The main source of information used by astronomers to learn about objects in space is.

The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha

Yuri [45]

Answer:

(a) 1.257 x 10^5 J

(b) 1.456 Watt

Explanation:

Volume of blood, v = 7500 L = 7.5 m^3

Height, h = 1.63 m

density of blood, d = 1.05 x 10^3 kg/m^3

(a) work done = m x g x h

W = v x d x g x h = 7.5 x 1.05 x 1000 x 9.8 x 1.63 = 1.257 x 10^5 J

(b) time = 1 day = 24 x 60 x 60 s = 86400 seconds

Power = Work / time = 1.257 x 10^5 / 86400 = 1.456 Watt

6 0

3 years ago

The energy band gap of GaAs is 1.4eV. calculate the optimum wavelength of light for photovoltaic generation in a GaAs solar cell

Viktor [21]

Answer:

λ = 8.88 x 10⁻⁷ m = 888 nm

Explanation:

The energy band gap is given as:

Energy Gap = E = 1.4 eV

Converting this to Joules (J)

E = (1.4 eV)(1.6 x 10⁻¹⁹ J/1 eV)

E = 2.24 x 10⁻¹⁹ J

The energy required for photovoltaic generation is given as:

E = hc/λ

where,

h = Plank's Constant = 6.63 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light = ?

Therefore,

2.24 x 10⁻¹⁹ J = (6.63 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.63 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.24 x 10⁻¹⁹ J)

7 0

3 years ago

A coal - fired power station produces 100MJ of electrical energy when it is supplied with 400MJ of energy from its fuel. The eff

Alexus [3.1K]

Answer:

100 divide by 400 times 100 percent

8 0

2 years ago

Number of valence electrons for calcium?

zimovet [89]

Calcium has 2 valence electrons

7 0

3 years ago

Read 2 more answers

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

$f=\frac{\omega}{2\pi}$

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

Learn more about pressure:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

5 0

3 years ago