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Helen [10]
2 years ago
13

If a player through a basketball to the target with an initial velocity of 17 m/s making an angle of 30 degrees with the horizon

tal, calculate the final position made with the vertical (Yf) after 1.3 seconds.
Physics
1 answer:
Svetllana [295]2 years ago
3 0

Answer:

The final position made with the vertical is 2.77 m.

Explanation:

Given;

initial velocity of the ball, V = 17 m/s

angle of projection, θ = 30⁰

time of motion, t = 1.3 s

The vertical component of the velocity is calculated as;

V_y = Vsin \theta\\\\V_y = 17 \times sin(30)\\\\V_y = 8.5 \ m/s

The final position made with the vertical (Yf) after 1.3 seconds is calculated as;

Y_f = V_yt  - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m

Therefore, the final position made with the vertical is 2.77 m.

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lesantik [10]
<span>The flying bully is a move used in the Superhero Movie "Hancock", it is not a real motion in our universe. However, the direction would be towards the target object and the acceleration would be maximal.</span>
7 0
3 years ago
A spring suspended vertically is 11.9 cm long. When you suspend a 37 g weight from the spring, at rest, the spring is 21.5 cm lo
Naddika [18.5K]

Answer:

Time period of oscillations is 0.62 s

Explanation:

Due to suspension of weight the change in the length of the spring is given as

\Delta L = L_f - L_i

\Delta L = 21.5 - 11.9 = 9.6 cm

now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force

mg = kx

0.037 (9.81) = k(0.096)

k = 3.78 N/m

Now the period of oscillation of spring is given as

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

Now plug in all values in it

T = 2\pi \sqrt{\frac{0.037}{3.78}}

T = 0.62 s

5 0
3 years ago
Number of vibrations (waves) in an amount of time
m_a_m_a [10]

Answer:

I think frequency not sure though

How frequently a wave or vibration occurs during a span of time, determines the waves frequency. Frequency is the number of waves per unit time. The unit for frequency if a Hertz ( 1/second). The speed a wave travels is the wavelength multiplied by this frequency. The amplitude of a wave is the maximum distance the wave is displaced.

5 0
2 years ago
A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is
AveGali [126]

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

<u>Given the following data;</u>

Radius, r = 2.6 km

Time = 360 seconds

<em><u>Conversion:</u></em>

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

Circular \; speed (V) = \frac {2 \pi r}{t}

Where;

  • r represents the radius and t is the time.

Substituting into the formula, we have;

Circular \; speed (V) = \frac {2*3.142*2600}{360}

Circular \; speed (V) = \frac {16338.4}{360}

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

Centripetal \; acceleration = \frac {V^{2}}{r}

Where;

  • V is the circular speed (velocity) of an object.
  • r is the radius of circular path.

Substituting into the formula, we have;

Centripetal \; acceleration = \frac {45.38^{2}}{2.6}

Centripetal \; acceleration = \frac {2059.34}{2600}

<em>Centripetal acceleration = 0.79 m/s²</em>

3 0
3 years ago
Calculate the gravitational potential energy of a 2 kg banana hanging 5
Hunter-Best [27]

Answer:

  100 J

Explanation:

The potential energy is given by the formula ...

  PE = mgh

  = (2 kg)(10 m/s^2)(5 m) = 100 J

7 0
3 years ago
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