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Gwar [14]
3 years ago
15

A wave travelling 140 m/sec with a wavelength of 4.0 m has what frequency? _______________Meters / Hz / seconds

Physics
1 answer:
valentina_108 [34]3 years ago
5 0

Answer:

The frequency of a wave is 35 Hz.

Explanation:

Speed of a wave is 140 m/s

Wavelength of a wave is 4 m

It is required to find the frequency. The relation between frequency, speed and the wavelength of a wave is given by :

v=f\lambda

f is frequency

f=\dfrac{v}{\lambda}\\\\f=\dfrac{140\ m/s}{4\ m}\\\\f=35\ Hz

So, the frequency of a wave is 35 Hz.

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Which of the following are examples of projectile motion?
liubo4ka [24]

Answer: A. a basketball being shot toward the basket

Explanation: The definition of projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. So, the basketball is the object being thrown and the person throwing the ball is aiming it to go into the basket making that the path of trajectory. Hope that makes sense and helps!

5 0
2 years ago
Light travels in a straight line at a constant speed of 300 000 km/s what is the lights acceleration
exis [7]

Answer:

it's acceleration is 0

Explanation:

since it is travelling at a constant speed it is not accelerating so its acceleration is 0

3 0
3 years ago
A 50 kg person receives an absorbed dose of gamma radiation of 20 millirads. What is the total energy absorbed?.
Masteriza [31]

For a 50 kg person receives an absorbed dose of gamma radiation of 20 millirads,  the total energy absorbed is mathematically given as

E=0.1457J

<h3>What is the total energy absorbed?</h3>

Generally, the equation for the total energy absorbed  is mathematically given as

E=mass*gamma radiation

Therefore

E=50*20*19^{-3}

E=0.1457J

In conclusion,  the total energy absorbed

E=0.1457J

Read more about Energy

brainly.com/question/13439286

4 0
2 years ago
Find Vxl and Vyl of a pumpkin launched at a velocity of 55 m/s at an angle of 20 degrees
Vinvika [58]

Answer:

             

Explanation:

is  A projectile is any object on which the only force acting is gravity and air resistance (drag).

Examples of projectiles are:

baseballs and softballs in the air after being hit by the bat

golf balls hit by a club

objects dropped from aircraft, such as people (skydivers), bombs, crates of food being dropped to refugees

objects launched by cannons, such as cannonballs, shells, and circus performers

Once the baseball, softball, golf ball, skydiver, bomb, crate, cannonball, shell, or clown are no longer touching the bat, club, aircraft, or cannon, and are in the air with only gravity and slight air resistance acting on it, then it is a projectile.

Here is an online projectile motion applets to play with, just for fun.

Unless otherwise stated in a particular problem or discussion, we will be ignoring the effects of air resistance.

The key to understanding the motion of projectiles is that the horizontal motion and the vertical motion of the projectile are independent of each other. So we can write separate equations for the displacement of the projectile in the horizontal (x) and vertical (y) directions.

                         

The only common variable between these two equations is t, the time. Because in projectile problems there is usually no acceleration (i.e. we ignore air resistance) in the horizontal direction, we can write

           

The velocity components follow the same equations we used for one-dimensional motion.

                             

Because there is usually no acceleration in the x direction, the x-velocity is constant.

3 0
3 years ago
Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc
expeople1 [14]

Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

Intensidade do campo elétrico (E) =?

1)

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

2)

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

K = 9,0 × 10 ^ 9

6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09

Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9

Q = 0,06 × 10 ^ (6-9)

Q = 0,06 × 10 ^ -3

Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC

7 0
3 years ago
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