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3 years ago

1. How much heat energy is required to raise the temperature of a 5 kg aluminium bar

Physics

1 answer:

Ray Of Light [21]3 years ago

5 0

Answer:

180 kJ

Explanation:

Given that:

Mass (m) = 5 kg

Initial temperature (T1) = 28°C

Final temperature (T2) = 68°C

The change in temperature (ΔT) = T2 - T1 = 68°C - 28°C = 40°C

Specific heat capacity of aluminium (c) = 900 J/kg°C

The quantity of heat energy required (q) is given by:

q = mcΔT

q = 5 kg × 900 J/kg°C × 40°C

q = 180000 Joules

q = 180 kJ

Therefore 180 kJ is required to raise the temperature of aluminium from 28°C to 68°C.

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Calculate the volume of this regular solid,

Oduvanchick [21]

Answer:

2144.7

Explanation:

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2 years ago

A small ferryboat is 4.70 m wide and 6.10 m long. When a loaded truck pulls onto it, the boat sinks an additional 5.00 cm into t

geniusboy [140]

Answer:

M = 1433.5 kg

Explanation:

This exercise is solved using the Archimedean principle, which states that the hydrostatic thrust is equal to the weight of the desalinated liquid,

B = ρ g V

with the weight of the truck it is in equilibrium with the push, we use Newton's equilibrium condition

Σ F = 0

B-W = 0

B = W

body weight

W = M g

the volume is

V = l to h

rho_liquid g (l to h) = M g

M = rho_liquid l a h

we calculate

M = 1000 4.7 6.10 0.05

M = 1433.5 kg

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3 years ago

You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

bezimeni [28]

Answer:

a) The resulting angular speed of platform is 1.38 rev/sec

b) The change in kinetic energy of the system is 53 J.

Explanation:

This question is incomplete. The complete question will be:

You stand on a frictional platform that is rotating at 1.1 rev/s. Your arms are outstretched, and you hold a heavy weight in each hand. The moment of inertia of you, the extended weights, and the platform is 8.8 kg · m2. When you pull the weights in toward your body, the moment of inertia decreases to 7.0 k g .m 2

a) What is the resulting angular speed of the platform? Answer in units of r e v / s .

b)What is the change in kinetic energy of the system? Answer in units of J.

<h3>ANSWER:</h3>

we know that:

Angular Momentum = L = Iω

From conservation of momentum:

Lo = Lf

(Io) (ωo) = (If) (ωf)

ωf = (Io) (ωo)/(If)

ωf = (8.8 kg.m²)(1.1 rev/s)/(7.0 kg.m²)

ωf = (1.38 rev/sec)(2π rad/ 1 rev) = 8.67 rad/sec

ωo = (1.1 rev/sec)(2π rad/ 1 rev) = 6.91 rad/sec

The kinetic energy for rotational motion is given as:

K.E = (1/2)Iω²

Thus, the change in kinetic energy will be:

ΔK.E = (K.E)f - (K.E)o

ΔK.E = (1/2)Ifωf² - (1/2)Ioωo²

ΔK.E = (1/2)(Ifωf² - Ioωo²)

ΔK.E = (1/2)[(7 kg.m²)(8.67 rad/sec)² - (8.8 kg.m²)(6.91 rad/sec)²

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2 years ago

What is the American psychology Association ethical principle of psychology and code of conduct based upon

solmaris [256]

The correct answer is B

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3 years ago

The wheel having a mass of 100 kg and a radius of gyration about the z axis of kz=300mm, rests on the smooth horizontal plane.a.

pickupchik [31]

Answer:

a) 20 rad/s

b) 6 m/s

Explanation:

b) Force acting on the wheel is 200 N

mass of the wheel is 100 kg

From Newton's second law of motion, F = m × a

Where F is the net force acting on the body

m is mass of the body

a is the acceleration of the body

By substituting the values we get, a = 2 m/s²

As acceleration is constant, we can use the below formula for calculating the final velocity of the object

v = u + a × t

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

u = 0 (∵ it starts from rest)

By substituting the values we get

v = 0 + 2 × 3 = 6 m/s

∴ Speed of center of mass after 3 seconds = 6 m/s

a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel

∴ Radius of the wheel = 300 mm

Torque acting on the wheel about axis of rotation = 300 mm × 200 N =

60 N·m

Torque = (Moment of inertia) × (angular acceleration)

Assuming that the mass of spokes of the wheel to be negligible,

Moment of inertia of the wheel about axis of rotation = 100 × 300² × $10^{-6}$ = 9 kg·m²

Then,

60 = 9 × (angular acceleration)

∴ angular acceleration ≈ 6·67 rad/s²

As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed

$w_{f}$ = $w_{i}$ + α × t

Where

$w_{f}$ is the final angular velocity

$w_{i}$ is the initial angular velocity

α is the angular acceleration

t is the time taken

$w_{i}$ is 0 (∵ initially it starts from rest)

By substituting the values we get

$w_{f}$ = 6·67 × 3 = 20 rad/s

∴ Angular velocity of the wheel after three seconds = 20 rad/s

3 0

3 years ago