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Alika [10]
1 year ago
12

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
Physics
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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Let us first know the given: Tennis ball has a mass of 0.003 kg, Soccer ball has a mass of 0.43 kg. Having the same velocity at 16 m/s. First the equation for momentum is P=MV P=Momentum M=Mass V=Velocity. Now let us have the solution for the momentum of tennis ball. Pt=0.003 x 16 m/s= (    kg-m/s ) I use the subscript "t" for tennis.  Momentum of Soccer ball Ps= 0.43 x 13m/s = (      km-m/s). If we going to compare the momentum of both balls, the heavier object will surely have a greater momentum because it has a larger mass, unless otherwise  the tennis ball with a lesser mass will have a greater velocity to be equal or greater than the momentum of a soccer ball.
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At an oceanside nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water th
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Answer is 30. just took it
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Four identical pallets of bricks, each with a mass of 40 kg with a square cross-sectional area of 0.50 m2, are stacked on the fl
STatiana [176]

Answer:

Explanation:

Modulus of elasticity of brick, Y = 28 x 10^9 Pa

mass of each, m = 40 kg

Area of each, A = 0.5 m²

number of bricks, n = 4

height, h = 5 m

Let the change in height is Δh.

Use the formula of Modulus of elasticity

E = stress/ strain

stress = n x m x g/A = 4 x 40 x 9.8 / 0.5 = 3136 Pa

So,

3136 = 28 \times  10^{9}\times \frac{\Delta h}{5}

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4 0
2 years ago
11. A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is finge
Anna35 [415]

Answer:

The frequency is  f_n  = 257.1 \ Hz

 

Explanation:

From the question we are told that

    The third harmonic frequency of the tight guitar string is  f_3 = 540 \ Hz

     

Let the original length be  L  

   Then the length at which it is fingered is  0.7 L

Generally the fundamental  is mathematically represented as

         f =  \frac{v_s}{ 2L}

Now when it finger at 70% it original length is

      f_n  =  \frac{v}{2 *  (0.7 L)}

      f_n  =  \frac{v}{1.4 L}

Here v  the velocity of sound

  So  

         \frac{f_n}{f}  =  \frac{\frac{v}{1.4L} }{\frac{v}{2L} }

Also the fundamental frequency for the original length can also be represented as

       f =  \frac{f_3}{3}

substituting values

          f =  \frac{540}{3}

          f = 180 \ Hz

So

       \frac{f_n}{180}  =  \frac{\frac{v}{1.4L} }{\frac{v}{2L} }

=>  f_n  =\frac{180}{0.7}

=>   f_n  = 257.1 \ Hz

 

     

3 0
1 year ago
A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b
Softa [21]

Answer:

  The net force on the block  F(net)  = mgsinθ).

   Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw  is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

                    F(net)  = mgsinθ

The net force on the block  F(net)  = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

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The horizontal component of normal force on the block is equal to force

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Substitute mgcosθ for n in the above equation;

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Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

6 0
1 year ago
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