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2 years ago

2 Name the device in a car that we use to change the speed of the car

Physics

1 answer:

Crank2 years ago

6 0

Answer:

Speedometer

Explanation:

Have a good day! :3

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The percent by which the fundamental frequency changed if the tension is increased by 30 percent is ? a)-20.04% b)-40.12% c)-30%

nika2105 [10]

Answer:

Percentage increase in the fundamental frequency is

d)-14.02%

Explanation:

As we know that fundamental frequency of the wave in string is given as

$f_o = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

now it is given that tension is increased by 30%

so here we will have

$T' = T(1 + 0.30)$

$T' = 1.30T$

now new value of fundamental frequency is given as

$f_o' = \frac{1}{2L}\sqrt{\frac{1.30T}{\mu}}$

now we have

$f_o' = \sqrt{1.3}f_o$

so here percentage change in the fundamental frequency is given as

$change = \frac{f_o' - f_o}{f_o} \times 100$

% change = 14.02%

3 0

2 years ago

Which two options are examples of waves reflecting?

Umnica [9.8K]

Answer:

transverse

Explanation:

none

5 0

2 years ago

How much work is done by the force lifting a

Ann [662]

The work done in lifting the hamburger is equal to the increase in gravitational potential energy of the hamburger, given by

$W=\Delta U=mg \Delta h$

where

m=0.1 kg is the mass of the hamburger

$g=9.81 m/s^2$ is the gravitational acceleration

$\Delta h=0.3 m$ is the increase in height of the hamburger

Substituting numbers into the equation, we find

$W=(0.1 kg)(9.81 m/s^2)(0.3 m)=0.3 J$

So, the correct answer is

(3) 0.3 J

3 0

3 years ago

15 POINTS PLEASE HELP! NO GUESSING

torisob [31]

He answer is A. <span>encourage agricultural usage in the watershed

if you want to read it for yourself go to
www.nature.org/ourinitiatives/regions/northamerica/unitedstates/indiana/journeywithnature/watersheds...

hope this helps you!!</span>

7 0

3 years ago

Read 2 more answers

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Pressure at reservoir = 10 atm

$T_1$ = Temperature at reservoir = 300 K

$P_2$ = Pressure at exit = 1 atm

$T_2$ = Temperature at exit

$R_s$ = Mass-specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

For isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

The density of the flow at the exit is 2.2721 kg/m³

4 0

2 years ago