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2 years ago

A rock falls for 1.43 seconds how far did it fall? The falls's velocity is an acceleration of -9.81 m/s2

Physics

1 answer:

BartSMP [9]2 years ago

7 0

Answer:

$s = -0.100 \ m$

Explanation:

Generally from kinematics equations we have that

$s = ut + \frac{1}{2}gt^2$

Here u is the initial velocity of the rock and the value is 0 m/s given that there was no information that it was in motion before the fall

substituting 1.43 s for t and $-9.8m/s^2$ for g we have that

$s = 0 *1.43 + \frac{1}{2}(-9.8)*(0.143)^2$

=> $s = -0.100 \ m$

The negative sign show that the displacement is in the direction of the negative y-axis

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What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropria

Likurg_2 [28]

Answer:

Approximately $\rm 11.2 \; N \cdot kg^{-1}$ at that distance from the center of the planet.

Option A) The low value of $g$ near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of $g$ on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

$\displaystyle \frac{G \cdot M \cdot m}{R^2}$ ,

where

$G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2$ .
$M$ is the mass of the planet, and
$m$ is the mass of the object.

To find an equation for $g$ , divide the equation for gravity by the mass of the object:

$\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}$ .

In this case,

$M = 5.68\times 10^{26}\; \rm kg$ , and
$R = 5.82 \times 10^7\; \rm m$ .

Calculate $g$ based on these values:

$\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}$ .

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for $g$ :

$\displaystyle g = \frac{G \cdot M}{R^2}$ .

The mass of the planet is in the numerator. If two planets are of the same size, $g$ would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since $R$ is in the denominator of $g$ , increasing the value of $R$ while keeping $M$ constant would reduce the value of $g$ . That explains why the value of $g$ near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately $9.81\; \rm N \cdot kg^{-1}$ .

As a side note, $5.82\times 10^7\rm \; m$ likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of $g$ near the cloud tops of Saturn is approximately $\rm 11.2 \; N \cdot kg^{-1}$ .

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2 years ago

Henri draws a wave that has a 4 cm distance between the midpoint and the trough. Geri draws a wave that has an 8 cm vertical dis

andre [41]

Answer:

Henri’s wave and Geri’s wave have the same amplitude and the same energy

Explanation:

The amplitude of a wave is the distance between the midpoint and the trough (or the crest). This is equivalent to half the distance between the trough and the crest. Therefore:

amplitude of Henri's wave: 4 cm

amplitude of Geri's wave: 8/2 = 4 cm

The energy of a wave is directly proportional to its amplitude.

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2 years ago

Read 2 more answers

Consider a capacitor made of two rectangular metal plates of length and width , with a very small gap between the plates. There

mezya [45]

Answer:

F= σ² L² /2ε₀

F = (L² ε₀/4π) ΔV² / r⁴

Explanation:

a) For this exercise we can use Coulomb's law

F = - k Q² / r²

where the negative sign indicates that the force is attractive and the value of the charge is equal to the two plates

Capacitance is defined by

C = Q / ΔV

Q = C ΔV

also the capacitance for a parallel plate capacitor is related to its shape

C = ε₀ A / r

we substitute

Q = ε₀ A ΔV / r

we substitute in the force equation

F = k (ε₀ A ΔV / r)² / r²

k = 1 / 4πε₀

F = ε₀ /4π L² ΔV² / r⁴4

F = L² ΔV² ε₀/ (4π r⁴)

F = (L² ε₀/4π) ΔV² / r⁴

b) Another way to solve the exercise is to use the relationship between the force and the electric field

F = q E

where we can calculate the field created by a plane using Gaussian law, where we use a cylinder with a base parallel to the plate as the Gaussian surface

Ф = ∫E .dA = $q_{int}$ / ε₀

the plate have two side

2E A = q_{int} / ε₀

E = σ / 2ε₀

σ = q_{int} / A

substituting in force

F = q σ / 2ε₀

the charge total on the other plate is

q = σ A

q = σ L²

F= σ² L² /2ε₀

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2 years ago

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

bixtya [17]

For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km

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2 years ago

Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 60 degrees. It reache

Nina [5.8K]

Answer:

R = 0.0503 m

Explanation:

This is a projectile launching exercise, to find the range we can use the equation

R = v₀² sin 2θ / g

How we know the maximum height

$v_{f}$ ² = $v_{oy}$ ² - 2 g y

$v_{f}$ = 0

$v_{oy}$ = √ 2 g y

$v_{oy}$ = √ 2 9.8 / 15

$v_{oy}$ = 1.14 m / s

Let's use trigonometry to find the speed

sin θ = $v_{oy}$ / vo

vo = $v_{oy}$ / sin θ

vo = 1.14 / sin 60

vo = 1.32 m / s

We calculate the range with the first equation

R = 1.32² sin(2 60) / 30

R = 0.0503 m

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2 years ago