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zysi [14]
3 years ago
8

PLEASE HELP!!

Physics
2 answers:
vesna_86 [32]3 years ago
8 0

It is A. Wind Blow from one location to another.

Blizzard [7]3 years ago
5 0
Wind blows, and water flows, from higher pressure to lower pressure.
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please help me with my question I will like and mark as brainliest for the first correct answer due tomorrow morning​
bazaltina [42]

Answer:

1845.26 ?

Explanation:

18.46 × 99.96= 1845.2616 = 1845.26

im not entirely sure though

6 0
3 years ago
Detailed research on comets indicates that they cannot be older than HOW MANY years.
Free_Kalibri [48]
It would be, 1.000. Hope that helps :)
6 0
3 years ago
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What is the equilibrium concentration of ibr?
asambeis [7]
IBR is the thermal decomposition of iodine(I) bromide to produce iodine and bromine. This reaction takes place at a temperature of over 40,5°C and is written as: 

 <span>2IBr ⇄ I2 + Br2
</span>
Equilibrium is a state of dynamic balance where the ratio of the product and reactant concentrations is constant.<span> You can calculate the equilibrium concentration if you know the equilibrium constant Kc (Kc=I^2*Br^2/IBR^2) and the initial concentration for the reaction. The initial concentration is obtained from ICE Table.</span>
3 0
3 years ago
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

  • If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
  • τ = F*r (2)
  • For a solid uniform disk, the rotational inertia regarding an axle passing through its center  is just I = m*r²/2 (3).
  • Replacing (2) and (3) in (1), we can solve for α, as follows:

       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

        \omega_{f}^{2}  - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)

  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
  • Replacing this value and (7) in (8), we get:

       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

b)    

  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

       a_{t} = \alpha * r (9)

  • where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
  • Replacing this value and (4), in (9), we get:

       a_{t}  = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)

  • Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
  • The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

       a_{c} = \omega^{2} * r  (11)

  • Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
  • Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
  • Replacing this value and (7) in (11) we get:

       a_{c} = \omega^{2} * r   = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)

  • The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
  • Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

6 0
3 years ago
What are the names of the 4 types of fronts? How are they created?
jeka57 [31]

Answer:

Stationary Front, warm front, cold front, Occluded Front.

Explanation:

Stationary Front. When the surface position of a front does not change (when two air masses are unable to push against each other; a draw), a stationary front is formed.

cold front is the leading edge of a cooler mass of air at ground level that replaces a warmer mass of air and lies within a pronounced surface trough of low pressure. It often forms behind an extratropical cyclone (to the west in the Northern Hemisphere, to the east in the Southern), at the leading edge of its cold air advection pattern—known as the cyclone's dry "conveyor belt" flow. Temperature differences across the boundary can exceed 30 °C (86 °F) from one side to the other. When enough moisture is present, rain can occur along the boundary. If there is significant instability along the boundary, a narrow line of thunderstorms can form along the frontal zone. If instability is weak, a broad shield of rain can move in behind the front, and evaporative cooling of the rain can increase the temperature difference across the front. Cold fronts are stronger in the fall and spring transition seasons and weakest during the summer.

A warm front is a density discontinuity located at the leading edge of a homogeneous warm air mass, and is typically located on the equator-facing edge of an isotherm gradient. Warm fronts lie within broader troughs of low pressure than cold fronts, and move more slowly than the cold fronts which usually follow because cold air is denser and less easy to remove from the Earth's surface. This also forces temperature differences across warm fronts to be broader in scale. Clouds ahead of the warm front are mostly stratiform, and rainfall gradually increases as the front approaches. Fog can also occur preceding a warm frontal passage. Clearing and warming is usually rapid after frontal passage. If the warm air mass is unstable, thunderstorms may be embedded among the stratiform clouds ahead of the front, and after frontal passage thundershowers may continue. On weather maps, the surface location of a warm front is marked with a red line of semicircles pointing in the direction of travel.

In meteorology, an occluded front is a weather front formed during the process of cyclogenesis. The classical view of an occluded front is that they are formed when a cold front overtakes a warm front, such that the warm air is separated (occluded) from the cyclone center at the surface. The point where the warm front becomes the occluded front is called the triple point; a new area of low-pressure that develops at this point is called a triple-point low. A more modern view of the formation process suggests that occluded fronts form directly during the wrap-up of the baroclinic zone during cyclogenesis, and then lengthen due to flow deformation and rotation around the cyclone.

3 0
2 years ago
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