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s2008m [1.1K]
3 years ago
7

Please help with vectors (will give BRAINLIEST answer)

Physics
1 answer:
zhenek [66]3 years ago
7 0

just analyze it in this way:

20cos30*=10( radical 3 )

20sin30*=10

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Which describes an object in projectile motion? Check all that apply.
Elodia [21]

Answer:

Gravity acts to pull the object down.

The object’s inertia carries it forward.

The path of the object is curved.

Explanation:

The motion of a projectile consists of two separate motions:

- A uniform motion along the horizontal direction, where the velocity is constant; since there are no forces along this direction, the velocity does not change, and so the object continues its motion for inertia --> so, the statement  "The object’s inertia carries it forward" is true.

- A uniformly accelerated motion along the vertical direction, with a constant downward acceleration (g=9.8 m/s^2, acceleration due to gravity). So, the vertical velocity changes, due to the presence of the gravity that acts to pull the object down.

- As a result of the combination of these two motions, the object follows a curved path (in particular, it is a parabolic path).

5 0
3 years ago
Um ladrão tenta fugir sozinho carregando em suas mãos uma mala cheia de barras de ouro. A densidade do ouro é igual a 20 g/cm³,
Roman55 [17]
I believe the answer to your question is A
4 0
3 years ago
Calculate the kinetic energy of Julie, a 60 kg biker, traveling at a velocity of 8 m/s to the right
Vadim26 [7]

Answer:

1,920 Joules

Explanation:

K.E. = 1/2  mv2

so  K.E. =  1/2 (60)(8x8) = 1,920 Joules

8 0
2 years ago
Why do some nucleus release electrons?
fredd [130]

Answer:

Electrons are not little balls that can fall into the nucleus under electrostatic attraction

Explanation:

5 0
3 years ago
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
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