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2 years ago

Krichoffs law of current questions

Physics

1 answer:

postnew [5]2 years ago

6 0

Answer:

Explanation:

Kirchhoff's Current Law, often shortened to KCL, states that “The algebraic sum of all currents entering and exiting a node must equal zero.

#I AM ILLITERATE

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Ice that formed thousands of years ago is often found to contain tiny bubbles of gas. this gas came from __________. ice that fo

motikmotik

Atmosphere

Atmospheric gas from prehistoric eras is found trapped in glaciers in the form of bubbles. These gas bubbles are the basis of studying ice cores as they provide us with accurate estimates of the conditions of past climates. The bubbles allow us to determine the composition of atmospheric air, such as the carbon dioxide and methane concentrations, as well as allow us to determine air temperatures in the past.

8 0

2 years ago

Mike walks 200 km in 6 hours.he then walks another 100km in 4 hours .what is his average speed?

kolbaska11 [484]

Average speed is defined as the ratio of total distance covered in total given time

$speed = \frac{distance}{time}$

here we know that total distance that man moved is

$d_1 = 200 km$

$d_2 = 100 km$

so total distance is

$d = d_1 + d_2$

$d = 200 km + 100 km$

$d = 300 km$

now here total time of the motion is

$t_1 = 6 hours$

$t_2 = 4 hours$

total time will be given as

$t = t_1 + t_2$

$t = 6 + 4 = 10 hours$

now by above formula

$v_{avg} = \frac{300}{10}$

$v_{avg} = 30 km/h$

so his average speed is 30 km/h

8 0

3 years ago

A child goes down the slide,starting from rest. If the length of the slide is 2m and it takes the child 3 seconds to go down the

lapo4ka [179]

Answer:

$0.44 m/s^2$

Explanation:

We have the following data:

- distance covered by the child: d = 2 m (length of the slide)

- time taken to cover this distance: t = 3 s

- initial velocity of the child: 0 m/s (he starts from rest)

So we can find the acceleration by using the equation:

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

Where a is the acceleration.

Substituting the values and solving for a,

$a=\frac{2d}{t^2}=\frac{2(2)}{3^2}=0.44 m/s^2$

3 0

3 years ago

It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used

stepladder [879]

Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

$a_{c}$ = rω²

ω² = $a_{c}$ / r

ω = √( $a_{c}$ / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a $a_{c}$ = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²

ω = 0.0500647 ≈ 0.05 rad/s

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM

1 rpm = 60 rph

ω = 0.478082 × 60

ω = 28.6849 revolutions per hour

Therefore, the required revolution per hour is 28.6849

7 0

2 years ago

I need homework help

KIM [24]

1) the weight of an object at Earth's surface is given by $F=mg$ , where m is the mass of the object and $g=9.81 m/s^2$ is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is
$F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N$

2) On Mars, the value of the gravitational acceleration is different: $g=3.7 m/s^2$ . The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: $F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N$

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus:
$g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2$

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg:
$g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2$

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as
$g= \frac{F}{m} = \frac{0.3 N}{0.5 kg} =0.6 m/s^2$

6) On Earth, the gravity acceleration is $g=9.81 m/s^2$ . The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is
 $F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N$

5 0

2 years ago

Krichoffs law of current questions​

Krichoffs law of current questions