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3 years ago

1. A 12 kg dog is running 5 m/s. How much kinetic energy does the dog have?

Physics

1 answer:

Neporo4naja [7]3 years ago

6 0

Answer:

Explanation:

1. KE = 1/2mv^2

= 1/2 (12)×5^2

= 6× 25

KE = 150j

2. KE = 1/2 mv^2

100j = 1/2m(5)^2

2(100j) = m×25

m = 200÷ 25

m = 8 kg

hope it helps

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A solenoid coil with 22 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 21.0 cm long an

horsena [70]

Answer:

The average magnetic flux through each turn of the inner solenoid is $11.486\times10^{-8}\ Wb$

Explanation:

Given that,

Number of turns = 22 turns

Number of turns another coil = 330 turns

Length of solenoid = 21.0 cm

Diameter = 2.30 cm

Current in inner solenoid = 0.140 A

Rate = 1800 A/s

Suppose For this time, calculate the average magnetic flux through each turn of the inner solenoid

We need to calculate the magnetic flux

Using formula of magnetic flux

$\phi=BA$

$\phi=\dfrac{\mu_{0}N_{2}I}{l}\times\pi r^2$

Put the value into the formula

$\phi=\dfrac{4\pi\times10^{-7}\times330\times0.140}{21.0\times10^{-2}}\times\pi\times(\dfrac{2.30\times10^{-2}}{2})^2$

$\phi=11.486\times10^{-8}\ Wb$

Hence, The average magnetic flux through each turn of the inner solenoid is $11.486\times10^{-8}\ Wb$

7 0

3 years ago

Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

3 0

3 years ago

Using what you know about air pressure and the water cycle, explain how hurricanes form. Why are there certain places on Earth m

Yuki888 [10]

Hurricanes form when hot air and cold air start colliding above water and they start to form a tornado but there is water in the tornado. Certain places on earth are more susceptible to hurricanes because they are closer to the coastlines. For example, in Iowa, and Illinois, they have a lower chance of a hurricane, while Florida, and Washington have a higher rate because they are along the ocean. Hope this helps.

4 0

3 years ago

You are working out on a rowing machine. Each time you pull the rowing bar (which simulates the "oars") toward you, it moves a d

tatyana61 [14]

Answer:

54.67 N

Explanation:

The total energy produced is the product of power and time duration:

E = Pt = 82 * 1 = 82 J

Which is converted from work, product of forced extended over a displacement

W = E = Fs = F*1.5 = 82

F = 82 / 1.5 = 54.67 N

So the magnitude of the force exerting on the handle is 54.67 N

8 0

3 years ago

Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli

Thepotemich [5.8K]

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

Explanation:

Given:

Mass of Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = $v_{f}=?$

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

$m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}$

Substituting the values we get

$2\times 14 + 5\times 0 =(2+5)\times v_{f}$

$v_{f}=\dfrac{28}{7}=4\ m/s$

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

8 0

3 years ago