234 cm …. _______________

A basketball player jumps straight up with a speed of 14 m/s. How high did the player jump?

alexira [117]

Answer:

<em>The basketball player jumped 10 m high.</em>

Explanation:

<u>Vertical Launch Upwards </u>

In a vertical launch upwards, an object is launched vertically up without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

$\displaystyle h_m=\frac{v_o^2}{2g}$

The basketball player jumps up with a speed of vo=14 m/s. The maximum height is:

$\displaystyle h_m=\frac{14^2}{2*9.8}$

$\displaystyle h_m=\frac{196}{19.6}=10$

The basketball player jumped 10 m high.

7 0

3 years ago

A certain car is capable of accelerating at a uniform rate of 0.85 m/s^2. What is the magnitude of the car's displacement as it

gogolik [260]

Solution: 85.11 m

Given:

acceleration of the car, $a= 0.85 m/s^{2}$

initial velocity, $u=83km/h=23.05 m/s$

final velocity, $v=94 km/h=26.11 m/s$

we need to find the displacement (s) of the car.

we would use the equation of motion:

$2as=v^{2}-u^{2}\\
\Rightarrow s=\frac{v^{2}-u^{2}}{2a} \\
\Rightarrow s=\frac{26.11^{2}-23.05^{2}}{2\times0.85}\\
\Rightarrow s=85.11 m$

hence, the displacement done by the car is = 85.11 m

7 0

3 years ago

Water flows past a flat plate that is oriented parallel to the flow with an upstream velocity of 0.4 m/s. (a) Determine the appr

Trava [24]

Answer:

1.12 m

0.08291 m

Explanation:

u = Upstream velocity = 0.4 m/s

Re = Reynold's number = $5\times 10^5$ (turbulent)

$\nu$ = Viscosity of water = $1.12\times 10^{-6}\ Pas$

Here the flow is turbulent so we have the relation

$Re_{xcr}=\frac{ux_{cr}}{\nu}\\\Rightarrow x_{cr}=\frac{Re_{xcr}\nu}{u}\\\Rightarrow x_{cr}=\frac{5\times 10^5\times 1.12\times 10^{-6}}{0.4}\\\Rightarrow x_{cr}=1.4\ m$

The approximate location downstream from the leading edge where the boundary layer becomes turbulent is 1.4 m

Boundary layer thickness relation is given by

$\delta={\frac{\nu x}{u}}^{\frac{1}{5}}\\\Rightarrow \delta={\frac{1.12\times 10^{-6}\times 1.4}{0.4}}^{\frac{1}{5}}\\\Rightarrow \delta=0.08291\ m$

The boundary layer thickness is 0.08291 m

4 0

4 years ago

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.

snow_lady [41]

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is $v = 3.79 *10^{5} \ m/s$