All categories

3 years ago

Which would be the most reliable source of information about the weather?

Physics

2 answers:

o-na [289]3 years ago

8 0

Answer:

C: The national weather service channel.

Explanation:

first, you can rule out A and B because they don't exactly provide exact information. D can be ruled out because it shows predicted weather while C, The national weather service channel, has 'round the clock info instead of predictions.

puteri [66]3 years ago

4 0

It is answer a a fishing magazine

You might be interested in

If a cliff jumper leaps off the edge of a 100m cliff, how long does she fall before hitting the water? (assume zero air resistan

andrew-mc [135]

<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

Assuming that the initial velocity of the jumper is zero, on Earth any freely falling object has an acceleration of 9.8 m/s².

Δt = (√2xgxh)/9,8
Δt = (14√10)/9,8
Δt ≈ 4,5 s

4 0

2 years ago

newton's second law states that when a net force acts on an object, it accelerates it.Explain how it would be possible for two o

Alika [10]

Answer:

See Explanation

Explanation:

According to Newton's second law of motion, the acceleration of a body is proportional to the net external force that acts on the body.

A body accelerated when it is acted upon by an unbalanced net external force.

When the external forces acting on a body are balanced, the effect of each force is cancelled by the other hence the body is not accelerated according to Newton's second law.

4 0

2 years ago

In volleyball, how is the server determined?

PtichkaEL [24]

the team rotates the serve

4 0

3 years ago

Read 2 more answers

Stress is a _____ to a perceived mental or physical stressor

ioda

I would say C because Instigator means <span>a person who brings about or initiates something. I hoped that helped </span>

6 0

2 years ago

What is the orbital velocity in km/s and period in hours of a ring particle at the outer edge of Saturn's A ring

mote1985 [20]

Answer:

The orbital velocity $v = 16.4 \ km/s$

The period is $T = 14.8 \ hours$

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational force between the ring particle and the planet , this is mathematically represented as

$\frac{GM_s * m }{r^2 } = m w^2 r$

=> $w = \sqrt{ \frac{GM}{r^3} }$

Here G is the gravitational constant with value $G = 6.67*10^{-11}$

$M_s$ is the mass of with value $M_s =5.683*10^{26} \ kg$

r is the is distance from the center of the to the outer edge of the A ring

i.e r = R + D

Here R is the radius of the planet with value $R = 60300 \ km$

D is the distance from the equator to the outer edge of the A ring with value $D = 80000 \ kg$

$r =80000 + 60300$

=> $r =140300 \ km = 1.4*10^{8} \ m$

=> $w = \sqrt{ \frac{ 6.67*10^{-11}* 5.683*10^{26}}{[1.4*10^{8}]^3} }$

=> $w = 1.175*10^{-4} \ rad/s$

Generally the orbital velocity is mathematically represented as

$v = w * r$

=> $v = 1.175*10^{-4} * 1.4*10^{8}$

=> $v = 1.64*10^{4} \ m /s = 16.4 \ km/s$

Generally the period is mathematically represented as

$T = \frac{2 \pi }{w }$

=> $T = \frac{2 * 3.142 }{ 1.175 *10^{-4} }$

=> $T = 53473 \ second = 14.8 \ hours$

Answer:

The orbital velocity $v = 16.4 \ km/s$

The period is $T = 14.8 \ hours$

Explanation:

Generally centripetal force acting ring particle is equal to the gravitational force between the ring particle and the , this is mathematically represented as

$\frac{GM_s * m }{r^2 } = m w^2 r$

3 0

2 years ago