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2 years ago

To protect her new two-wheeler, Iroda Bike

Physics

1 answer:

goldfiish [28.3K]2 years ago

6 0

Answer:

The length of chain she is allowed is 1.169 ft

Explanation:

The given parameters are;

The linear density of the chain = 0.83 lb/ft

The weight limit of the chain she wants = 1.4 lb

The formula for linear density = Weight/length

Therefore, in order to keep the chain below 1.4 lb, we have;

Linear density = Weight/length

Therefore;

The maximum length she wants = Weight/(Linear density)

Which gives;

The maximum length she wants = 1.4 lb/(0.83 lb/ft) =1.169 ft

Therefore;

The length of chain she is allowed = 1.169 ft.

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Label and describe what is happening in this picture

SOVA2 [1]

Something is reproducing.

7 0

3 years ago

The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve

Katena32 [7]

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

$\varepsilon _t=\dfrac{D-d}{d}$

$\varepsilon _t=\dfrac{32-30}{30}$

$\varepsilon _t=0.0667$

We know that

$n=-\dfrac{\epsilon _t}{\varepsilon _{long}}$

$\varepsilon _{long}=-\dfrac{\epsilon _t}{n}$

$\varepsilon _{long}=-\dfrac{0.0667}{0.45}$

$\varepsilon _{long}=-0.148$

So the axial pressure

$P=E\times \varepsilon _{long}$

$P=5\times 0.148$

P=740 KPa

The movement in the sleeve

$\Delta =\varepsilon _{long}\times L$

$\Delta =0.148\times 50$

Δ=7.4 mm

6 0

3 years ago

A 1451 kg car is traveling at 48.0 km/h. How much kinetic energy does it possess? K.E. =

jarptica [38.1K]

   Kinetic energy = (1/2) (mass) (speed)²

BUT . . . in order to use this equation just the way it's written,
the speed has to be in meters per second. So we'll have to
make that conversion.

KE = (1/2) · (1,451 kg) · (48 km/hr)² · (1000 m/km)² · (1 hr/3,600 sec)²

   = (725.5) · (48 · 1000 · 1 / 3,600)² (kg) · (km·m·hr / hr·km·sec)²

   = (725.5) · ( 40/3 )² · ( kg·m² / sec²)

   = 128,978 joules (rounded)

8 0

2 years ago

A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w

amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

10 x 3 + 25 x 0 = v(10 + 25)

30 = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

$K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\$

the final kinetic energy of the package;

$K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J$

the fraction of the initial energy lost;

$= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71$

7 0

3 years ago

I need help with question 2 and 2 please

vitfil [10]

Alright here the answer to number 2

5 0

3 years ago