Find the magnitude of pressure in Pascal of 150N that acts normally on area. Of 0.25m2

two objects were lifted by machine one object had a massive 2 kg it was lifted at a speed of 2 m/s the other had a massive 4 kg

rosijanka [135]

Kinetic energy =(1/2) (mass) (speed²)

First object: KE = (1/2) (2 kg) (2m/s)² = 4 joules during the lift.

Second object: KE = (1/2) (4kg) (3 m/s)² = 18 joules during the lift.

The second object has more kinetic energy while it's being lifted
than the first object has while it's being lifted. Once they reach their
final heights and stop, neither object has any kinetic energy.

8 0

4 years ago

Does anybody have any ideas for Wall art in a blue room.

Iteru [2.4K]

Answer: Mabye like an ocean with dolphins swiming/jumping? Or even use the blue as a sky and then put green grass and do foxes or and a phoenix flying with a fox under it?

Explanation:

Just some ideas!

8 0

2 years ago

Read 2 more answers

A go cart starts from REST and accelerates uniformly over a time of 7 seconds for a distance of 100m. Determine the acceleration

Hunter-Best [27]

Answer:

please find attached pdf

Explanation:

Download pdf

6 0

3 years ago

Summerize: Newton's first law of motion by telling how an object in motions inertia changed and how an object at rest is affecte

valina [46]

Answer:

Object in motion: By the Newton's first law about inertia, an object in motion tends to keep in motion, unless an external force act on it.

Object at rest: By the Newton's first law about inertia, an object at rest tends to keep at rest, unless an external force act on it.

4 0

3 years ago

One point of the circuit is grounded (V = 0). What are the (a) size and (b) direction (up or down) of the current through resist

Svetach [21]

<h3>Answer:</h3>

(a) <u>i₁ = 0.03818 A = 38.18 mA</u>

(b) downward

(c) <u>i₂ = 0.01091 A = 10.91 mA</u>

(d) rightward

(e) <u>i₃ = 0.02727 A = 27.27 mA</u>

(f) leftward

(g) <u>Eₐ = 3.818 Volts</u>

<h3>Question:</h3>

The complete question is stated below and the figure is provided in the attachment:

In Fig. 27-47, E 1 = 6.00 V, E 2 = 12.0 V, R1 = 100 Ω,

R2 = 200 Ω, and R3 = 300 Ω. One point of the circuit is grounded

(V = 0). What are the (a) size and (b) direction (up or down) of the

current through resistance 1, the (c) size and (d) direction

(left or right) of the current through resistance 2, and the

(e) size and (f) direction of the current through resistance 3?

(g) What is the electric potential at point A?

<h3></h3><h3>Explanation:</h3>

Applying Kirchoff's voltage law in the loops of botg E₁ and E₂, in the clockwise and anti clockwise direction:

E₁ - i₂R₂ - i₁R₁ = 0

E₂ - i₃R₃ - i₁R₁ = 0

If, we apply Kirchhoff's current law at junction A, we get:

i₁ = i₂ + i₃

Using these relations in loop equations, and re-arranging:

E₁ - i₂R₂ - (i₂ + i₃) R₁ = 0 ___________ eqn (1)

E₂ - i₃R₃ - (i₂ + i₃) R₁ = 0 ___________ eqn (2)

Eqn (1) implies:

6 - 200 i₂ - 100 i₂ - 100 i₃ = 0

i₂ = (6 - 100i₃)/300

Eqn (2) implies:

12 - 300 i₃ - 100 i₂ - 100 i₃ = 0

12 - 400 i₃ = 100 i₂

using value of i₃ from eqn (1)

12 - 400 i₃ = (1/3)(6 - 100 i₃)

36 - 1200 i₃ = 6 - 100 i₃

1100 i₃ = 30

using this value in eqn of i₂:

i₂ = [6 - 100(0.02727)]/300

i₂ = (6 - 2.727)/300

Since:

i₁ = i₂ + i₃

i₁ = 0.01091 A + 0.02727 A

<u></u>

(a)

(b)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>downward</u>

(c)

(d)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>rightward</u>

(e)

(f)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>leftward</u>

With respect to the grounded portion, the potential drop at the resistance 1 will be equal to the potential at A Eₐ.

Therefore,

Eₐ = i₁R₁

Eₐ = (0.03818 A)(100 Ω)

<u>Eₐ = 3.818 Volts</u>

3 0

4 years ago