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olga2289 [7]
3 years ago
7

40 Newtons

Physics
1 answer:
Kobotan [32]3 years ago
6 0
Although the weight of the spears a different the gravitational forces are the same
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You and a friend each carry a 15 kg suitcase up two flights of stairs, walking at a constant speed. Take each suitcase to be the
AlekseyPX

Answer:

Both of you did the same work but you expended more power.

Explanation:              

<em>Work done</em> by an object is calculated by force applied multiplied by the distance.

  W=F*d

From the figure given below let us calculate force applied bith you and yopur friend.

Let us take the stairs in positive x direction,

Work done by you W₁ ,

The force applied Fₓ = F - mgsinθ =maₓ

here aₓ = 0, because both of you move with constant speed

F - mgsinθ = 0

F=  mgsinθ

The work done by you on the suitcase is

W = F L cos0°  ,    where L is he length of the staircase.

W = FL = mgsinθL ,  by substituting value of F

Work done by you is W₁ = mgLsinθ

Similarly work done by your friend is W₂ = mgLsinθ.

Because both of you carry suitcase of same weight and in staircase is in same angle the force applied is same .

Therefore <em>work done by both of you is same</em> . Both of you did equal work.

The power , is defined as amount of energy converted or transfered per second or rate at which work is done .

P =\frac{W}{t} =\frac{FL}{t}

Power spend by you P₁ = mgLsinθ/t

P₁ = 15*9.8*Lsinθ/30

P₁ = 4.9L sinθ  eqn 1

Power spend by your friend is P₂ = mgLsinθ/t

P₂ =15*9.8*Lsinθ/60

P₂ = 2.45Lsinθ    eqn 2

Dividing eqn 1 and eqn 2

P₁ = 2P₂

You have spend more power than your friend .

Hence Both of you did equal work but you spend more power.

7 0
4 years ago
What is the correct name for this formula: SiO2?
Brrunno [24]
Silicon Dioxide is the answer
6 0
4 years ago
Read 2 more answers
Two objects A and B (both masses are m) are connected by a spring of elastic constant k and placed on a frictionless surface. At
umka2103 [35]

Answer:

a)   Em_f = ½ m vₐ²,  b)    v = vₐ /√2

Explanation:

a) For this part we use conservation of energy

   starting point .. block A moving, spring unstretched

          Em₀ = K = ½ m vₐ²

end point. Stretched spring

          Em_f = K_e = ½ k x²

energy is conserved

          Em₀ = Em_f

          Em_f = ½ m vₐ²

b) Let's analyze the movement a little, block A began to move at a speed va, I stretch the spring an amount Dx, it exerts a force on block b that begins to move and the elongation of the spring decreases.

In all this process there is no friction force, therefore the energy is conserved, therefore the maximum energy stored in the spring must be distributed among the bodies.

           Em₀ = K_e = E₀

where E₀ is the initial energy of block a

            E₀ = ½ m vₐ²

At the point where we are in equilibrium

          Em_f = Kₐ + K_b = ½ m vₐ² +1/2 m v_{b}^2

so that the spring does not stretch or shrink, the two bodies must go at the same speed.

          Em_f = m v²

energy is conserved

          Em₀ = Em_f

          ½ m vₐ² = m v²

           v = vₐ /√2

therefore both blocks must go at this speed

7 0
3 years ago
Greg’s mom baked 24 brownies for the bake sale. Vinnie’s mom
AysviL [449]

(24 + 36) x 2 = 120

simple math. you're welcome.

"college student"

4 0
3 years ago
Read 2 more answers
There are (one can say) three coequal theories of motion for a single particle: Newton's second law, stating that the total forc
PtichkaEL [24]

Answer:

vf = 14.2176 m/s

Explanation:

Given

m = 4 Kg

viy = 7.00 ĵ m/s

Fx = 11.0 î N

t = 4.5 s

vf = ?

Using the Impulse - Momentum Theorem, we have

F*Δt = m*Δv    ⇒  F*Δt = m*(vf - vi)

⇒    vf = (F*Δt + m*vi) / m

⇒    vf = (F*Δt + m*vi) / m

For <em>x-component</em>

⇒    vfx = (Fx*Δt + m*vix) / m = (11 N*4.5 s + 4 Kg*0 m/s) / (4 Kg)

⇒    vfx = 12.375 î m/s

For <em>y-component</em>

⇒    vfy = (Fy*Δt + m*viy) / m = (0 N*4.5 s + 4 Kg*7 m/s) / (4 Kg)

⇒    vfy = 7 ĵ m/s

Finally:

vf = √(vfx² + vfy²)

⇒   vf = √((12.375 m/s)² + (7 m/s)²)

⇒   vf = 14.2176 m/s

8 0
3 years ago
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