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IgorLugansk [536]
3 years ago
7

Calculate the average velocity of a car that travels 555 km north east in 3.7 hours

Physics
1 answer:
Vera_Pavlovna [14]3 years ago
7 0

Your answer:

150 kilometer's per hour.

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Explain the necessity of smoothing the output voltage before applying it to a transistor
ra1l [238]

In simple words, smoothing can be defined as the circuit which is done to eliminate the ripple from the yield of a direct current power supply.

<u>Explanation:</u>

<u>Necessity of smoothing:</u>

  • Smoothing can be also applied in the form of a capacitor that acts to decrease or level out variations in a signal. And these capacitors are mostly used after power supply in voltage.
  • The yield DC voltage of a half-wave rectifier provided in the figure of a sinusoidal wave.
  • In a method to provide a constant DC voltage from a corrected AC source, a filter or smoothing circuit is required.

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6 0
3 years ago
A hypothetical planet has a mass 2.81 times that of Earth, but the same radius.
patriot [66]

The acceleration due to gravity near the surface of the planet is 27.38 m/s².

<h3>Acceleration due to gravity near the surface of the planet</h3>

g = GM/R²

where;

  • G is universal gravitation constant
  • M is mass of the planet
  • R is radius of the planet
  • g is acceleration due to gravity = ?

g = (6.626 x 10⁻¹¹ x 2.81 x 5.97 x 10²⁴) / (6371 x 10³)²

g = 27.38 m/s²

Thus, the acceleration due to gravity near the surface of the planet is 27.38 m/s².

Learn more about acceleration due to gravity here: brainly.com/question/88039

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4 0
1 year ago
Which law describes how the Earth applies a gravitational force on the Moon, the Moon applies a gravitational force on Earth? *
nikdorinn [45]
The answer is Newton's 3rd Law. The reason why is because a force is a push or a pull that acts upon an object as a results of its interaction with another object. ... These two forces are called action and reaction forces and are the subject ofNewton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction.
3 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
On Earth, a brick has a mass of 10 kg and a weight of 5 lbs. What predictions could we make about the mass and weight of the bri
belka [17]

Answer:

Mass remains constant but weight reduces

Explanation:

Mass is the amount of matter in an object so whether on moon or any other planet, it does not change despite the changes in acceleration.

Weight is a product of mass and acceleration due to gravity, expressed as W=mg where m is the mass, W is weight and g is acceleration. From the above formula, it is evident that when you decrease g, then W also decreases while m is constant. Similarly, when m is constant and g is increased then W also increases.

Therefore, for this case, since g decreases, the weight decreases but mass remains constant.

8 0
3 years ago
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