150 ml of water is heated with a burning marshmallow From 20 c to 93.5 C

A string is wrapped around a uniform cylinder of mass M and radius R. The cylinder is released from rest with the string vertica

VARVARA [1.3K]

Answer:

a.) W/3, b.)2g/3 c.) (4gh/3)^0.5

Explanation:

First we have to find tension in terms of torque. To do that we have to find the moment of inertia of a rigid cylinder. From Wikipedia I get:

$I = 0.5mr^2$

We also know the equation for torque. Let T be the tension, r be the distance, I is the moment of inertia and alpha is angular acceleration (ignore theta because it is perpendicular)

$torque = I*\alpha = T*r$

We can then substitute a/r for α

Therefore we get:

$I*\frac{a}{r}=T*r$

Isolating T and substitute the moment of inertia in for I we get

$T = [0.5mr^2]a/r^2= 0.5ma$

There are two known forces acting on the cylinder, gravity and tension. The sum of these two forces gives us mass times acceleration (Newton's second law)

$F = ma => mg - 0.5ma = ma\\g=\frac{3}{2}a\\a = \frac{2}{3}g$

This allows us to plug acceleration back into Newton's Second Law:

$mg - T = m[\frac{2}{3}g]\\ T = \frac{1}{3} mg = \frac{1}{3}w$

w = the weight

For part b, we solved in part a:

$a = \frac{2}{3}g$

For part c, we use the conservation of energy. We know that the sum of energy in the system is zero.

$mgh + \frac{1}{2} mv^2 + \frac{1}{2}I(omega)^2 = 0\\(omega) = (\frac{v}{r})^2\\ I = \frac{1}{2} mr^2\\-gh +\frac{1}{2} v^2+\frac{1}{4} v^2 = 0\\gh = \frac{3}{4} v^2\\v = \sqrt{ \frac{4}{3}gh }$

8 0

3 years ago

Any help is appreciated

sdas [7]

image distance,di=10 cm

object distance,do=20cm

magnification, m=di/do

=10/20

=0.5

since the image is virtual, magnification is negative.

therefore m=-0.5

8 0

2 years ago

A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge p

Rufina [12.5K]

Answer:

0.0072 m³/s

Explanation:

Using Bernoulli's law

P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂ since the pipe is horizontal

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

flow rate is constant

A₁v₁ = A₂v₂

A₁ = πr₁² = π (0.06/2)² = 0.0028278 m²

A₂ = πr₂² = π (0.0225)² = 0.00159 m²

v₁ = (A₂ / A₁)v₂

v₁ = (0.00159 m²/ 0.0028278 m²) v₂ = 0.562 v₂

substitute v₁ into the Bernoulli's equation

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

500 ( 1 - 0.3161 ) v₂² = (31.0 - 24 ) × 10³ Pa

341.924 v₂² = 7000

v₂² = 20.472

v₂ = √ 20.472 = 4.525 m/s

volume follow rate = 0.00159 m² × 4.525 m/s = 0.0072 m³/s

8 0

3 years ago

How much heat would be absorbed by 75.20 g of iron when heated from 22 C to 28 C

Bumek [7]

Mass x SH x °C (or K) ΔT

= 75g x 0.45J/g/K x 6.0 ΔT

= 202.5 Joules of heat absorbed.

(202.5J / 4.184J/cal = 48.4 calories).

I guess that is the answer

8 0

3 years ago

Read 2 more answers

Rocks A and B are located at the same height on top of a hill. The mass of rock A is twice the mass of rock B. How does the pote

Ronch [10]

Potential energy is defined by formula

$U = mgh$

here

m = mass

g = acceleration due to gravity

h = height

Now here two different stones are located at same height

while mass of stone A is twice that of stone B

so here we can say potential energy of A is

$U_a = (2m)gh$

Similarly potential energy of B is

$U_b = mgh$

now if we take the ratio of two energy

$\frac{U_a}{U_b} = 2$

so we can say potential energy of stone A is two times the potential energy of B

7 0

3 years ago