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Maslowich
3 years ago
5

On a smooth horizontal floor an object

Physics
2 answers:
Vika [28.1K]3 years ago
8 0
Would love to help you but your question is a little unclear what do you mean by this? Please let me know! :)
lukranit [14]3 years ago
7 0

Answer:

what do you mean by this?

Explanation:

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Name of compounds<br>a. A1C1<br> 3​
Orlov [11]

Explanation:

Aluminum Chloride

Hope it helps u

8 0
3 years ago
How do electromagnetic waves differ from mechanical waves?
stich3 [128]

They don't require a medium to propagate, meaning that they can travel through air, solid objects, and space.

6 0
4 years ago
Which best Describes How the function of the Cell Membrane differs in plant and animal Cells
kramer

Answer:

C

Explanation:

Animals do not have cell wall thus membrane controls movement in and out while in plants it helps store genetic material inside

8 0
3 years ago
Pleaseeeee help me with b, c, and d. There are no angles.
taurus [48]

Answer:

a. 150 J

b. 150 J

c. 0 J

d. 0 J

Explanation:

The given parameters are;

The horizontal force with which the man pulls the canister, F = 50 N

The distance he moves the vacuum cleaner, d = 3.0 m

a. Work done, W = Force applied, F × Distance moved by the force, d

Therefore, for the work done by the 50 N force on the canister, we have;

W = 50 N × 3.0 m = 150 N·m = 150 J

b. Given that he pulls the canister at a constant speed, we have;

The acceleration of the canister, a = 0 m/s²

Therefore, the net force on the canister, F_{NET} = F - F_{Friction}  = m × a

Where;

m = The mass of the canister

a = The acceleration of the canister

F = The applied force = 50 N

F_{Friction} = The force of friction

∴ F_{NET} = m × a = m × 0 m/s² = 0 N

Therefore;

F_{NET} =  F - F_{Friction} = 0 N

From which we have;

F = F_{Friction} = 50 N (The applied force, F is equal to the force of friction,

The work done by friction = The force of friction × The distance in which the force of friction acts

∴ The work done by friction = F_{Friction} × d - 50 N × 3.0 m = 150 J

The work done by friction = 150 J

c. The normal force, N acts perpendicular to the force of friction

The distance the canister moves in the perpendicular direction, d_p = 0 m

∴ The work done by the normal direction = N × d_p = N × 0 m = 0 J

The work done by the normal direction = 0 J

d. The vacuum weight, W, acts on the same line as the normal force but in the opposite direction to the normal force, N

Therefore, the weight, W, acts perpendicular to the line of motion of canister

The distance the canister moves in the direction of the weight, d_{wieght} = 0 m

Therefore, the work done by the weight = W × d_{wieght} = W × 0 m = 0 J

The work done by the weight = 0 J

7 0
3 years ago
A plane is flying east at 115 m/s. The wind accelerates it at 2.88 m/s^2 directly northwest. After 25.0s, what is the magnitude
kondaur [170]

Answer:

81.8 m/s

Explanation:

The initial velocity of the plane is:

v_0=115 m/s (toward east)

So, decomposing along the x- and y- directions:

v_{x0} = 115 m/s\\v_{y0} = 0

(we took east as positive x-direction and north as positive y-direction)

The acceleration is

a=2.88 m/s^2 (northwest, so the angle with the positive x-direction is 135 degrees)

Decomposing it along the two directions:

a_x = a cos 135^{\circ} = (2.88 m/s^2)(cos 135^{\circ})=-2.04 m/s^2\\a_y = a sin 135^{\circ} = (2.88 m/s^2)(sin 135^{\circ})=2.04 m/s^2

So the two components of the velocity after a time t = 25.0 s will be

v_x = v_{x0} + a_x t = 115 m/s + (-2.04 m/s^2)(25.0 s)=64 m/s\\v_y = v_{y0} + a_y t = 0 m/s + (2.04 m/s^2)(25.0 s)=51 m/s

So, the magnitude of the velocity of the plane will be

v=\sqrt[v_x^2+v_y^2}=\sqrt{(64 m/s)^2+(51 m/s)^2}=81.8 m/s

6 0
3 years ago
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