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Viktor [21]
3 years ago
6

Heat from the Sun reaches Earth thanks to _____.

Physics
2 answers:
Aliun [14]3 years ago
7 0
Definitely not radiation. That crosses one answer choice out.
kotegsom [21]3 years ago
4 0
Heat from the Sun reaches Earth thanks to __Radiation___.
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What is the mass of a stone moving at a speed of 15 meters/second and having a momentum of 7.5 kilogram - meters / seconds
Elanso [62]

Answer:

0.5 kg

Explanation:

The momentum of an object is defined as

p = mv

where

m is the mass

v is the velocity

In this problem we have,

v = 15 m/s is the velocity of the stone

p = 7.5 kg m/s is the momentum

Solving for m, we can find the mass of the stone:

m=\frac{p}{v}=\frac{7.5 kg m/s}{15 m/s}=0.5 kg

4 0
3 years ago
Read 2 more answers
A horizontal force of 350N is exerted on a 2.5 kg ball as it rotates uniformly in a horizontal circle of radius of 0.90m. Calcul
harkovskaia [24]
F=mv^2/R
----> V^2=FR/m=(350x0.9)/2.5=126
----- V=11.22 m/s
5 0
3 years ago
How do the discovery of gravity and the invention of electronic satellites most likely relate to the processes of scientific inv
Leya [2.2K]

Answer:

c

Explanation:

because i said so

5 0
3 years ago
Which part of a river would have animals with very muscular bodies and adaptations that let survive in turbulent water? Source z
lidiya [134]
The part of a river that would have animals with muscular bodies and adaptations that let survive in turbulent water is in the transition zone, the mid-transition zone to be precise. Water at the source zone possesses a lot of potential energy and as it flows from the upper reaches the potential energy is turned into kinetic energy when the course of the river begins to gradually level out and this translates into increase in velocity. By the time river water reaches the middle of the transition zone, most of the potential energy would have been turned into kinetic energy and thus water velocity would be quite high here. Animals living here would develop muscles because of constantly fighting against the strong current to avoid being swept downstream.
8 0
3 years ago
Read 2 more answers
what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame
Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

  • There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
  • This friction force has a maximum value, that can be written as follows:

       F_{frmax} = \mu_{s} *F_{n} (1)

       where  μs is the coefficient of static friction, and Fn is the normal force,

       perpendicular to the wall and aiming to the center of rotation.

  • This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
  • This force has the following general expression:

       F_{c} =  m* \omega^{2} * r (2)

       where ω is the angular velocity of the riders, and r the distance to the

      center of rotation (the  radius of the circle), and m the mass of the

      riders.

      Since Fc is actually Fn, we can replace the right side of (2) in (1), as

      follows:

     F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)

  • When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

       m* g = m* \mu_{smin} * \omega^{2} * r (4)

  • (The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)
  • Cancelling the masses on both sides of (4), we get:

       g = \mu_{smin} * \omega^{2} * r (5)

  • Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

      60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)

  • Replacing by the givens in (5), we can solve for μsmín, as follows:

       \mu_{smin} = \frac{g}{\omega^{2} *r}  = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)

5 0
2 years ago
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