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3 years ago

The answer to this digram, the question is in the photo.

Physics

1 answer:

Alina [70]3 years ago

5 0

there will be no noise with the ball

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Calculate the heat gained by 100 grams of ice at -20°C in order to become water at 50°C. ( C = .5 for ice and C = 1 for water, Q

Korvikt [17]

Answer:

$6008$ cal

Explanation:

$m_{i}$ = mass of ice = 100 g = 0.1 kg

$c_{i}$ = specific heat of ice = 0.5 cal/(kg°C)

$c_{w}$ = specific heat of water = 1 cal/(kg°C)

$L$ = Latent heat of fusion of ice = 80 J/g

$T_{i}$ = initial temperature of ice = - 20 °C

$T_{f}$ = final temperature of ice = 50 °C

Q = Heat gained

Heat gained is given as

$Q = m_{i} c_{i}(0 - (T_{i}))+ m_{i}L + m_{i}c_{w}(T_{i} - 0)$

$Q = (100) (0.5) (0 - (- 20))+ (0.1)(80) + (100) (1)(50 - 0)$

$Q = 6008$ cal

8 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

After the switch in is closed on point a, there is current i through resistance r. gives the current for four sets of values of

vovangra [49]

Ehbfwjfvh3r;iuuuuuuuuuuuu1f ce

8 0

3 years ago

A projectile is launched at some angle to the horizontal with some initial speed vi, and air resistance is negligible.(a) Is the

Paul [167]

Answer:

Explanation:

When a projectile is launched at some angle to the horizontal with some speed vi , and air resistance is negligible , it is definitely a freely falling body .

It is so because it is free to accelerate towards the earth with acceleration of g . Air has no resistance , hence no force is acting on it except the gravitational force . Hence it is a freely falling body .

b )

The acceleration in the vertical direction is due to force exerted by the earth that is gravitational force on it . Hence its acceleration is equal to g in vertically downward direction .

c )

It has zero acceleration in horizontal direction . It is so because no force is acting on it in horizontal direction . So no acceleration will be present in horizontal direction . It will move in horizontal direction with constant speed of vi cos θ where θ is the angle vi make with the horizontal .

8 0

3 years ago

Which statements best describe energy ? check all that apply

Ainat [17]

<u>The statements that describes energy best as follows: </u>

The total amount of energy in the universe remains constant.
Energy can be converted from one form to another.
Energy cannot be created or destroyed.

<u>Explanation :</u>

The Universe is a closed system of energy from where the thermodynamics first law is quoted i.e. the energy in the closed system remains constant.

The energy in the closed system can neither be created nor destroyed in the absence of external energy supply. However, it keeps on changing from one state to another. These are the absolute statements about the total energy existing in the Universe.

There are many energy forms existing in the Universe such heat, mass, vibrations etc. that consistently change their form from one to another but the total energy amount is always constant until external energy is applied to the system.

According to the quantum theory, even an electron absorbs or emits energy while jumping from one energy band to the other. There is never a case where energy can create or destroy in a closed system.

6 0

4 years ago