Ask question

Ask question

All categories

3 years ago

6

PLEASE HELP.

2 answers:

Lelu [443]3 years ago

5 0

The frequency of the wave will not change

zalisa [80]3 years ago

4 0

Answer:

the frequency of the wave wont change because it just wont and i got it right

Explanation:

You might be interested in

How far is tau hercules in light years from earth?

Nady [450]

Hipparcos estimated its distance at roughly 96 parsecs from Earth, or 310 ± 20 light years away.

8 0

3 years ago

A 5.00-g bullet is fired into a 900-g block of wood suspended as a ballistic pendulum. The combined mass swings up to a height o

Thepotemich [5.8K]

Answer:

The value is $KE_b =0.710 \ J$

Explanation:

From the question we are told that

The mass of the bullet is $m_b = 5.00 \ g = 0.005 \ kg$

The mass of the wood is $m_w = 900 \ g = 0.90\ kg$

The height attained by the combined mass is $h = 8.0 \ cm = 0.08 \ m$

Generally according to the law of energy conservation

$KE _b = PE_c$

Here $KE_b$ is the kinetic energy of the bullet before collision.

and $PE_c$ is the potential energy of the combined mass of bullet and wood at the height h which is mathematically represented as

$PE_m = [m_b + m_w] * g * h$

So

$KE_b =PE_c = [0.005 + 0.90] * 9.8 *0.08$

=> $KE_b =0.710 \ J$

3 0

3 years ago

What is the <br>newton's first law of motion

koban [17]

Answer:

Anybody which is in state of rest ,will be in rest if we don't apply any external force ...

8 0

3 years ago

Read 2 more answers

HELP HELP HELP HELP IIII NEED HELP!!!!

romanna [79]

Answer:

Heavy Rains.

These heavy rains can lead to severe flooding by oversaturating the ground, overfilling storm drains, or causing rivers to spill over their banks or levees

6 0

2 years ago

Read 2 more answers

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

7 0

3 years ago