What is the function of the phytoplankton in ocean ecosystems?

A projectile is fired from the ground at a velocity of 30.0 m/s, 35.0 º from the horizontal. What is the maximum height the proj

Norma-Jean [14]

Answer:

Vy = V sin theta = 30 * ,574 = 17.2 m/s

t1 = 17.2 / 9.8 = 1.76 sec to reach max height

Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m

H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m

Time to fall from zero speed to ground = rise time = 1.76 sec

Vx = V cos 35 = 24.6 m / sec horizontal speed

Time in air = 1.76 * 2 = 3.52 sec before returning to ground

S = 24.6 * 3.52 = 86.6 m

8 0

2 years ago

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr

slava [35]

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

W = $F_{total} .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} .d =\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}$

$F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_{sprinter} = F_{total} + F_{wind} = 39.7 + 30 = 69.68 N$

7 0

3 years ago

Read 2 more answers

Conduction transfers Thermal energy by

Vesnalui [34]

Answer:

Conduction is the transfer of heat between substances that are in direct contact with each other. The better the conductor, the more rapidly heat will be transferred. Metal is a good conduction of heat. Conduction occurs when a substance is heated, particles will gain more energy, and vibrate more.

5 0

3 years ago

A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be for a

andriy [413]

To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,

$sin\theta = 1.22\frac{\lambda}{d}$

Here,

$\lambda$ = Wavelength

d= Diameter of aperture

$\theta$ = Angular resolution or diffraction angle

Our values are given as,

$\theta = 11\°$

The frequency of the sound is $f = 9100 Hz$

The speed of the sound is $v = 343 m/s$

The wavelength of the sound is

$\lambda = \frac{v}{f}$

Here,

v = Velocity of the wave

f = Frequency

Replacing,

$\lambda = \frac{(343 m/s)}{(9100 Hz)}$

$\lambda = 0.0377 m$

The diffraction condition is then,

$sin\theta = 1.22\frac{\lambda}{d}$

Replacing,

$sin(11\°) = 1.22\frac{(0.0377 m)}{(d)}$

d = 0.24 m

Therefore the diameter should be 0.24m

6 0

3 years ago

A liquid is used to make a mercury-type barometer. The barometer is intended for space-faring astronauts. At the surface of the

Anarel [89]

Answer:

Density of liquid = 4730 kg/m³

Atmospheric pressure on planet X = 8401.7 N/m²

Explanation:

Pressure, P = ρgh where ρ = density of liquid, g =9.8 m/s² and h = height of column at earth's surface = 2185 mm. Since P = atmospheric pressure, for mercury, P = ρ₁gh₁ where ρ₁ = 13.6 g/cm³ and h₁ = 760 mm

So, ρgh = ρ₁gh₁

ρ = ρ₁h₁/h = 13.6 g/cm³ × 760/2185 = 4.73 g/cm³ = 4730 kg/m³

The atmospheric pressure on planet X

P = ρg₁h₃ g₁ = g/4 and h₃ = 725 mm = 0.725 m

on planet X

P = ρg₁h₃ = (4730 kg/m³ × 9.8 m/s² × 0.725 m)/4 = 8401.7 N/m²

6 0

3 years ago