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steposvetlana [31]
2 years ago
7

Which line represents a stationary object? Position Time

Physics
1 answer:
AlladinOne [14]2 years ago
7 0

Answer:

Green one. Red is acc. Blue is Dec and other pink one is constant speed I think

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Which sector is not part of the circular flow model of the economy ?
morpeh [17]

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a. government sector

Explanation:

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A helicopter is hovering at a constant height of 35 m. The upward lift force on the helicopter is 85500. What is the weight of t
Vika [28.1K]
2,442! I think..right ok
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2 years ago
Two steel balls, of masses m1=1.00 kg and m2=2.00 kg, respectively, are hung from the ceiling with light strings next to each ot
Zolol [24]

Answer:

(a) The maximum height achieved by the first ball, m₁ is 0.11 m

(a) The maximum height achieved by the second ball, m₂ ball is 0.44 m

Explanation:

Given;

mass of the first ball, m₁ = 1 kg

mass of the second ball, m₂ = 2 kg

The velocity of the first when released from a height of 1 m before collision;

u₁² = u₀² + 2gh

u₀ = 0, since it was released from rest

u₁² =  2gh

u₁² = 2 x 9.8 x 1

u₁² = 19.6

u₁ = √19.6

u₁ = 4.427 m/s

The velocity of the second ball before collision, u₂ = 0

Apply the principle of conservation of linear momentum, to determine the velocity of the balls after an elastic collision.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

v₁ is the final velocity of the first ball after an elastic collision

v₂ is the final velocity of the second ball after an elastic collision

m₁u₁ + m₂(0) = m₁v₁ + m₂v₂

m₁u₁ =  m₁v₁ + m₂v₂

1 x 4.427 = v₁ + 2v₂

v₁ + 2v₂ = 4.427

v₁  = 4.427 - 2v₂  ----- equation (1)

one directional velocity;

u₁ + v₁ = u₂ + v₂

u₂ = 0

u₁ + v₁ = v₂

v₁ = v₂ - u₁

v₁ = v₂ - 4.427 ------ equation (2)

Substitute v₁ into equation (1)

v₂ - 4.427 = 4.427 - 2v₂

3v₂ = 4.427 + 4.427

3v₂  = 8.854

v₂ = 8.854 / 3

v₂  = 2.95 m/s (→ forward direction)

v₁ = v₂ - 4.427

v₁ = 2.95 - 4.427

v₁  = - 1.477 m/s

v₁  = 1.477 m/s ( ← backward direction)

Apply the law of conservation of mechanical energy

mgh_{max} = \frac{1}{2}mv_{max}^2

(a) The maximum height achieved by the first ball (v₁  = 1.477 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(1.477^2)\\\\ h_{max}  = 0.11 \ m

(b) The maximum height achieved by the second ball (v₂  = 2.95 m/s)

mgh_{max} = \frac{1}{2}mv_{max}^2 \\\\gh_{max} = \frac{1}{2}v_{max}^2\\\\ h_{max}  =  \frac{1}{2g}v_{max}^2\\\\ h_{max}  = \frac{1}{2*9.8}(2.95^2)\\\\ h_{max}  = 0.44 \ m

6 0
3 years ago
in a lever, a load of 600N is lifted by using 400 effort. If the load is at the distance of 20cm and the effort at the distance
koban [17]

Explanation:

Load (l) = 680N

Effort (E) = 500N

Length slope (l) = 12m

Height slope (h) = 8 m

Output = load * height

680 *8 = 5.44 *103 J

The Input = effort * length = 500 *12 = 6000J

the Mechanical advantage (M.A) = load effort= 600500=1.36

the Velocity ratio (V.R) =lh=128 = 1.5

the Efficiency =M.A100%V.R= 90.6%

8 0
2 years ago
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