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Leokris [45]
3 years ago
5

Constructive interference occurs when the crests of one wave overlap the crests of another wave. What is the result of construct

ive interference? A. decreased frequency B. increased frequency C. decreased amplitude D. increased amplitude
Physics
2 answers:
Zigmanuir [339]3 years ago
4 0
That would be increased amplitude
GarryVolchara [31]3 years ago
4 0

Answer:

D.

hope this helps

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The world speed record on water was set on October
Luda [366]

Answer: Remember speed is distance divided by time, so if he travels 1000 m in 7.045 s, his speed is

(1000 m)/(7.045 s) = 141.9 m/s.

Note there are 1609 metres in a mile, or 1 mi = 1609 m, so m = 1/1609 mi, or

141.9/1609 mi/s = 0.08822 mi/s. Now, note that 1 h = 3600 s, so the speed is

0.08822*3600 mi/h = 317.6 mi/h.

8 0
3 years ago
A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition
Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So mg=Kx

K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m

Now we know that \omega =\sqrt{\frac{K}{m}}

So \frac{2\pi }{T} =\sqrt{\frac{K}{m}}

\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}

T=2.1371sec

8 0
2 years ago
A box is sitting on a 2 m long board at one end. A worker picks up the board at the end with the box so it makes an angle with t
Vlada [557]

Answer:

Explanation:

When the box is on the ramp , component of its weight along the ramp

= mg sinθ

Friction force acting on it in upward direction

=μ mg cosθ

For sliding

μ mg cosθ < mg sinθ

μ cosθ < sinθ

.5 x cos35 < sin35

.41 < .57

So the box will slide

When sliding starts , kinetic friction acts

Net force in downward direction

mgsinθ - μ mg cosθ

acceleration

= gsinθ - μ g cosθ

= 5.62 - .3 x 9.8 x cos35

= 5.62 - 2.4

= 3.22 m /s²

3 0
2 years ago
A scaffold of mass 57 kg and length 6.5 m is supported in a horizontal position by a vertical cable at each end. A window washer
Mrac [35]

Explanation:

The given data is as follows.

     m_{1} = 57 kg,        m_{2} = 79 kg

      l_{1} = 6.5 m,         l_{2} = (6.5 - 1.9) m = 4.6 m

(a)  The sum of torque ends about far end is as follows.

    m_{1}g \frac{l_{1}}{2} + m_{2}g \times l_{2} - T \times l_{1} = 0

     57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (6.5 - 1.9) - T \times 6.5 = 0

                     T = 828 N

Therefore, 828 N is the tension in the cable closer to the painter.

(b)  Now, we will calculate the sum about close ends as follows.

m_{1}g \frac{l_{1}}{2} + m_{2}g \times (1.9) - T \times l_{1}    

  57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (1.9) - T \times 6.5 = 0

                            T= 506 N

Therefore, 506 N is the tension in the cable further from the painter.  

8 0
3 years ago
A boxed 12.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang
satela [25.4K]

Nah a a yuloure

Explanation:

5 0
3 years ago
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