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lidiya [134]
3 years ago
14

What is the speed of a wave that has a frequency of 45 Hz and a wavelength of 0.1 meters?

Physics
2 answers:
Yuri [45]3 years ago
6 0
<span>λν=c
(wavelength x frequency = speed)

speed = 45 x 0.1
= 4.5 m/s</span>
Xelga [282]3 years ago
5 0

Answer:

4.5

Explanation:

yes

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What are the forces like in a collision?
Anuta_ua [19.1K]

Answer:

The forces are a convergent

Explanation:

6 0
2 years ago
A child whirls a 3.00 kg ball on a string .50 m from the axis of rotation in a horizontal circle. The ball makes 1 revolution in
melamori03 [73]

Answers:

a) 0.5 m/s^{2}

b) 1.5 N

Explanation:

a) The centripetal acceleration a_{c} of an object moving in a uniform circular motion is given by the following equation:  

a_{c}=\omega^{2} r  

Where:

\omega=1 \frac{rev}{s} is the angular velocity of the ball

r=0.5 m is the radius of the circular motion, which is equal to the length of the string

Then:

a_{c}=(1 \frac{rev}{s})^{2} 0.5 m  

a_{c}=0.5 m/s^{2} This is the centripetal acceleration of the ball

b) On the other hand, in this circular motion there is a force (centripetal force F) that is directed towards the center and is equal to the tension (T) in the string:

F=T=m. a_{c}

Where m=3 kg is the mass of the ball

Hence:

T=(3 kg)(0.5 m/s^{2})

T=1.5 N This is the tension in the string

7 0
3 years ago
1. An object (m = 500 g) with an initial speed of 0.2 m/s collides with another object (m = 1.5 kg) which was at rest before the
Anettt [7]

Answer:

Explanation:

1 )

We shall apply conservation of momentum law to solve the problem.

mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .

.5 x .2 = (1.5 + .5)V

V = .05 m /s

2 ) We shall use formula for velocity of object after elastic collision as follows

v₁ = \frac{(m_1-m_2)}{(m_1+m_2)} u_1+\frac{2m_2u_2}{(m_1+m_2)}

m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.

Putting the values

= \frac{(200-1000)}{(1000+200)} \times 1 +\frac{2 \times1000\times0}{(1000+200)}

= - .66 m /s

Since the sign is negative so it will be in opposite direction .

4 0
3 years ago
You are piloting a helicopter which is rising vertically at a uniform velocity of 14.70 m/s. When you reach 196.00 m, you see Ba
Cloud [144]

Answer:

The ball reaches Barney  head in  t = 8 \ s

Explanation:

From the question we are told that

 The rise velocity is  v  =  14.70 \  m/s

  The height considered is h =  196 \  m

   The horizontal velocity of the large object is  v_h  =  8.50 \  m/s

   

Generally from kinematic equation  

   s = ut + \frac{1}{2} gt^2

Here s is the distance of the object from Barney head ,

        u is the velocity of the object along the vertical axis which is equal but opposite to the velocity of the helicopter

So  

     u = -14.7 m/s

So

    196  = -14.7 t  + \frac{1}{2} * 9.8 * t^2

=  4.9 t^2 - 14.7t - 196 = 0

Solving the above equation using quadratic formula  

    The value of  t obtained is  t = 8 \ s

6 0
2 years ago
Two men standing on the same side of wall and at the same distance from It, such that they are 4oom apart when one fires a gun t
Mamont248 [21]

Answer:

1. 571.43m/s

2. 142.9m and 342.9m

Explanation:

1.Take the difference in time.

1.2-0.7=0.7 seconds

Take the distance between them and divide with differnce in time.

400÷0.7=571.43 seconds.

2.Take the time of the two men and divide by two.

0.5÷2= 0.25 secs

1.2÷2= 0.6 secs

multiply each with the velocity.

0.25×571.43=142.9m

0.6×571.43=342.9m

8 0
1 year ago
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