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3 years ago

What is the best estimate of the frequency of the wave shown below

Physics

1 answer:

GrogVix [38]3 years ago

5 0

Answer:

D. 2.5 Hz

Explanation:

Frequency = speed of wave / wavelength

= 335 /140 ( from graph)

= 2.4

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I was driving along at 20 m/s , trying to change a CD and not watching where I was going. When I looked up, I found myself 46 m

vampirchik [111]

Answer:

6.46393559312 m/s²

Explanation:

Time taken to cover 56 m

$t=\dfrac{56}{29}\ s$

Distance covered in 0.42 seconds

$0.42\times 20=8.4\ m$

From equation of linear motion

$s=ut+\frac{1}{2}at^2$

$8.4+20(\dfrac{56}{29}-0.42)+\dfrac{1}{2}a(\dfrac{56}{29}-0.42)^2=46\\\Rightarrow a=(46-8.4-20(\dfrac{56}{29}-0.42))\times\dfrac{2}{(\dfrac{56}{29}-0.42)^2}\\\Rightarrow a=6.46393559312\ m/s^2$

The minimum acceleration is 6.46393559312 m/s²

5 0

3 years ago

In which state of matter are particles held together but still able to move around

Aloiza [94]

The answer is B because the water molecules

5 0

3 years ago

Read 2 more answers

A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro

tekilochka [14]

Answer:

distance between object and image = 18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm ......................3

so from equation 2 and 3

distance between object and image = 5.4 + 13.5

distance between object and image = 18.9 cm

6 0

3 years ago

. A ball is thrown downward at a speed of 20 m/s. Choosing

a_sh-v [17]

The final velocity is $v = 20 - gt$

The distance traveled by the ball at time t is $y = y_o + (-20)t - \frac{1}{2} gt^2$

The maximum distance traveled by the object is $0 = (20)^2 - 2g(y-y_0)$

The given parameters;

initial velocity of the ball, u = 20 m/s

acceleration due to gravity, g = 9.8 m/s²

The final velocity can be calculate as;

$v = 20 - gt$

The distance traveled by the ball at time t;

$y = y_o + (-20)t - \frac{1}{2} gt^2$

The maximum distance traveled by the object is calculated as;

$v^2 = u^2 - 2g(y - y_0)\\\\0 = (20)^2 - 2g(y-y_0)$

Learn more here: brainly.com/question/16878713

3 0

2 years ago

A car weighing 14,700 N is speeding down a highway with a velocity of 99 km/h. What is the

tankabanditka [31]

Answer: 148348.6239 kg•m/s

Explanation: Firstly, we need to convert the 14700 N into kilograms, and to do so, use the formula net force is equal to mass times acceleration and rearrange the formula to find mass like shown below...

F = ma

F/a = m

14700/9.81 = 1498.470948 kg, this is your mass

Now that we convert it into kilograms, plug all the numbers into the variable of the momentum formula.

Momentum formula is P = mass x velocity

Like this:

P = 1498.470948 x 99

p = 148348.6239 kg•m/s.

I believe that is your answer, hope that helps you even a bit out.

Thanks.

7 0

2 years ago