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jonny [76]
3 years ago
13

In this activity, you will analyze the motion of a cart being pulled up an inclined plane at a constant speed. The angle of the

incline can be modified by 10 degrees increments between the values of 30 degrees and 90 degrees. Three different masses can be selected - 2.0kg, 3.0kg, and 4.0kg in each simulation the cart is pull d to the same height - 1.0 meter above the original starting position. For each simulation, the force that must be applied is reported on the screen. The displacement of the cart can be measured using the cm-ruler that is displayed for each trial. Can some help me

Physics
1 answer:
Verizon [17]3 years ago
8 0

Answer:

darn this too much lololpop

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A beam of sunlight falling on a prism refracts and forms seven color bands. This illustrates that
ExtremeBDS [4]

That isn"t the right answer the correct answer is B.

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What form of energy provides power for a gas stove?
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A horizontal force of 12N is applied to 1.5kg of block which rests on a horizontal surface. If the coefficient friction is 0.3,
Rama09 [41]

Answer:

7 m/s²

Explanation:

From the question given above, the following data were:

Force applied (Fₐ) = 15 N

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Acceleration (a) of block =?

Next, we shall determine the frictional force. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Acceleration due to gravity (g) = 10 m/s²

Coefficient of friction (μ) = 0.3

Normal reaction (R) = mg = 1.5 × 10 = 15 N

Frictional force (Fբ) =?

Fբ = μR

Fբ = 0.3 × 15

Fբ = 4.5 N

Next, we shall determine the net force acting on the block. This can be obtained as follow:

Force applied (Fₐ) = 15 N

Frictional force (Fբ) = 4.5 N

Net force (Fₙ) =.?

Fₙ = Fₐ – Fբ

Fₙ = 15 – 4.5

Fₙ = 10.5 N

Finally, we shall determine the acceleration produced in the block. This can be obtained as follow:

Mass (m) of block = 1.5 Kg

Net force (Fₙ) = 10.5 N

Acceleration (a) of block =?

Fₙ = ma

10.5 = 1.5 × a

Divide both side by a

a = 10.5 / 1.5

a = 7 m/s²

Therefore, the acceleration produced in the block is 7 m/s²

8 0
3 years ago
A 25-kg child sits at the top of a 4-meter slide. After sliding down, the child is traveling at 5 m/s. How much PE does he start
Semmy [17]

Daniddmelo says it right there, don't know why he got reported.

The potential energy (PE) is mass x height x gravity. So it would be 25 kg x 4  m x 9.8 = 980 joules. The child starts out with 980 joules of potential energy. The kinetic energy (KE) is (1/2) x mass x velocity squared. KE = (1/2) x 25 kg x 5 m/s2 = 312.5 joules. So he ends with 312.5 joules of kinetic energy. The Energy lost to friction =  PE - KE. 980- 312.5 = 667.5 joules of energy lost to friction.

Please don't just copy and paste, and thank you Dan cause you practically did it I just... elaborated more? I dunno. 

4 0
3 years ago
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

8 0
3 years ago
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