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AleksandrR [38]
4 years ago
6

When an object of weight w is suspended from the center of a massless string?

Physics
2 answers:
pickupchik [31]4 years ago
6 0
<span>An object suspended from its own center will not have any motion or rotation unless acted on by an outside force. In the question the idea of a massless string is introduced. The mass of the string is pointless. In this trivial example the weight w did not need to be defined as "w" as there is only one object considered.</span>
igomit [66]4 years ago
4 0
I attached a picture of the diagram associated with this question.

Now,
When we check the vertical components of the tension in the rope, we will find that we have two equal components acting upwards.
These two components support the weight and each of them has a value of TcosΘ

The net force acting on the body is zero.
Fnet=Force of tension acting upwards-Force due to weight acting downwards
0 = 2TcosΘ -W
W = 2TcosΘ
T = W / 2cosΘ

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Depends

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A data table shows the height of a person on his birthday each year for ten years. What is the the dependent variable? Provide e
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3 years ago
Read 2 more answers
Please Help. I would Really Appreciate it
Readme [11.4K]

Answer:

a) 0.60 kg cart has final velocity 3.0 m/s [E]

0.80 kg cart has final velocity 4.0 m/s [W]

b) 0.12 m

Explanation:

Take east to be positive.

a) Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.60) (-5.0) + (0.80) (2.0) = (0.60) v₁ + (0.80) v₂

-1.4 = 0.6 v₁ + 0.8 v₂

Kinetic energy is conserved in elastic collisions.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(0.60) (-5.0)² + (0.80) (2.0)² = (0.60) v₁² + (0.80) v₂²

18.2 = 0.6 v₁² + 0.8 v₂²

Solve the system of equations.

-1.4 = 0.6 v₁ + 0.8 v₂

-1.4 − 0.6 v₁ = 0.8 v₂

-1.75 − 0.75 v₁ = v₂

18.2 = 0.6 v₁² + 0.8 (-1.75 − 0.75 v₁)²

18.2 = 0.6 v₁² + 0.8 (3.0625 +2.625 v₁ + 0.5625 v₁²)

182 = 6 v₁² + 8 (3.0625 + 2.625 v₁ + 0.5625 v₁²)

182 = 6 v₁² + 24.5 + 21 v₁ + 4.5 v₁²

0 = 10.5 v₁² + 21 v₁ − 157.5

0 = v₁² + 2 v₁ − 15

0 = (v₁ − 3) (v₁ + 5)

v₁ = 3 or -5

Since u₁ = -5.0 m/s, v₁ must be 3.0 m/s.

Solving for v₂:

v₂ = -0.75 v₁ − 1.75

v₂ = -4.0 m/s

b) The compression of the spring is a maximum when the carts have the same velocity.

Momentum is conserved.

m₁u₁ + m₂u₂ = (m₁ + m₂) v

(0.60) (-5.0) + (0.80) (2.0) = (0.60 + 0.80) v

-1.4 = 1.4 v

v = -1.0

Energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ (m₁ + m₂) v² + ½ kx²

m₁u₁² + m₂u₂² = (m₁ + m₂) v² + kx²

(0.60) (-5.0)² + (0.80) (2.0)² = (0.60 + 0.80) (-1.0)² + (1200) x²

18.2 = 1.4 + 1200 x²

16.8 = 1200 x²

x² = 0.014

x = 0.12

5 0
4 years ago
I will pick the branliest if answered these questions.
Bond [772]

1.

n₁ = index of refraction of diamond = 2.4

n₂ = index of refraction of water = 1.33

θ₁ = angle of incidence = 24 deg

θ₂ = angle of refraction = ?

using snell's law

n₁ Sinθ₁ = n₂ Sinθ₂

inserting the values

(2.4) Sin24 = (1.33) Sinθ₂

θ₂ = 47.22 deg


2.

c = specific heat = 4.18

m = mass of water = 3.5 kg

ΔT = change in temperature = 55 - 25 = 30 C

Q = heat taken

heat taken is given as

Q = m c ΔT

inserting the values

Q = (3.5) (4.18) (30)

Q = 439 J


3.

n = number of moles = 1

m = molar mass = 0.0399 kg

T = temperature = 27 K

v = average velocity

average velocity is given as

v = sqrt(3RT/m)

v = sqrt(3 x 8.314 x 275/0.0399)

v = 414.6m/s


4.

n = number of moles = 2

T = temperature = 35 C  = 35 + 273 K = 308 K

U = internal energy

internal energy is given as

U = 2.5 n RT

U = 2.5 (2) (8.314) (308

U = 1.28 × 10⁴ J


5.

F₁ = 3300 N

F₂ = ?

A₁ = 0.060 m²

A₂ = 0.18 m²

Using pascal's law

F₁/A₁ = F₂/A₂

3300/0.060 = F₂/0.18

F₂ = 9900 N


6.

R = resistance of each resistance in series = 10 ohm

R' = equivalent resistance in series

equivalent resistance is given as

R' = 3 R

R' = 3 x 10

R' = 30 ohm


7.

R = resistance of each resistance in parallel = 10 ohm

R'' = equivalent resistance in parallel

equivalent resistance is given as

R'' = R/3

R'' = 10/3

R'' = 3.3 ohm


8.

for series circuit

R' = equivalent resistance in series = 30 ohm

V = Voltage applied = 60 Volts

i' = current in each resistor in series

current in each resistor in series is given as

I' = V/R'

i' = 60/30

i' = 2 A


for parallel circuit :

i = current in each resistor = V/R = 60/10 = 6 A








6 0
3 years ago
Read 2 more answers
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