Accurate forecasts require careful study of the location, size, movement, and characteristics of large ____

Suppose a soccer ball is kicked from the ground at an angle 20.0º above the horizontal at 8.00 m/s. The y-velocity is determined

julia-pushkina [17]

initial speed of the ball by which it is kicked is 8 m/s

now the angle at which it is kicked is 20 degree

here we will have

$v_y = v sin20$

$v_y = 8 sin20 = 2.74 m/s$

now we will have

$\Delta y = v_y t + \frac{1}{2}at^2$

so if the ball again land on the ground at same level then we have

$\Delta y = 0$

$0 = 2.74 t - \frac{1}{2}(9.8) t^2$

$0 = 2.74 - 4.9 t$

$t = 0.56 s$

so total time will be 0.56 s

8 0

3 years ago

The cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.7 m/s in 2.50 s and

viktelen [127]

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

$1\ m=\frac{1}{1609.344}\ miles$

1 hour = 60×60 seconds

$1\ s=\frac{1}{3600}\ hours$

$28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h$

Top speed of the cheetah is 64.2 mi/h

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2$

Acceleration of the cheetah is 8.68 m/s²

2)

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s$

It takes a cheetah 3.31 seconds to reach its top speed.

3)

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m$

It travels 47.5 m in that time

4) When s = 120 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s$

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds

8 0

3 years ago

In a research facility, a person lies on a horizontal platform which floats on a film of air. When the person's heart beats, it

anastassius [24]

Answer: 3.48g

Explanation:

here, we will be using conservation of momentum to solve the problem. i.e the total momentum remains unchanged, unless an external force acts on the system. We'll in thus question, there is no external force acting in the system.

Remember, momentum = mass * velocity, then

mass of blood * velocity of blood = combined mass of subject and pallet * velocity of subject and pallet

Velocity of blood = 56.5cm = 0.565m

mass of blood * 0.565 = 54kg * (0.000063/0.160)

mass of blood * 0.565 = 54 * 0.00039375

mass of blood * 0.565 = 0.001969

mass of blood = 0.00348kg

Thus, the mass of blood that leaves the heart is 3.48g

7 0

3 years ago

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