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Anestetic [448]
3 years ago
6

2

Physics
1 answer:
luda_lava [24]3 years ago
4 0

Answer:

500000N/m²

5250N

Explanation:

Given parameters:

Depth(H) = 50m

Density of water  = 1000kg/m³

Acceleration of free fall  = 10m/s

Unknown:

Pressure the water exerts on the diver = ?

Solution:

Pressure is the force per unit area on a body. In fluids, pressure is the product of density, gravity and height

  Pressure in fluids  = Density x acceleration due to gravity x height

Input the variables and solve;

    Pressure in fluids  = 1000 x 10 x 50  = 500000N/m²

B.

width of window = 150mm

height of window = 70mm

Force water exerts on the window = ?

To solve this problem;

         Pressure  = \frac{Force}{Area}

Area of the window = width x height  = 150 x 10⁻³ x 70 x 10⁻³

                                                           = 1.05 x 10 ⁻²m²

Force  = pressure x area

Input the variables;

             = 500000N/m²   x  1.05 x 10 ⁻²m²

             = 5250N

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our friend is constructing a balancing display for an art project. She has one rock on the left (ms=2.25 kgms=2.25 kg) and three
Licemer1 [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The torque produced by the pile of rocks is \tau = 35.63\ N \cdot m  

b

The distance of the single for equilibrium to occur is r_s =1.62 \ m

Explanation:

From the question we are told that

     The mass of the left rock is  m_s = 2.25 \ kg

     The mass of the rock on the right m_p = 10.1 kg

    The distance from  fulcrum to the center of the pile of rocks is  r_p = 0.360 \ m

   

Generally the torque produced by the pile of rock is mathematically represented as

           \tau = m_p * g * r_p

Substituting values

         \tau = 10.1 * 9.8  * 0.360                  

          \tau = 35.63\ N \cdot m      

Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows

   The torque due to the single rock is

           \tau = m_s  * g * r_s

At equilibrium the both torque are equal

            35.63 = m_s * r_s * g

Making r_s the subject of the formula

             r_s = \frac{35.63 }{m_s * g}

Substituting values

            r_s = \frac{35.63 }{2.25 * 9.8}

            r_s =1.62 \ m

6 0
3 years ago
A ball rolls down a hill and hits a box. The momentum of the ball decreases. Which happens to the momentum? Select one of the op
Irina-Kira [14]
<span>So we want to know what happens to the momentum of the ball that rolls down hill and hits a box. So we need to use the law of conservation of momentum which states that the momentum must be conserved. It cant be transformed into inertia or mass. It can only be transferred to other object via some interactions like collisions. So it has to be a. transferred to the box and that is the correct answer. </span>
7 0
3 years ago
Un trineo de 20 kg descansa en la cima de una pendiente de 80 m de longitud y 30° de inclinación. Si µ = 0.2, ¿cuál es la veloci
Mariulka [41]

Answer:

v= 26.70 m/seg

Explanation:  Ver anexo ( diagrama de cuerpo libre)

De acuerdo a la segunda ley de Newton

∑ F  =  m*a

∑ Fx  =  m* a(x)             ∑ Fy  =  m* a(y)

También sabemos que el coeficiente de roce dinámico es:

  μ  = 0.2 = F(r)/N            siendo N la fuerza normal.

Si descomponemos la fuerza P = mg  =  20Kg* 9.8m/seg²

P =  196 [N]    en sus componentes sobre los ejes x y y tenemos

Py  =  P* cos30  =  196* √3/2  =  98*√3

Px  = P* sen30   =  196*1/2  =  98

La sumatoria sobre el eje y es :

∑ F(y)  =  m*a         Py  - N  = 0          98*√3  = N       ( no hay movimiento en la dirección y)

∑ F(x)  = m*a    P(x)  -  Fr  =  m*a

Fr  =   μ *N  =  0.2* 98*√3

Fr  =  19.6*√3  [N]

98 -  19.6*√3  =  m*a

98  -  33.52  = m*a

a =  (98  -  33.52 ) / 20

a = 3.22 m/seg²

Para calcular la velocidad del trineo al pié del plano, sabemos que al pié del plano el trineo ha recorrido 80 m, y que de cinemática

v²  =  v₀²  +  2*a*d             ( se pueden chequear unidades para ver la consistencia de la ecuación  v  y  v₀    vienen dados en m/seg  entonces  v²  y  v₀²  vienen en m²/seg²,  el producto de a (m/seg²) por la distancia d (m) resulta en m²/seg²  entonces es consistente la relación

v²   =  0   +  2*3.22*80       ( la velocidad inicial es cero)

v²  = 515.2  m²/seg²

v  =  √515.2  m/seg

v= 26.70 m/seg

6 0
3 years ago
Which of the following describes a referee's job?
Serhud [2]

Answer:

C. Supervising the game to make sure teams are playing fairly

5 0
3 years ago
Brycen can cover half a basketball court (about 14 meters) in 4.0 seconds flat! How fast can he run?
Vedmedyk [2.9K]

Speed is equal to distance traveled divided by the time. So it's 3.5 m/s

6 0
3 years ago
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