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valina [46]
3 years ago
6

A continental polar air mass forms in ____

Physics
1 answer:
Sladkaya [172]3 years ago
7 0
I believe it is A bit not completely sure
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An object is dropped and falls freely to the ground with an acceleration of g. If it is thrown upward at an angle instead, its a
Savatey [412]

Answer:

g

Explanation:

if an object is thrown upward or at any angle, the acceleration acting on that object is the same as acceleration due to gravity which always acts towards the vertically downwards direction because there is no acceleration or the force acting on the object in horizontal direction.

Thus, the acceleration is same as acceleration due to gravity g.  

5 0
3 years ago
You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after
kotykmax [81]

Answer: d = 33 cm or 0.33 m

Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:

W = F.d.cosθ

F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.

For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:

W = F.d.cos(0)

W = F.d

To determine d:

d = \frac{W}{F}

d = \frac{3}{9}

d = 0.33 m

The distance d the block ice moved is 33 cm.

7 0
3 years ago
Why does changing the shape of an object have no effect on the density?
iren [92.7K]
Because the object is still made of the same material 
Density is not affected by the weight and shape of an object its affected by how concentrated the atoms are in a given volume 
7 0
3 years ago
Using the free-body diagram, calculate the net force
yuradex [85]

Answer:yes it is

Explanation:

5 0
3 years ago
Read 2 more answers
In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point
Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

=  (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0
3 years ago
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