Give an example of when a conscientious objector might exercise their right:

If a wave hits a smooth surface at an angle of incidence of 40 degrees, the angle of reflection is

ruslelena [56]

A, 40 degrees. Reflexion laws state that both incidence and reflection angles are the same.

8 0

3 years ago

An 8.75 kg point mass and a 14.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be

marysya [2.9K]

Answer

given,

mass of the = m₁ = 8.75 Kg

another mass of the object = m₂ = 14 Kg

distance between them = 50 cm

R₁ = 17 cm

R₂ = 50 -17 = 33 cm

a) Force applied due to the Mass 8.75 in +ve x- direction

$F_1 = \dfrac{GM_1 m}{R_1^2}$

$F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}$

$F_1 = 2.019\times 10^{8}\ m$

Force applied due to mass 14 Kg in -ve x-direction

$F_2 = \dfrac{GM_2 m}{R_2^2}$

$F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}$

$F_2 = 0.857\times 10^{8}\ m$

net force

F = F₁ + F₂

$F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m$

$F = 1.162\times 10^{8}\ m$

Using newton second law

$a = \dfrac{F}{m}$

$a = \dfrac{ 1.162\times 10^{8}\ m}{m}$

$a =1.162\times 10^{8} \ m/s^2$

b) As the acceleration of mass comes out to be +ve hence, the direction will be toward the mass of 8.75 Kg

6 0

3 years ago

You and a friend are playing with a bowling ball to demonstrate some ideas of Rotational Physics. First, though, you want to cal

RideAnS [48]

Answer:

K_{total} = 19.4 J

Explanation:

The total kinetic energy that is formed by the linear part and the rotational part is requested

$K_{total} = K_{traslation} + K_{rotation}$

let's look for each energy

linear

$K_{traslation}$ = ½ m v²

rotation

$K_{rotation}$ = ½ I w²

the moment of inertia of a solid sphere is

I = 2/5 m r²

we substitute

$K_{total}$ = ½ mv² + ½ I w²

angular and linear velocity are related

v = w r

we substitute

K_{total} = ½ m w² r² + ½ (2/5 m r²) w²

K_{total} = m w² r² (½ + 1/5)

K_{total} = $\frac{7}{10}$ m w² r²

let's calculate

K_{total} = $\frac{7}{10}$ 6.40 16.0² 0.130²

K_{total} = 19.4 J

6 0

2 years ago

Suppose that a meter stick is balanced at its center. A 0.24 kg mass is then positioned at the 6-cm mark. At what cm mark mus

Scilla [17]

Answer:

80.17 cm

Explanation:

Taking moments of forces about the center, the total clockwise moments is equal to the total counter clockwise moment:

Force * distance (counter clockwise) = force * distance (clockwise)

0.24 * 9.8 * (50 - 6) = 0.35 * 9.8 * (x - 50)

0.24 * 44 = 3.43x - 171.5

103.5 = 3.43x - 171.5

=> 3.43x = 103.5 + 171.5

3.43x = 275

x = 275/3.43 = 80.17 cm

8 0

3 years ago

The net forced experienced by a box is 500 N. If the force of friction opposing the motion is 100 N, what is the force exerted b

Alexeev081 [22]

The force exerted by the person is 600 N because to find the original force you would add back 100 from friction

8 0

3 years ago