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3 years ago

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Lified True or False

1 answer:

Nana76 [90]3 years ago

5 0

Trueeeeeeeeeeeeeeeeeeeeeee

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An energetic child runs forward and backward. It's motion is shown on the following graph of horizontal position x vs time t

Klio2033 [76]

Answer: 0.50 m/s

Explanation: it’s on khan academy this was the correct answer

6 0

3 years ago

If you want to double the kinetic energy of a gas molecule, by what factor must you increase its momentum?A) 16B) 2^(1/2)

worty [1.4K]

Answer: $\sqrt{2}$

The linear momentum $p$ is given by the following equation:

$p=m.v$ (1)

Where $m$ is the mass and $v$ the velocity.

On the other hand, the kinetic energy $K$ is given by:

$K=\frac{p^{2}}{2m}$ (2)

Which is the same as:

$K=\frac{1}{2}m.v^{2}$

Now, if we double the kinetic energy, equation (2) changes to:

$2K=2\frac{p^{2}}{2m}$

$2K=\frac{p^{2}}{m}$ (3)

So, if we want to obtain the kinetic energy as shown in (3), the only option that works is increasing momentum by a factor of $\sqrt{2}$ or $2^{1/2}$ :

Applying this in (2):

$K=\frac{(\sqrt{2}p)^{2}}{2m}$

$K=\frac{(2p)^{2}}{2m}$

$K=\frac{p^{2}}{m}$ >>>As we can see, this equation is the same as equation (3)

Therefore, the correct answer is B

3 0

3 years ago

How is energy involved in physical changes and in chemical changes?

GaryK [48]

During phase changes, energy changes are usually involved. For example, when solid dry ice vaporizes (physical change), carbon dioxide molecules absorb energy. Meanwhile, when liquid water becomes ice energy is released.

hope this helps :)

4 0

3 years ago

A flexible loop has a radius of 10.5 cm and it is in a magnetic field of B = 0.117 T. The loop is grasped at points A and B and

MAXImum [283]

Explanation:

Given that,

Radius = 10.5 cm

Magnetic field = 0.117 T

Time = 0.243 s

After stretched, area is zero

(I). We need to calculate the magnetic flux through the loop before stretched

Using formula of magnetic flux

$\phi=B\times A$

$\phi=B\times \pi r^2$

Where, B = magnetic field

r = radius

Put the value into the formula

$\phi=0.117\times3.14\times(10.5\times10^{-2})^2$

$\phi=4.05\times10^{-3}\ Tm^2$

(II). We need to calculate the magnetic flux through the loop after stretched

$\phi=B\times A$

Here, A = 0

$\phi=0$

So, The magnetic flux through the loop after stretched is zero.

(III). We need to calculate the magnitude of the average induced electromotive force

Using formula of the induced electromotive force

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{\phi_{after}-\phi_{before}}{t}$

$\epsilon=-\dfrac{0-4.05\times10^{-3}}{0.243}$

$\epsilon =16.67\times10^{-3}\ V$

Hence, This is the required solution.

3 0

3 years ago

A body of mass 450g changes it speed from 5ms¹ to 25ms¹. what is the work done by the body?

Andrej [43]

Answer:

135J

Explanation:

So we know ΔKinetic Energy= ΔWork

Kinetic energy=1/2mv²

So Kf-Ki=ΔK

ΔK=1/2*0.45(25²-5²)=135J

135J=ΔWork

3 0

3 years ago