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3 years ago

The element oxygen, represented by the symbol O, is classified as

Physics

1 answer:

Yakvenalex [24]3 years ago

7 0

Answer:

Explanation:

Oxygen is the chemical element with the symbol O and atomic number 8. It is a member of the chalcogen group in the periodic table, a highly reactive nonmetal, and an oxidizing agent that readily forms oxides with most elements as well as with other compounds.

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The distance of the force from the axis of rotation is called?

kondor19780726 [428]

Answer:

explain you qns. briefly

7 0

3 years ago

A newly discovered planet has a mean radius of 7380 km. A vehicle on the planet\'s surface is moving in the same direction as th

Butoxors [25]

Answer:

292796435 seconds ≈ 300 million seconds

Explanation:

First of all, the speed of the car is 121km/h = 33.6111 m/s

The radius of the planet is given to be 7380 km = 7380000 m

From the relationship between linear velocity and angular velocity i.e., v=rw, the angular velocity of the car will be w=v/r = 33.6111/7380000 = 0.000000455 rad/s = 4.55 x 10⁻⁶ rad/sec

If the angular velocity of the vehicle about the planet's center is 9.78 times as large as the angular velocity of the planet then we have

w(vehicle) = 9.78 x w(planet)

w(planet) = w(vehicle)/9.78 = 4.55 x 10⁻⁶ / 9.78 = 4.66 x 10⁻⁷ rad/sec

To find the period of the planet's rotation; we use the equation

w(planet) = 2π÷T

Where w(planet) is the angular velocity of the planet and T is the period

From the equation T = 2π÷w = 2×(22/7) ÷ 4.66 x 10⁻⁷ = 292796435 seconds

Therefore the period of the planet's motion is 292796435 seconds which is approximately 300, 000, 000 (300 million) seconds

8 0

4 years ago

What is the natural frequency (wo) of a circuit that has an inductor of 0.048 henery's, and a capacitor of 0.0004 farads?

adell [148]

Answer:

The natural frequency will be 228.11 rad/sec

Explanation:

We have given Inductance L = 0.048 Henry

And the capacitance C = 0.0004 farad

We have to find natural frequency

When only inductor and capacitor is present in the circuit then it is known as LC circuit and Natural frequency of LC circuit is given by $\omega _0=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.048\times 0.0004}}=228.21\ rad/sec$

So the natural frequency will be 228.21

4 0

3 years ago

A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.

SVETLANKA909090 [29]

A) The position at t = 2.0 sec is 43.0 m east

B) The position is 55 m east

Explanation:

In order to solve the problem, we take the east direction as positive direction.

We know that:

- at t = 0, the motorcyclist is at a position of $x_0 = 5.0 m$

- at t = 0, the initial velocity of the motorcyclist is $v_0 = 15.0 m$ east

- The acceleration of the motorcyclist is constant and it is $a=4.0 m/s^2$

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

$x(t)=x_0 + v_0t + \frac{1}{2}at^2$

where t is the time.

Substituting t = 2.0 s, we find the position:

$x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m$

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

$v(t)=x'(t)=v_0 + at$

where:

$v_0=15.0 m/s$ is the initial velocity

$a=4.0 m/s^2$ is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

$t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s$

And therefore, the position at t = 2.5 s is:

$x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m$

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3 0

3 years ago

a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r

dimaraw [331]

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

spring constant k= 2250 Nm

now according to principal of conservation of energy we observe,

the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.

mgh= μ (mgl) +1/2 kx²

10 x 10 x 3= μ(600) +(1125) (0.09)

μ(600) =300 - 101.25

μ = 198.75÷600

μ =0.33125

The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

Learn more about kinetic friction here-

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4 0

2 years ago