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yulyashka [42]
3 years ago
12

Cu(s) + AgNO3(aq) → Cu(NO3)2(aq) + Ag(s) The unbalanced equation for the reduction of silver nitrate by copper is shown here. Wh

at is the coefficient for Cu(NO3)2 in the balanced equation?
Physics
1 answer:
cricket20 [7]3 years ago
3 0

Answer:

1

Explanation:

Cu + 2AgNO3 ---> Cu(NO3)2 + 2Ag

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What will it be ? Plz help me
IgorC [24]

Answer:

For number 2, it will be 8.

3 0
3 years ago
Find the minimum radius of a helium balloon that will lift her off the ground. the density of helium gas is 0.178 kg/m3, and the
marysya [2.9K]

Volume of balloon = \frac{4}{3} \pi r^{3}

Where r is the radius of balloon.

Here mass of woman = 68 kg

Mass of air displaced by a balloon with volume V = 1.29*V

Mass of helium inside balloon = 0.178*V

Total mass to be lifted by balloon = 68 +0.178*V

Buoyant force = 1.29V-0.178V=1.112V

So we have 1.112 V =  68+ 0.178*V

   0.934 V = 68

     V = 72.81 m^3

    \frac{4}{3} \pi r^{3}[/tex]= 72.81

     r = 2.59 m

So radius of helium balloon = 2.59 m

4 0
3 years ago
Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellit
GrogVix [38]

Answer:

1) C

2) E

Explanation:

3 0
3 years ago
One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo
yuradex [85]

Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

3 0
3 years ago
A projectile is fired with a velocity of 22 m/s at an angle of 25°. What is the vertical component of the velocity?
7nadin3 [17]

Answer:

Vertical component of velocity is 9.29 m/s

Explanation:

Given that,

Velocity of projection of a projectile, v = 22 m/s

It is fired at an angle of 22°

The horizontal component of velocity is v cosθ

The vertical component of velocity is v sinθ

So, vertical component is given by :

v_y=v\ sin(25)

v_y=22\ m/s\times\ sin(25)

v_y=9.29\ m/s

Hence, the vertical component of the velocity is 9.29 m/s

3 0
3 years ago
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