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Vadim26 [7]
3 years ago
9

Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 60 degrees. It reache

s the maximum height of 15 m. The acceleration due to gravity is 30 m/s2. Find the horizontal range of the projectile in meters.
Physics
1 answer:
Nina [5.8K]3 years ago
3 0

Answer:

R = 0.0503 m

Explanation:

This is a projectile launching exercise, to find the range we can use the equation

       R = v₀² sin 2θ / g

How we know the maximum height

      v_{f}² =v_{oy}² - 2 g y

      v_{f}= 0

      v_{oy} = √ 2 g y

      v_{oy} = √ 2 9.8 / 15

      v_{oy} = 1.14 m / s

Let's use trigonometry to find the speed

    sin θ = v_{oy} / vo

    vo = v_{oy} / sin θ

    vo = 1.14 / sin 60

    vo = 1.32 m / s

We calculate the range with the first equation

     R = 1.32² sin(2 60) / 30

    R = 0.0503 m

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DochEvi [55]
Power  =  Work done  /  Time taken.

Work done =  mgh
Mass, m = 33kg    ( Am presuming it is 33 kg).
h =  85 m.
Work done =  33 * 9.81* 85 =  27517.05  J.

Time taken.
Since object was dropped from height, it fell under gravity.
Using    H =  ut  +  (1/2) * gt^2.              u = 0.
               H =  1/2 gt^2.
               t  =   (2H/g) ^ (1/2)
               t  =  (2*85/9.81) ^ 0.5 =  4.1628 s.

Power  =    27517.05 / 4.1628 = 6610.23 Watts.
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3 years ago
Which of the following is NOT a characteristic of an inner planet?. . A.. rocky. . B.. solid surface. . C.. near the sun. . D..
sveticcg [70]

Being made mostly of gas is NOT a characteristic of an inner planet. The correct answer between all the choices given is the last choice or letter D. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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3 years ago
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An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o
Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

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3 years ago
Capacitors, C1 = 1.0 F and C2 = 1.0 F, are connected in parallel to a 6.0 volt battery (ΔV = 6.0V). If the battery is disconnect
Iteru [2.4K]

Answer:

66.2 sec

Explanation:

C₁ = 1.0 F

C₂ = 1.0 F

ΔV = Potential difference across the capacitor = 6.0 V

C = parallel combination of capacitors

Parallel combination of capacitors is given as

C = C₁ + C₂

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Time constant is given as

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T = 33 x 2

T = 66 sec

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V = final potential difference = 2.2 volts

Using the equation

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2.2 = 6 e^{\frac{-t}{66}}

t = 66.2 sec

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You can't. Velocity and acceleration measure two different things, so their units are incompatible. It's like asking, "How many meters does this book weigh?"

Maybe you mean "find" acceleration using given velocities, or a velocity function?

4 0
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