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2 years ago

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A yo-yo can be thought of as a solid cylinder of mass m and radius r that has a light string wrapped around its circumference (s

ee below). One end of the string is held fixed in space. If the cylinder falls as the string unwinds without slipping, what is the acceleration of the cylinder? (Enter the magnitude. Use the following as necessary: g, m, and r.)

1 answer:

lbvjy [14]2 years ago

4 0

Answer:

6.5 m/s^2

Explanation:

The net force acting on the yo-yo is

F_net = mg-T

ma=mg-T

now T= mg-ma

net torque acting on the yo-yo is

τ_net = Iα

I= moment of inertia (= 0.5 mr^2 )

α = angular acceleration

τ_net = 0.5mr^2(a/r)

Tr= 0.5mr^2(a/r)

(mg-ma)r=0.5mr^2(a/r)

a(1/2+1)=g

a= 2g/3

a= 2×9.8/3 = 6.5 m/s^2

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According to the work-energy theorem, if work is done on an object, its potential and/or kinetic energy changes. Consider a car

Leni [432]

Answer:

The force of the car engine.

Explanation:

The work- energy theorem states that the work done on an object is equal to the change in its kinetic energy. Its expression is given by :

$W=\dfrac{1}{2}m(v^2-u^2)$

Also, W = F.d

$Fd=\dfrac{1}{2}m(v^2-u^2)$

Where

F is the force applied by the engine of car

d is the displacement

m is the mass of an object

u is the initial speed

v is the final speed

So, the force of the car engine increased the car’s kinetic energy. Hence, this is the required solution.

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2 years ago

If 56 grams of carbon monoxide burns in oxygen to produce 88 grams of carbon dioxide, the mass of oxygen involved in the reactio

pentagon [3]

Answer : The mass of oxygen involved in the reaction is, 32 grams.

Explanation : Given,

Mass of carbon monoxide = 56 g

Mass of carbon dioxide = 88 g

Molar mass of carbon monoxide $(CO)$ = 28 g/mole

Molar mass of oxygen $(O_2)$ = 32 g/mole

First we have to calculate the moles of carbon monoxide.

$\text{Moles of CO}=\frac{\text{Mass of CO}}{\text{Molar mass of CO}}=\frac{56g}{28g/mole}=2moles$

Now we have to calculate the moles of oxygen gas.

The balanced chemical reaction will be,

$2CO+O_2\rightarrow 3CO_2$

From the balanced chemical reaction, we conclude that

2 moles of CO react with 1 mole of $O_2$

So, the moles of $O_2$ = 1 mole

Now we have to calculate the mass of oxygen.

$\text{Mass of }O_2=\text{Moles of }O_2\times \text{Molar mass of }O_2$

$\text{Mass of }O_2=1mole\times 32g/mole=32g$

Therefore, the mass of oxygen involved in the reaction is, 32 grams.

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2 years ago

Which shows the correct lens equation? The inverse of f equals the inverse of d Subscript o Baseline times the inverse of d Subs

musickatia [10]

Answer:

The inverse of f equals the inverse of d Subscript o Baseline plus the inverse of d Subscript I Baseline.

Explanation:

The lens equation shows the relation among focal length of the lens, image distance and object distance. It can be expressed as:

$\frac{1}{f}$ = $\frac{1}{d_{o} }$ + $\frac{1}{d_{I} }$

where: f is the focal length of the lens, $d_{o}$ is the object distance to the lens and $d_{I}$ is the image distance to the lens.

The lens equation can be used to determine the unknown value among the variables f , $d_{o}$ and $d_{o}$ .

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2 years ago

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Two objects exert a gravitational force on 8 N on one another. What would that force be if the mass of BOTH objects were doubled

seropon [69]

<span>Based on Newton's law of universal gravitation, the equation for the gravitational force exerted by an object on another object is given by:
F = Gm1m2/(r^2)
where G is the universal gravitational constant, F is the gravitational force exerted, m1 is the mass of the first object, m2 is the mass of the second object, and r is the separation distance between the two objects.
If the mass of both objects were doubled, then we would have: m1' * m2' = (2m1) * (2m2) = 4m1m2. Assuming r stays constant (G is a constant so that won't change anyway), then this means that the new force will be 4 times greater, ie 8N * 4 = 32N of gravitational force. </span>

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2 years ago

The perihelion of the comet TOTAS is 1.69 AU and the aphelion is 4.40 AU. Given that its speed at perihelion is 28 km/s, what is

dybincka [34]

Answer:

The speed at the aphelion is 10.75 km/s.

Explanation:

The angular momentum is defined as:

$L = mrv$ (1)

Since there is no torque acting on the system, it can be expressed in the following way:

$t = \frac{\Delta L}{\Delta t}$

$t \Delta t = \Delta L$

$\Delta L = 0$

$L_{a} - L_{p} = 0$

$L_{a} = L_{p}$ (2)

Replacing equation 1 in equation 2 it is gotten:

$mr_{a}v_{a} =mr_{p}v_{p}$ (3)

Where m is the mass of the comet, $r_{a}$ is the orbital radius at the aphelion, $v_{a}$ is the speed at the aphelion, $r_{p}$ is the orbital radius at the perihelion and $v_{p}$ is the speed at the perihelion.

From equation 3 v_{a} will be isolated:

$v_{a} = \frac{mr_{p}v_{p}}{mr_{a}}$

$v_{a} = \frac{r_{p}v_{p}}{r_{a}}$ (4)

Before replacing all the values in equation 4 it is necessary to express the orbital radius for the perihelion and the aphelion from AU (astronomical units) to meters, and then from meters to kilometers:

$r_{p} = 1.69 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $2.528x10^{11} m$

$r_{p} = 2.528x10^{11} m x \frac{1km}{1000m}$ ⇒ $252800000 km$

$r_{a} = 4.40 AU x \frac{1.496x10^{11} m}{1 AU}$ ⇒ $6.582x10^{11} m$

$r_{p} = 6.582x10^{11} m x \frac{1km}{1000m}$ ⇒ $658200000 km$

Then, finally equation 4 can be used:

$v_{a} = \frac{(252800000 km)(28 km/s)}{(658200000 km)}$

$v_{a} = 10.75 km/s$

Hence, the speed at the aphelion is 10.75 km/s.

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2 years ago