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Sholpan [36]
3 years ago
14

In the popular game "Angry Birds" (refer Figure), for the red bird to hit the green bird with their range 55.0 m, what should be

the launch angle (with respect to the horizontal) of the slingshot if the time of flight is 2.50 s? ​

Physics
1 answer:
Inga [223]3 years ago
4 0

Answer:

Explanation:

What are we to use as gravity in this imaginary world? I will ASSUME 9.81 m/s² and assume no air resistance.

horizontal velocity component

vx = 55.0/2.50 = 22 m/s

vertical initial velocity component

(12.0 - 10.0) = 0 + vy₀(2.50) + ½(-9.81)2.50²

vy₀ = 13.0625 m/s

θ = arctan(vy₀/vx) = arctan(13.0625/22) = 30.6997225... = 30.7°

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The following is current scientific evidence supporting the nebular theory on the formation of the solar system.
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A force did 80 j of work on an object in 4 m how big was the force
yanalaym [24]

Work = (force) x (distance)

80 J = (force) x (4 m)

Force = (80 J) / (4 m)  =  20 N

That's IF the force was in the same direction as the 4m of motion.
If the force was kind of slanted, then it had to be stronger, and
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3 years ago
Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l
Masja [62]

Answer:

a)   I2 = 3 (o-10) / (o- 30) , b)   h ’/h=  3 (o-10) / o (o-30)

Explanation:

The builder's equation is

          1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

         1 / i1 = 1 / f - 1 / o

         1 / i1 = 1/15 - 1 / o

         i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

        o2 = d-i1

We write the builder equation

       1 / f2 = 1 / o2 + 1 / i2

       1 / i2 = 1 / f2 -1 / o2

       1 / i2 = 1 / f2 - 1 / (d-i1)

       1 / i2 = 1/20 - 1 / (d-i1)            (1)

Let's evaluate the last term

      d-i1 = d - 15 o / (o-15)

      d-i1 = (d (o-15) - 15 o) / (o-15)

      d- i1 = (30 or -30 15 -15 o) / (o-15)

      d-i1 = (15 or - 450) / (o- 15)

      d-i1 = = (15 or -450) / (o-15)

replace in 1

      1 / i2 = 1/20 - (or - 15) / (15 or -450)

      1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

      1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

      1 / i2 = (-5 or -150) / (15 or -150)

      1 / i2 = (or -30) / (3 or - 30)

      I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

     m = h ’/ h = - i / o

     h ’= - h i / o

In our case

     h ’= h i2 / o

     h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

5 0
3 years ago
A 129-kg horizontal platform is a uniform disk of radius 1.61 m and can rotate about the vertical axis through its center. A 65.
LUCKY_DIMON [66]

Answer:

Moment of inertia of the system is 289.088 kg.m^2

Explanation:

Given:

Mass of the platform which is a uniform disk = 129 kg

Radius of the disk rotating about vertical axis = 1.61 m

Mass of the person  standing on platform = 65.7 kg

Distance from the center of platform = 1.07 m

Mass of the dog on the platform = 27.3 kg

Distance from center of platform = 1.31 m

We have to calculate the moment of inertia.

Formula:

MOI of disk = \frac{MR^2}{2}

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia (I_y_y )  of the system with respect to the axis yy.

⇒ I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g

⇒ I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2

⇒ I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2

⇒ I_y_y=289.088\ kg.m^2

The moment of inertia of the system is 289.088 kg.m^2

7 0
3 years ago
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