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FinnZ [79.3K]
1 year ago
13

In a game of Tug-of-War,

Physics
1 answer:
NemiM [27]1 year ago
8 0
Eventually one of the teams will became tired and its pull will be smaller than that of the other team. When the force acting on the robe are unbalanced the rope will start to move in the direction of the great force.
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Blocks A and B of unknown masses m1 and m2, respectively, are set up on an inclined plane as shown. Block A is attached to block
Korvikt [17]

Newton's second law we can find that the correct answer is:  

 E)  It cannot be determiner whick block has more masses from the information provided

Newton's second law establishes the relationship between force, mass, and acceleration of a body. Since force and acceleration are vector quantities, their components must be added on each axis

For this problem we have two bodies, let's write Newton's second law for the body B, we assume that the body B descends

            W_b - T = m_b a

            W_b  = m_b g

            m_b - T = m_b a

Where W_b is the weight of block B, T the tension of the string, mb the mass of block b and the acceleration

Now let's find the relation for block A

let's set a datum with the x axis parallel to the ramp

           T - Wₓ = mₐ a

           sin θ = Wₓ / W

            Wₓ = Wₐ sin θ

             Wₐ = mₐ g

Where Wₓ is the component of the weight, Wₐ the weight of the body A and θ the angle of the plane

Let's write our system of equations

           m_b g - T = m_b a

           T - mₐ g sin θ = mₐ a

let's add the equations

            g (m_b - mₐ sin θ) = (m_b + mₐ) a

            a =   \frac{m_b - m_a \ sin  \ \theta}{m_b+m_a} \ g

Let's analyze this expression

  • The numerator is positive the body B descends, this occurs when

          m_b - mₐ sin θ > 0

           

  • The numerator is negative, body B rises

           m_b - mₐ  sin θ <0

We can observe that the acceleration is positive or negative depending on the relation of the masses and the angle of the plane.

In conclusion using Newton's second law we find that the correct answer is  

 E )   It cannot be determiner whick block has more masses from the information provided

learn more about Newton's second law here:

brainly.com/question/9099891

8 0
2 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
2 years ago
A pingpong ball has 2 kg m/s of momentum when thrown 8 m/s find the mass of the ball
avanturin [10]

Answer:

0.25 kg

Explanation:

p = mv

2 = m(8)

2/8 = m(8)/8 *cancels

m = 1/4 OR 0.25 kg

4 0
2 years ago
A −4.00 μC charge sits in static equilibrium in the center of a conducting spherical shell that has an inner radius 3.13 cm and
Mariulka [41]

Answer:

(a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

Explanation:

Given that,

Charge q₁ = -4.00 μC

Inner radius = 3.13 m

Outer radius = 4.13 cm

Net charge q₂ = -6.43 μC

We need to calculate the charge on the outer surface

Using formula of charge

q_{out}=q_{2}-q_{1}

q_{out}=-6.43-(-4.00)

q_{out}=-2.43\ \mu C

The charge on the inner surface is q.

q+(-2.43)=-6.43

q=-6.43+2.43= 4.00\ \mu C

We need to calculate the electric field outside the shell

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times6.43\times10^{-6}}{(4.13\times10^{-2})^2}

E=33927618.73\ N/C

E=3.39\times10^{7}\ N/C

Hence, (a). The charge on the outer surface is −2.43 μC.

(b). The charge on the inner surface is 4.00 μC.

(c). The electric field outside the shell is 3.39\times10^{7}\ N/C

5 0
2 years ago
A car turns a certain curve of radius 24.98 m with constant linear speed of
Anastaziya [24]

Answer:

3525.19 kg

Explanation:

The computation of the mass of the car is shown below:

As we know that

Fc = m × V^2 ÷ R

m = Fc × R ÷ V^2

Provided that:

Fc = 34.652 kN = 34652 N

R = Radius = 24.98 m

V = speed = 15.67 m/s

So,

m = 34652 × 24.98 ÷ 15.67^2

 = 3525.19 kg

7 0
3 years ago
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