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1 year ago

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A 69.0 kg ice skater moving to the right with a velocity of 2.61 m/s throws a 0.22 kg snowball to the right with a velocity of 2

5.2 m/s relative to the ground. (a) What is the velocity of the ice skater after throwing the snowball

1 answer:

Luda [366]1 year ago

4 0

Answer:

0.08m/s

Explanation:

Given data

M1= 69kg

v1= 2.61m/s

M2= 0.22kg

v2= 25.2m/s

Before snowball is thrown:

Total mass of skater + snowball = 69+ 0.22 = 69.22kg

Total Momentum of skater + snowball = mv = 69.22 x 2.61 = 180.7 kgm/s

After snowball is thrown:

Let's call the velocity of the skater V.

Total momentum = momentum of skater + momentum of snowball

=69.22V + (5.544)

= 69.22V + 5.544

So:

180.7 = 69.22V+5.544

180.7- 5.544= 69.22V

175.156= 69.22V

V= 175.156/69.22

V = 2.53m/s

The total momentum after catching the snowball is mV or:

(69.0 + 0.22) x V

So:

5.544= 69.22V

V= 5.544/69.22

V=0.08m/s

The velocity of the ice skater after throwing the snowball is 0.08m/s

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Answer:

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Explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I = $\frac{M}{3} (a^{2} +b^{2} )$ ---------------------------------1

(a) The kinetic energy of a parallel pipe is also given as;

k = $\frac{1}{2} Iw^{2}$ --------------------------------2

Putting equation 1 into equation 2, we have;

k = $\frac{M}{6} (a^{2} +b^{2} )w^{2}$

k = $\frac{Mw^{2} }{6} (a^{2} +b^{2} )$

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τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ = $\frac{Mw}{3} (a^{2} +b^{2} )$

But

τ = dτ/dt = $\frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}$ ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ = $\frac{M}{3} (a^{2} +b^{2} )$ ∝

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1 year ago

The temperature, T, of a gas is jointly proportional to the pressure P of the gas and the volume V occupied by the gas. Use C as

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Answer:

T=C*P*V

Explanation:

It is said that a variable - let's call 'y' -, is proportional to another - let's call it 'x' - if x and y are multiplicatively connected to a constant 'C'. It means that their product (x*y) can be always equaled to the constant 'C' or their division ( $\frac{x}{y}$ ) can be always equaled to 'C'. The first case is the case of the inverse proportionality: It is said that x and y are inversely proportional if

$x*y=C$

The second case is the case of the direct proportionality: It is said that x and y are directly proportional if

$\frac{x}{y} =C$ : x is directly proportional to y.

or

$\frac{y}{x} =C$ : y is directly proportional to x.

Always that any text does not specify about directly or inversely proportionality, it is assumed to mean directly automatically.

For our case, we are said that the temperature T is proportional to the pressure P and the volume V (we assume that it means directly); it is a double proportionality but follows the same rules:

If T were just proportional to P, we would have:

$\frac{T}{P} =C$

If T were just proportional to V, we would have:

$\frac{T}{V} =C$

As T is proportional to both P and V, the right equation is:

$\frac{T}{P*V}=C$

In order to isolate the temperature, let's multiply (P*V) at each side of the equation:

$\frac{T}{P*V}*(P*V)=C*(P*V)\\T=C*P*V$

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1 year ago

A 3.00-kg model airplane has velocity components of 5.00 m/s due east and 8.00 m/s due north. What is the plane’s kinetic energy

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Answer:

Kinetic energy, E = 133.38 Joules

Explanation:

It is given that,

Mass of the model airplane, m = 3 kg

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$v=\sqrt{v_1^2+v_2^2}$

$v=\sqrt{5^2+8^2}=9.43\ m/s$

Let E is the kinetic energy of the plane. It is given by :

$E=\dfrac{1}{2}mv^2$

$E=\dfrac{1}{2}\times 3\ kg\times (9.43\ m/s)^2$

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The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati

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Answer:

The tension in the rope is 229.37 N.

Explanation:

Given:

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Coefficient of static friction is, $\mu = 0.214$

Angle of inclination is, $\theta = 31.5$ °

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are $mg\cos \theta$ and normal $N$ . Now, there is no motion in the direction perpendicular to the inclined plane. So,

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The forces acting along the plane are $mg\sin \theta$ and frictional force, $f$ , down the plane and tension, $T$ , up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

$T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N$

Therefore, the tension in the rope is 229.37 N.

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1 year ago

Cass is walking her dog (Oreo) around the neighborhood. Upon arriving at Calina's house (a friend of Oreo's), Oreo turns part mu

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Answer:

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