Hey guys who is on.......

telo118 [61]

Answer:

The answer is B.

Explanation:

Given that the <em>current </em>(Ampere) in a series circuit is same so we can ignore it. We can assume that the total voltage is 60V and all the 3 resistance are different, 20Ω, 40Ω and 60Ω. So first, we have to find the total resistance by adding :

Total resistance = 20Ω + 40Ω + 60Ω

= 120Ω

Next, we have to find out that 1Ω is equal to how many voltage by dividing :

120Ω = 60V

1Ω = 60V ÷ 120

1Ω = 0.5V

Lastly, we have to calculate the voltage at R1 so we have to multiply by 20 (R1) :

1Ω = 0.5V

20Ω = 0.5V × 20

20Ω = 10V

8 0

3 years ago

If a cylindrical space station 275 m in diameter is to spin about its central axis, at how many revolutions per minute (rpm) mus

sergij07 [2.7K]

The centripetal acceleration is responsible for the artificial gravity because the acceleration of an object moving in constant circular motion causing from net external force is called centripetal acceleration. It defines to the center or seeking the center.

Given the following:

Cylindrical space station diameter = 275 meters; 137.5 meters for the radius

Standard gravity = 9.80665 m/s²

Using the formula:

w² x r =g

w² = g / r

w² = 9.80665 m/s² / 137.5 m

w² = 0.0713 s²

Then take the roots

w = 0.267 this is radians per second / 2 x (3.1416 which is the pi)

w = 0.0424 rps convert to rpm

w = 0.0424 r/s (1minute / 60 seconds)

w = 7.08 x 10⁻⁴ revolutions per minute

4 0

3 years ago

How much pressure is applied to the ground

statuscvo [17]

Answer:

Pressure applied by the man= 285103.125 $Pa$ or 41.35 $lb/in^{2}$

Explanation:

Pressure is defined as the perpendicular force applied per unit area.

i.e. $Pressure=\frac{Force}{Area}$

Now, $Force= mg$

where, $m$ = mass of the body(man) = 93 kg

$g$ = acceleration due to gravity of Earth = 9.81 $m/{s^{2}}$

$Area$ covered is equal to the area of both stilts(a man generally stands on two feet)

therefore $Area=2(0.04)^{2}$ $m^{2}$

and putting in the values, we get,

$Pressure=\frac{93\times9.81}{2\times(0.04)^{2}}Nm^{-2}=285103.125Nm^{-2}$

Now we need to convert to our required units:

$1Nm^{-2}=1Pa\\1Pa=0.000145038lb/in^{2}$

(We can get the above result by individually converting kg to lb and meters to inches respectively)

Using the above relations we get,

$Pressure=285103.125Pa=0.000145038\times285103.125lb/in^{2}=41.35lb/in^{2}$

7 0

3 years ago

A train whistle has a sound intensity level of 70. dB, and a library has a sound intensity level of about 40. dB. How many times

kodGreya [7K]

Answer:

The sound intensity of train is 1000 times greater than that of the library.

Explanation:

We have expression for sound intensity level,

$L=10log_{10}\left ( \frac{I}{I_0}\right )$

A train whistle has a sound intensity level of 70 dB

We have

$70=10log_{10}\left ( \frac{I_1}{I_0}\right )$

A library has a sound intensity level of about 40 dB

We also have

$40=10log_{10}\left ( \frac{I_2}{I_0}\right )$

Dividing both equations

$\frac{70}{40}=\frac{10log_{10}\left ( \frac{I_1}{I_0}\right )}{10log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\\frac{7}{4}=\frac{log_{10}\left ( \frac{I_1}{I_0}\right )}{log_{10}\left ( \frac{I_2}{I_0}\right )}\\\\10^7\frac{I_2}{I_0}=10^4\frac{I_1}{I_0}\\\\\frac{I_1}{I_2}=10^3=1000$

The sound intensity of train is 1000 times greater than that of the library.

3 0

2 years ago

A(n) 12500 lb railroad car traveling at 7.8 ft/s couples with a stationary car of 7430 lb. The acceleration of gravity is 32 ft/

navik [9.2K]

To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then

$\text{Initial Momentum} = \text{Final Momentum}$