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laila [671]
3 years ago
10

a projectile is launched straight up at 141 m/s . How fast is it moving at the top of its trajectory? suppose it is launched upw

ard at 45 degree above the horizontal plane. how fast is it moving at the top of its curved trajectory?
Physics
1 answer:
BARSIC [14]3 years ago
7 0

The velocity of projectile has 2 components, horizontal component vcosθ and vertical component vsinθ, where v is the velocity of projection and θ is the angle between +ve X-axis and projectile motion.

In case 1, θ = 90⁰

    So horizontal component is vcos90 = 0

         Vertical component at maximum height = 0

So velocity at maximum height = 0 m/s


In case 2, θ = 45⁰

    So horizontal component is 141cos45 = 100m/s

         Vertical component at maximum height = 0

So velocity at maximum height = 100 m/s


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g You shine orange laser light that has a wavelength of 600 nm through a narrow slit. The slit forms a diffraction pattern on a
zimovet [89]

Answer:

 λ = 3 10⁻⁷ m,   UV laser

Explanation:

The diffraction phenomenon is described by the expression

         a sin θ = m λ

let's use trigonometry

         tan θ = y / L

as in this phenomenon the angles are small

        tan θ = \frac{sin \ \theta}{cos \ \theta} = sin θ

        sin θ = y / L

we substitute

      a y / L = m  λ

let's apply this equation to the initial data

       a  0.04 / L = 1 600 10⁻⁹

       a / L = 1.5 10⁻⁵

now they tell us that we change the laser and we have y = 0.04 m for m = 2

      a 0.04 / L = 2  λ

       a / L = 50  λ

we solve the two expression is

         1.5 10⁻⁵ = 50  λ

          λ = 1.5 10⁻⁵ / 50

          λ = 3 10⁻⁷ m

    UV laser

3 0
3 years ago
According to the rayleigh criterion, what is the shortest object we could resolve at the 25.0 cm near point with light of wavele
Blizzard [7]

Answer:

6.71 *10^{-5} rad

Explanation:

∅ = \frac{1.22*wavelength}{D} = \frac{1.22*550 * 10^{-9} }{25 * 10^{-2} }

∅ = 6.71 *10^{-5} rad

The minimum resolvable angle = 6.71 *10^{-5} rad

7 0
3 years ago
A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba
Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

      The mass of the rubber ball is m_r   =  2 \ kg

      The  initial  speed of the rubber ball is  u =  3 \ m/s

      The final speed at which it bounces bank v  - 3 \ m/s

      The mass of the clay ball  is  m_c =  2  \ kg

       The  initial  speed of the clay  ball is u = 3 \ m/s

       The final speed of the clay ball is  v = 0 \  m/s

Generally Impulse is mathematically represented as

       I  =  \Delta p

where \Delta  p is the change in the linear momentum so  

       I  =  m(v-u)

For the rubber  is  

        I_r  =  2(-3 -3)

       I_r  = -12\ kg \cdot  m/s

=>     |I_r|  = 12\ kg \cdot  m/s

For the clay ball

       I_c  =  2(0-3)

        I_c =  -6 \ kg\cdot \ m/s

=>    | I_c| =  6 \ kg\cdot \ m/s

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

       

8 0
4 years ago
A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant s
Monica [59]
Hey There,

Question: "<span>A student gives a brief push to a block of dry ice. A moment later, the block moves across a very smooth surface at a constant speed. When drawing the free body diagram for the block of dry ice moving at a constant speed, the forces that should be included are: (select all that apply)"

Answer: C. Force Of Friction
              B. Force

If This Helps May I Have Brainliest?</span>
7 0
3 years ago
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Does the diagram show a spring or neap tide?How do you know?
vampirchik [111]
It shows a neap tide 
6 0
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