All categories

2 years ago

Helpppp!!!

Physics

1 answer:

Korvikt [17]2 years ago

5 0

Answer:

A) 0.25 Hz

B) 6 m

C) 1.5 m/s

Explanation:

The number of waves that pass a certain spot every 20 seconds = 5 waves

The distance between the crest = 6 m

A) The frequency of the wave, f = The number of cycles per second = 5/20 = 0.25 per second = 0.25 Hz

B) The wavelength of the wave, λ = The distance between crests = 6 m

C) The velocity of the wave, v = f × λ = 0.25/s × 6 m = 1.5 m/s.

You might be interested in

A block of mass m = 1.5 kg is released from rest at a height of H = 0.81 m on a curved frictionless ramp. At the foot of the ram

kogti [31]

Answer:

0.31 m

Explanation:

m = mass of the block = 1.5 kg

H = height from which the block is released on ramp = 0.81 m

k = spring constant of the spring = 250 N/m

x = maximum compression of the spring

using conservation of energy

Spring potential energy gained by spring = Potential energy lost by block

(0.5) k x² = mgH

(0.5) (250) x² = (1.5) (9.8) (0.81)

x = 0.31 m

7 0

3 years ago

Although all sports require a person to be in shape to some degree, some sports demand more from the body. Which sport does not

RSB [31]

Answer:

I think golf does not demand high level of physical fitness. A golfer just need be skillful and able to play under pressure. In terms of physical exercise i think he/she just need to do some muscle stretches.

Explanation:

3 0

3 years ago

Read 2 more answers

The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

Gennadij [26K]

Answer:

Part a)

$k = 6.06 \times 10^4 N/m$

Part b)

$b = 1795.4 kg/s$

Explanation:

Part a)

as the mass of the suspension system is given as

$m = 2100 kg$

also we have

$x = 8.5 cm$

so now for force balance we have

$mg = kx$

$(525)(9.81) = k(0.085)$

$k = 6.06 \times 10^4 N/m$

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

$A = 0.37 A_o$

also we know

$A = A_o e^{-bt/2m}$

$0.37 A_o = A_o e^{-bt/2m}$

$\frac{bt}{2m} = 1$

$b = \frac{2m}{t}$

here t = time period of one oscillation

so it is

$t = 2\pi\sqrt{\frac{m}{k}}$

$t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}$

$t = 0.58 s$

now damping constant is

$b = \frac{2(525)}{0.58}$

$b = 1795.4 kg/s$

7 0

3 years ago

What’s the answer please?

photoshop1234 [79]

Answer:

Explanation:

Hope this helps!

8 0

3 years ago

In an oil drop experiment, a drop with a weight of 1.9 x 10-14 N was suspended when the potential difference between 2 plates th

AlekseyPX

Answer:

Study More You will get it :L

5 0

2 years ago