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2 years ago

The velocity of a free falling ball at different time intervals on unknown planet is shown in Figure 1.24.

Physics

1 answer:

dsp732 years ago

4 0

Answer:

a1 = (V2 - V1) / 1 s = 3.75 m/s^2

a2 = (V2 - V1) / (t2 - t1) = (7.5 - 3.75) / 1 s = 3.75 m/s^2

a3 = (V2 - V1) / (t2 - t1) = (7.5 - 3.75) / 1 s = 3.75 m/s^2

A) a = (a1 + a2 + a3) / 3 = 3.75 m/s^2

v = a (t2 - t1)

B) after 1 sec v = 3.75 m / sec^2 * 1 sec = 3.75 m/s

after 2 sec v = v0 + a t = 3.75 m/s + 3.75 m/s^2 * 1 s = 7.5 m/s

C) The velocity increases proportionaly to the time

D) One factor depends on the variation of the acceleration due to gravity because of different positions of measurement (on earth)

A much larger factor is due to air resisance of the air on the falling object

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A model used for the yield y of an agricultural crop as a function of the nitrogen level n in the soil (measured in appropriate

pychu [463]

The maxima of an equation can be obtained by taking the 1st derivative of the equation then equate it to 0.The value of N that result in best yield is when dy/dn = 0.

Taking the 1st derivative of the equation y=(kn)/(9+n^2) :

By using the quotient rule the form of the equation is:
y = g(n) / h(n)
where:

g(n) = kn ---> g'(n) = k

h(n) = 9 + n^2 ---> h'(n) = 2n 
dy/dn is defined as:
dy/dn = [h(n) * g'(n) - h'(n) * g(n)] / h(n)^2
dy/dn = [(9 + n^2)(k) - (kn)(2n)] / (9 + n^2)^2
dy/dn = (9k + kn^2 - 2kn^2) / (9 + n^2)^2
dy/dn = (9k - kn^2) / (9 + n^2)^2
dy/dn = k(9 - n^2) / (9 + n^2)^2

Equate dy/dn = 0, then solve for n
k(9 - n^2) / (9 + n^2)^2 = 0
k(9 - n^2) = 0
9 - n^2 = 0
n^2 = 9
n = sqrt(9)
n = 3

Answer: The nitrogen level that gives the best yield of agricultural crops is 3 units.

5 0

3 years ago

Thorium^+2

Bad White [126]

Answer:

chemical symbol: Th

atomic number:90

protrons :90

neutrons:142

group#:4

period#: 9

Explanation:

you take the atomic weight (232.038)and subtract the atomic number to get (90) which is your neutrons

6 0

3 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the

Rama09 [41]

1) $1.11\cdot 10^{-7} J$

The capacitance of a parallel-plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

$r=\frac{2.0 cm}{2}=1.0 cm=0.01 m$

so the area is

$A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2$

While the separation between the plates is

$d=0.50 mm=5\cdot 10^{-4} m$

So the capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F$

And now we can find the energy stored,which is given by:

$U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J$

2) 0.71 J/m^3

The magnitude of the electric field is given by

$E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m$

and the energy density of the electric field is given by

$u=\frac{1}{2}\epsilon_0 E^2$

and using

$E=4\cdot 10^5 V/m$ , we find

$u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3$

7 0

4 years ago

An object from which neither matter nor light can escape is called a _________.

vichka [17]

The correct answer is a blackhole.

A blackhole is the only object in which nothing, not even light can escape.

Hope this helps!

4 0

4 years ago

Read 2 more answers

A small steel wire of diameter 1.0mm is connected to an oscillator and is under a tension of 5.7N . The frequency of the oscilla

kotegsom [21]

Answer:

The power output of the oscillator is 0.350 watt.

Explanation:

Given that,

Diameter = 1.0 mm

Tension = 5.7 N

Frequency = 57.0 Hz

Amplitude = 0.54 cm

We need to calculate the power output of the oscillator

Using formula of the power

$P=\dfrac{1}{2}\times\mu\times\omega^2\times a^2\times v$

Put the value into the formula

$P=\dfrac{1}{2}\times A\times\rho\times\omega^2\times a^2\times\dfrac{\sqrt{T}}{\mu}$

$P=\dfrac{1}{2}\times3.14\times(0.0005)^2\times7850\times(2\times\pi\times57.0)^2\times(0.54\times10^{-2})^2\times\sqrt{\dfrac{5.7}{3.14\times(0.0005)^2\times7850}}$

$P=0.350\ Watt$

Hence, The power output of the oscillator is 0.350 watt.

3 0

3 years ago