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serg [7]
3 years ago
8

One strategy in a snowball fight is to throw

Physics
1 answer:
blondinia [14]3 years ago
7 0

Answer:

Range formula:  R = v^2 sin (2 theta) / g

If theta = 69 deg and v = 15.1

R = 15.1^2 sin 138 / 9,8 = 15.6 m

sin 138 = .669 = sin 42

So a snowball thrown at 21 deg will travel

R = 15.1 * .669^2 / 9.8 = 15.6 m

The second snowball can be thrown at 21 deg to travel the same distance

Vx = V cos theta = 15.1 * cos 69 = 5.41     first snowball

t1 = 15.6 / 5.41 = 2.88 sec

Vx = V cos theta = 15.1 cos 21 = 14.1 m/s

t2 = 15.6 / 14.1 = 1.11 sec

Difference = t1 - t2 = 1.77 sec     time delay for second snowball

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5 0
3 years ago
Explain your problem of<br>Federalism<br>​
telo118 [61]

That is more of a History of English question.

6 0
3 years ago
Water of density 1000 kg/m3 falls without splashing at a rate of 0.373 L/s from a height of 40.5 m into a 0.64 kg bucket on a sc
Sphinxa [80]

Answer:

       F_scale = 20.18 N

Explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

* Let's find the amount of liquid in the container for a time of t = 2.93 s

Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there

        V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l

        V_{water} = 1.09 10⁻³ m³

the amount of water is

       ρ = m / V

       m = ρ V

       m = 1000 1.09 10⁻³

       m = 1.09 kg

so the weight of the liquid in the container for this time is

       W = mg

       W = 1.09 9.8

       W = 10.68 N

* Let's look for the force of the falling jet

Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h

        P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2  ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

         P₂ = P_{atm}

since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0

let's use kinematics to find the speed of a drop when it reaches the container y = 0

         v² = v₀² - 2 g (y-y₀)

         v = \sqrt{0 -2 g ( 0-y_o)}

let's calculate

         v = √(2 9.8 40.5)

         v = 28.17 m / s

this is the speed in the container v₁ = 28.17 m / s

the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0

we substitute in Bernoulli's equation

         P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂

         P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂

         P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²

         P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²

         P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵

         P₁ = 1.28 10⁵ Pa

The definition of Pressure is

         P = F / A

         F = P A

We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving

          m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³

the volume is V = A l

if the length of l = 1 m

A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²

the force of this jet is

            F = P A

            F = 1.28 10⁵  7.46 10⁻⁵

            F = 9.5 N

with these data let's use the equilibrium equation

           F_ scale -W - F = 0

           F_scale = W + F

           F_scale = 10.682 + 9.5

           F_scale = 20.18 N

4 0
3 years ago
Light rays in a material with index of refrection 1.35 1.35 can undergo total internal reflection when they strike the interface
nekit [7.7K]

Answer:

1.30

Explanation:

To calculate the critical angle we have ti use the formula:

sin\theta_c=\frac{n_2}{n_1}

where theta_c is the critical angle, n1 is the index of refraction of the material where the light is totally reflected, and n2 is the refractive index of the other material.

By taking n_2 and replacing we obtain:

n_2=n_1sin\theta_c=(1.35)sin75.1\°=1.30

hope this helps!!

6 0
3 years ago
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