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3 years ago

Which planet will take the shortest revolution around the sun?

2 answers:

6 0

MERCURY is the all-around winner.

It's the planet closest to the sun. It has the fastest orbital speed,
the shortest distance to go, and the shortest period of revolution.

LUCKY_DIMON [66]3 years ago

3 0

The planet closest to the sun; Mercury.

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In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor re

Leya [2.2K]

Answer:

$I=2.6363\ kg.m^2$

Explanation:

Given:

dimension of uniform plate, $(0.673\times 0.535)\ m^2$

mass of plate, $m=10.7\ kg$

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

$I_{cm}=\frac{1}{12} \times m(L^2+B^2)$

where:

$L=$ length of the plate

$B=$ breadth of the plate

$I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)$

$I_{cm}=0.6591\ kg.m^2$

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

$s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}$

$s=0.4299\ m$

Now using parallel axis theorem:

$I=I_{cm}+m.s^2$

$I=0.6591+10.7\times 0.4299^2$

$I=2.6363\ kg.m^2$

6 0

3 years ago

What is the energy in joules of a photon with a frequency of 3.16e 12 s-1?

erica [24]

We have: Energy(E) = Planck's constant(h) × Frequency(∨)
Here, Planck's constant(h) = 6.626 × 10⁻³⁴ J/s
Frequency (∨) = 3.16 × 10¹² /s

Substitute the values into the expression:
E = (6.626 × 10⁻³⁴)(3.16 × 10¹²) J
E = 2.093 × 10⁻²¹ Joules

In short, Your Final answer would be 2.093 × 10⁻²¹ J

Hope this helps!

5 0

3 years ago

Read 2 more answers

With the switch open, roughly what must be the resistance of the resistor on the right for the current out of the battery to be

Delvig [45]

Answer:

The resistance must be 6.67 $\Omega$

Solution:

Resistance, $R_{1} = 20\Omega$

Resistance, $R_{2} = 10\Omega$

For the current to be the same when the switch is open or closed, the resistances must be connected in parallel as current is distributed in parallel with the same voltage across the circuit:

Thus in parallel:

$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

$\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{10}$

$R_{eq} = 6.67\ \Omega$

4 0

3 years ago

1. What is the formula for the period of a pendulum and what is the main determining factor in its period?

Brut [27]

Answer:

$T=2\pi \sqrt{\frac{L}{g}}$

Explanation:

A simple pendulum is a system consisting of a mass attached to a string, and oscillating in a periodic motion, back and forth, along an equilibrium position.

The period of a pendulum is the time it takes for the pendulum to complete one oscillation.

The period of a pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration due to gravity

From the formula, we see that the period of a pendulum does not depend on the mass.

Therefore, the only 2 factors affecting the period of a pendulum are:

- The length of the pendulum: the longer it is, the longer the period of oscillation

- The acceleration due to gravity: the greater it is, the shorter the period of the pendulum

7 0

3 years ago

A spring that has a spring constant of 1400 N/m is stretched to a length of 2.5 m. If the normal length of the spring is 1.0 m,

Elena L [17]

The answer is C. 1575 J. I just took this review.

5 0

4 years ago

Read 2 more answers