The initial speed of the shot is 15.02 m/s.
The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.
Pl refer to the attached diagram.
Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.
Write an expression for R.
Therefore,
In the time t, the net displacement of the shotput is y in the downward direction.
Use the equation of motion,
Substitute the value of t from equation (1).
Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.
The shot put was thrown with a speed 15.02 m/s.
Answer:
less
Explanation:
Sliding friction is always less than static friction. This is because in sliding friction, the bodies slide with each other and thus the effect of friction is not more. However, it does not happen in the case of static friction.
Pressure=hrg
pressure is great at the bottom because the weight lf water exerts pressure below the container due to the gravitational pull of the earth.
Answer:
560 m
Explanation:
The speed of sound in air is approximately:
v ≈ v₀ + 0.6T
where v₀ is the speed of sound at 0°C (273 K) in m/s, and T is the temperature in Celsius.
The speed of sound at 20°C at that altitude is:
v ≈ 327 + 0.6(20)
v ≈ 339 m/s
The sound travels from the hikers to the mountain and back again, so it travels twice the distance.
339 m/s = 2d / 3.3 s
2d = 1118.7 m
d = 559.35 m
Rounding, the mountain is approximately 560 m away.
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