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2 years ago

Can someone help answers these questions? (I attached the pictures)

Physics

1 answer:

RoseWind [281]2 years ago

3 0

Nothing works if Switch-3 is open.

-- None

-- A, E, F, G

-- None

-- A, B, E, F, G

-- A, C, D, E, F, G (everything except B)

You might be interested in

Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.39 kW/m2. (a) Assuming that Earth (and

Zigmanuir [339]

Answer:

$F=5.8\times 10^{8}\ N$

$F=35.57\times 10^{21}\ N$

Explanation:

Given that

Intensity I

$I= 1.39\ KW/m^2$

$Speed\ of \ light = 2.99\times 10^8\ m/s$

Radius of earth,R = 6370 Km

We know that surface area of earth, A

$A=\pi R^2$

$A=\pi (6370\times 10^3)^2\ m^2$

$A=1.27\times 10^{14}\ m^2$

As we know that pressure due to intensity given as

$P=\dfrac{I}{V}$

V =Velocity of light

$V=3\times 10^8\ m/s$

$P=\dfrac{1.39\times 1000}{=3\times 10^8}$

$P=4.6\times 10^{-6}\ Pa$

We know that force F

F = P .A

$F=4.6\times 10^{-6}\times 1.27\times 10^{14}\ N$

$F=5.8\times 10^{8}\ N$

b)Gravitational force F

$F=G\dfrac{m.M}{r^2}$

$M = mass\ of\ sun = 2\times 10^{30} kg\\m = mass\ of\ earth = 6\times 10^{24}kg$

$r =1.5\times 10^{11}\ m$

$G =6.67\times 10^{-11}Nm^2/kg^2$

So F

$F=6.67\times 10^{-11}\times \dfrac{2\times 10^{30}\times 6\times 10^{24}kg}{1.5\times 10^{11}}$

$F=35.57\times 10^{21}\ N$

7 0

3 years ago

Read 2 more answers

How to do this problem

marusya05 [52]

The answer I'm pretty sure is 3

4 0

3 years ago

Read 2 more answers

A steel rope is used to lift a 26 kg slab of concrete from the ground to a height of 20 m. Assume that the rope moves the concre

Ne4ueva [31]

Answer:

5200 Joules

Explanation:

Work Formula:

W = F . D

W = (26.10) . 20

W = 260 . 20

W = 5200 Joules

5 0

3 years ago

If two people are running at the same speed in the same direction. One person is one meter ahead of the other. The person in fro

Gre4nikov [31]

Answer:yes

Explanation:

5 0

3 years ago

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:

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)

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

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

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

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

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

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

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

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

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