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AlekseyPX
2 years ago
12

A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediat

ely after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction relative to the x-axis does the third piece move
Physics
2 answers:
mel-nik [20]2 years ago
8 0

Answer:

8.3 m/s, 2196 degree from + X axis

Explanation:

m = 320 g , u = 2 m/s along X axis

m'  = 355 g, u' = 1.5 m/s along Y axis

m'' = 100 g, u'' = v

Let the speed of the third piece is v makes an angle A from the X axis.

use conservation of momentum along X axis

0 = 320 x 2 + 100 x v cos A

v cos A =  - 6.4 ..... (1)

Use conservation of momentum along Y axis

0 = 355 x 1.5 + 100 x v sin A

v sinA = - 5.3 ... (2)

Squaring and adding

v^2 = (-6.4)^2 +(-5.3)^2\\\\v= 8.3 m/s

The angle is given by

tan A = \frac{-5.3}{-6.4}\\\\A = 219.6 degree from + X axis

Olegator [25]2 years ago
6 0

Answer:

Explanation:

There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.

momentum of first piece = .320 kg x 2 m/s

= 0.64 kg m/s along x -axis.

momentum of second piece = .355 kg x 1.5 m/s

= 0.5325 kg m/s along y- axis .

Let the velocity of third piece be v and it is making angle of θ with x -axis .

Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ

vertical  component of its velocity = .100 kg x v sinθ = .1 v sinθ

For making total momentum in the plane zero

.1 v cosθ = 0.64 kg m/s

.1 v sinθ = 0.5325 kg m/s

Dividing

Tanθ = .5325 / .64 = .83

θ = 40⁰.

The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.

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3 years ago
Steam at 4 MPa and 400C enters a nozzle steadily with a velocity of 60 m/s, and it leaves at 2 MPa and 300C. The inlet area of
frez [133]

Answer:

(A) 4.09 kg/s

(B) 589.9 m/s

(C)   0.0008707 m^{3} =  8.71 cm^{2}

Explanation:

inlet pressure of steam (P1) = 4 MPa

inlet temperature of steam (T1) = 400 degree celcius

inlet velocity (V1) = 60 m/s

outlet pressure (P2) = 2 MPa

outlet temperature (T2) = 300 degree celcius

inlet area (A1) = 50 cm^{2} = 0.005 m^{2}

rate of heat loss (Q) = 75 kJ/s

(A) mass flow rate (m) = \frac{A1 x V1}{α1}

where the initial specific volume (α1) for the given temperature and pressure is gotten from tables A-6 = 0.07343 m^3/kg

m = \frac{0.005 x 60}{0.07343}

m = 4.09 kg/s

(B) we can get the outlet velocity using the energy balance equation

  E in = E out

   m(h1 + \frac{(V1)^{2}}{2}) =  m(h2 + \frac{(V2)^{2}}{2})

V2 = \sqrt{2(h1 - h2) +(V1)^{2} - 2\frac{Q}{m}

where h1 and h2 are the enthalpies and are gotten from table A-6

V2 = \sqrt{2 x 1000 x(3214.5 - 3024.2) +(60)^{2} - 2\frac{75 x 1000}{4.09}

V2 = 589.9 m/s  

(C) the outlet area is gotten from mass flow rate (m) = \frac{A2 x V2}{α}

  A2 = (α2 x m) / V2

where the initial specific volume (α2) for the given temperature and pressure is gotten from tables A-6 = 0.12552 m^3/kg

A2 = (0.12552 x 4.09) / 589.5 = 0.0008707 m^{3} =  8.71 cm^{2}

4 0
2 years ago
A 0.50-kg croquet ball is initially at rest on the grass. When the ball is struck by a mallet, the average force exerted on it i
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The impulse given to the ball is equal to the change in its momentum:

J = ∆p = (0.50 kg) (5.6 m/s - 0) = 2.8 kg•m/s

This is also equal to the product of the average force and the time interval ∆t :

J = F(ave) ∆t

so that if F(ave) = 200 N, then

∆t = J / F(ave) = (2.8 kg•m/s) / (200 N) = 0.014 s

7 0
1 year ago
Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig
HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

y=-\frac{1}{2} gt^2+v_0t+y_0

With the given values in the question the equation has one unknown v₀:

v_0=\frac{y-y_0}{t}+\frac{1}{2}gt

Solving for t=1:

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To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

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If h is the height of the tower, the energy on top of the tower:

3) E=mgh

Combining equation 2 and 3 and solving for h:

4) h=\frac{{v_0}^2}{2g}+y_0

Combining equation 1 and 4:

h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0

4 0
3 years ago
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Answer:

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Explanation:

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where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

 

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indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

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         v₂ = \sqrt {2g \ y_1}

let's calculate

         v₂ = √( 2 9.8 250)

         v₂ = 70 m / s

6 0
3 years ago
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