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ziro4ka [17]
3 years ago
6

What causes differences in air pressure around the Earth?

Physics
1 answer:
Olegator [25]3 years ago
7 0
Wind is primarily driven by differences in air pressure. These variations in air pressure are due to temperature differences caused by variations in solar energy received at the surface of the Earth.
You might be interested in
List the steps you would use to reach one of your personal growth goals.
Zinaida [17]
Im not sure, but here are a few of mine : Learn to be silent, and listen, take massive action and be proactive, listen, focus, and lastly, persist. Hope this helped! (:
6 0
3 years ago
The membrane that surrounds a certain type of living cell has a surface area of 7.1 x 10-9 m2 and a thickness of 1.5 x 10-8 m. A
hoa [83]

Answer:

Q = +1.4 pC

Explanation:

  • The capacitance of any capacitor, by definition, is as follows:
  •  C = \frac{Q}{V} (1)
  • Applying Gauss'Law and the definition of electric potential, it can be showed that the capacitance of a parallel-plate capacitor can be expressed as follows:
  •  C = \frac{\epsilon_{r}*\epsilon_{0} * A}{d} (2)
  • Where εr is the dielectric constant of the material that fills the space between the plates, A is the area of one of the plates, and d, is the separation between them.
  • Replacing by the givens in (2) we can find the value of the capacitance C, as follows:

       C = \frac{\ 4.4*8.85e-12C2/N*m2*7.1e-9m2}{1.5e-8m} = 18.4e-12 F = 18.4 pF

  • Replacing the values of C and V in (1), we can solve for Q, as follows:

       Q = C*V = 18.4e-12 F* 75.9e-3 V = 1.4e-12 C = +1.4 pC

  • As the outer surface is at a higher potential that the inside surface, the charge on it must be positive, and is equal to +1.4 pC.
7 0
3 years ago
you push a shopping cart with 2 packs of water bottles. you stop the cart and add two more packs. you push the cart again. does
HACTEHA [7]

Answer:

Yes

Explanation:

The cart will definitely behave differently because more weight is added to it.

In order to successfully push a cart, the applied force must exceed the force of the weight of the cart along with its constituents and the frictional force between the tyres and the floor.

<em>When more weights are added (the 2 packs of water bottles), it means that more force would be needed to overcome the opposing forces to the movement of the cart in order to successfully push the cart again.</em>

4 0
2 years ago
An 18.8 kg block is dragged over a rough, horizontal surface by a constant force of 156 N acting at an angle of angle 31.9◦ abov
Sav [38]

Answer:

a) W = 2635.56 J

b) Wf = 423.27 J

c) c)  The Sign of the work done by the frictional force (Wf) is negative (-)

d) W=0

Explanation:

Work (W) is defined as the Scalar product of the force (F) by the distance (d) that the body travels due to this force .  

The formula for calculate the work is :

W = F*d*cosα  

Where:

W : work in Joules (J)

F : force in Newtons (N)

d: displacement in meters (m)

α  :angle that form the force (F) and displacement (d)

Known data

m =  18.8 kg : mass of the block

F= 156 N,acting at an angle θ = 31.9◦°: angle  above the horizontal

μk= 0.209 : coefficient of kinetic friction between the cart and the surface

g = 9.8 m/s²: acceleration due to gravity

d = 19.9 m : displacement of the block

Forces acting on the block

We define the x-axis in the direction parallel to the movement of the cart on the floor  and the y-axis in the direction perpendicular to it.

W: Weight of the cart  : In vertical direction  downaward

N : Normal force :  In vertical direction the upaward

F : Force applied to the block

f : Friction force: In horizontal direction

Calculated of the weight  of the block

W= m*g  =  ( 18.8 kg)*(9.8 m/s²)= 184.24 N

x-y components of the force F

Fx = Fcosθ = 156 N*cos(31.9)° = 132.44 N

Fy = Fsinθ = 156 N*sin(31.9)°  = 82.44 n

Calculated of the Normal force

Newton's second law for the  block in y direction  :

∑Fy = m*ay    ay = 0

N-W+Fy= 0

N-184.24+82,44= 0

N = 184.24-82,44 

N = 101.8 N

Calculated of the kinetic friction force (fk):

fk = μk*N = (0.209)*( 101.8)

fk = 21.27 N

a) Work done by the F=156N.

W = (Fx) *d  *cosα

W = (132.44 )*(19.9)(cos0°) (N*m)

W = 2635.56 J

b) Work done by the force of friction

Wf = (fk) *d *cos(180°)

Wf = (21.27 )*(19.9) (-1) (N*m)

Wf = - 423.27 J

Wf = 423.27 J  :magnitude

c)  The Sign of the work done by the frictional force is negative (-)

d) Work done by the Normal force

W = (N) *d *cos(90°)

W = (101.8 )*(19.9) (0) (N*m)

W = 0

4 0
3 years ago
HELP PLEASE AND HURRRRRY
kykrilka [37]

Answer:

the path the object follows is called its trajectory so A. trajectory

6 0
3 years ago
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