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3 years ago

A pulley with the radius of 10.0 cm was connected to a motor with a massless

Physics

1 answer:

kogti [31]3 years ago

8 0

Answer:

(i) -556 rad/s²

(ii) 17900 revolutions

(iii) 11250 meters

(iv) -55.6 m/s²

(v) 18 seconds

Explanation:

(i) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

α = (10000 − 15000) / 9

α ≈ -556 rad/s²

(ii) Constant acceleration equation:

θ = θ₀ + ω₀ t + ½ αt²

θ = 0 + (15000) (9) + ½ (-556) (9)²

θ = 112500 radians

θ ≈ 17900 revolutions

(iii) Linear displacement equals radius times angular displacement:

s = rθ

s = (0.100 m) (112500 radians)

s = 11250 meters

(iv) Linear acceleration equals radius times angular acceleration:

a = rα

a = (0.100 m) (-556 rad/s²)

a = -55.6 m/s²

(v) Angular acceleration is change in angular velocity over time.

α = (ω − ω₀) / t

-556 = (0 − 15000) / t

t = 27

t − 9 = 18 seconds

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A long distance runner sees the finish line and accelerates at a rate in 1.2 m/s2 for

Nuetrik [128]

Answer: he did travel 15 meters.

Explanation:

We have the data:

Acceleration = a = 1.2 m/s^2

Time lapes = 3 seconds

Initial speed = 3.2 m/s.

Then we start writing the acceleration:

a(t) = 1.2 m/s^2

now for the velocity, we integrate over time:

v(t) = (1.2 m/s^2)*t + v0

with v0 = 3.2 m/s

v(t) = (1.2 m/s^2)*t + 3.2 m/s

For the position, we integrate again.

p(t) = (1/2)*(1.2 m/s^2)*t^2 + 3.2m/s*t + p0

Because we want to know the displacementin those 3 seconds ( p(3s) - p(0s)) we can use p0 = 0m

Then the displacement at t = 3s will be equal to p(3s).

p(3s) = (1/2)*(1.2 m/s^2)*(3s)^2 + 3.2m/s*3s = 15m

6 0

3 years ago

Which is a characteristic of thermal energy transfer through convection

Lynna [10]

Answer: The thermal energy transfer is When a fluid, such as air or a liquid, is heated and then travels away from the source, it carries the thermal energy along.

Explanation: heat transfer is called convection. hopefully this was helpful.

6 0

3 years ago

a 2.0-mole sample of an ideal gas is gently heated at constant temperature 330 k. it expands from initial volume 19 l to final v

shutvik [7]

Isothermal Work = PVln(v₂/v₁)

PV = nRT = 2 mole * 8.314 J/ (k.mol) * 330 k = 5487.24 J

Isothermal Work = PVln(v₂/v₁) v₂ = ? v₁ = 19L,

1.7 kJ = (5487.24)In(v₂/19)

1700 = (5487.24)In(v₂/19)

In(v₂/19) = (1700/5487.24) = 0.3098

In(v₂/19) = 0.3098

(v₂/19) = $e^{0.3098}$

v₂ = 19* $e^{0.3098}$

v₂ = 25.8999

v₂ ≈ 26 L Option b.

6 0

3 years ago

Two concrete spans of a 380 m long bridge are

Mazyrski [523]

Answer:

4.163 m

Explanation:

Since the length of the bridge is

L = 380 m

And the bridge consists of 2 spans, the initial length of each span is

$L_i = \frac{L}{2}=\frac{380}{2}=190 m$

Due to the increase in temperature, the length of each span increases according to:

$L_f = L_i(1+ \alpha \Delta T)$

where

$L_i = 190 m$ is the initial length of one span

$\alpha =1.2\cdot 10^{-5} ^{\circ}C^{-1}$ is the temperature coefficient of thermal expansion

$\Delta T=20^{\circ}C$ is the increase in temperature

Substituting,

$L_f=(190)(1+(1.2\cdot 10^{-5})(20))=190.0456 m$

By using Pythagorean's theorem, we can find by how much the height of each span rises due to this thermal expansion (in fact, the new length corresponds to the hypothenuse of a right triangle, in which the base is the original length of the spand, and the rise in heigth is the other side); so we find:

$h=\sqrt{L_f^2-L_i^2}=\sqrt{(190.0456)^2-(190)^2}=4.163 m$

4 0

3 years ago

A real gas will behave most like an ideal gas under conditions of ________.

KengaRu [80]

Answer: high temperature and low pressure

Explanation:

The Ideal Gas equation is:

$P.V=n.R.T$

Where:

$P$ is the pressure of the gas

$V$ is the volume of the gas

$n$ the number of moles of gas

$R=0.0821\frac{L.atm}{mol.K}$ is the gas constant

$T$ is the absolute temperature of the gas in Kelvin

According to this law, molecules in gaseous state do not exert any force among them (attraction or repulsion) and the volume of these molecules is small, therefore negligible in comparison with the volume of the container that contains them.

Now, real gases can behave approximately to an ideal gas, under the conditions described above and taking into account the following:

When <u>temperature is high</u> a real gas approximates to ideal gas, because the molecules move quickly, preventing the repulsion or attraction forces to take effect. In addition, at <u>low pressures</u>, the volume of molecules is negligible.

4 0

3 years ago