of conservation of mechanical energy as shown below;

ΔK.E = ΔP.E

¹/₂m(v²- u²) = mg(hi - hf)

¹/₂(v²- u²) = g(0 - hf)

v² - u² = -2ghf

v² = u² - 2ghf

where;

v is the final velocity at upper level
u is the initial velocity
hf is final height
g is acceleration due to gravity

when u² = 2gh, then v² = 0,

when gravity reduces, u² > 2gh, and v² > 0

Thus, at a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).

<h3>Final speed</h3>

v² = u² - 2ghf

where;

u is the initial speed = 5 m/s
g is acceleration due to gravity and its less than 9.8 m/s²
v is final speed
hf is equal height

Since g on Epislon is less than 9.8 m/s² of Earth;

5² - 2ghf > 3 m/s

Thus, if the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.

Learn more about conservation of mechanical energy here: brainly.com/question/6852965

5 0

2 years ago

what will be the moment of inertia of a body if it rotates at a uniform rate of 10rad/sec^2 by a torque of 120Nm?

Naya [18.7K]

Answer:

12 kgm²

Explanation:

here angular acceleration = 10rad/sec²

torque= 120Nm

moment of inertia=?

we know,

torque= angular acceleration× moment of Inertia

or, moment of inertia = torque/angular acceleration

= 120/10

= 12kgm²

7 0

3 years ago

Match the organs to the functions they perform during digestion. passes food to esophagus