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sergiy2304 [10]

3 years ago

10

A 0.42 kg soccer ball is moving down field with a velocity of 12 m/s. A player kicks the ball so that it has a final velocity of

18 m/s down field. What is the change in the balls momentum? and find the constant force exerted by the players foot on the ball if the two are in contact for 0.020 s?

1 answer:

KATRIN_1 [288]3 years ago

8 0

Answer:

the change in momentum is 2.52 N·s

the constant force on the ball while in contact with the players foot is 126 N

Explanation:

p=momentum

m=mass

v=change in velocity

p=(0.42)(6)

p=2.52

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The length of a certain wire is kept same while its radius is doubled. what is the change in the resistance of this wire?

kaheart [24]

The resistance of a wire is given by
$R= \frac{\rho L}{A}$
where $\rho$ is the resistivity of the material, L the length of the wire and A its cross-sectional area.

In the problem, $\rho$ and L remain the same, while A changes because the radius changes. The area is given by:
$A=\pi r^2$
This means that if we double the radius (2r), the area becomes
$A_{new}= \pi (2r)^2 = 4 \pi r^2 =4A$
And therefore, the new value of the resistance is
$R_{new} = \frac{\rho L}{4 A}= \frac{1}{4}R$

So, when the radius is doubled, the resistance becomes $\frac{1}{4}$ of its original value.

6 0

4 years ago

If the block is subjected to the force of F = 500 N, determine its velocity at s = 0.5 m. When s = 0, the block is at rest and t

STatiana [176]

Answer:

The velocity is $4.6 m/s^2$

Explanation:

Given:

Force = 500N

Distance s= 0

To find :

Its velocity at s = 0.5 m

Solution:

$\sum F_{x}=m a$

$F\left(\frac{4}{5}\right)-F_{S}=13 a$

$500\left(\frac{4}{5}\right)-\left(k_{s}\right)=13 a$

$400-(500 s)=13 a$

$a = \frac{400 -(500s)}{13}$

$a = (30.77 -38.46s) m/s^2$

Using the relation,

$a=\frac{d v}{d t}=\frac{d v}{d s} \times \frac{d s}{d t}$

$a=v \frac{d v}{d s}$

$v d v=a d s$

Now integrating on both sides

$\int_{0}^{v} v d v=\int_{0}^{0.5} a d s$

$\int_{0}^{v} v d v=\int_{0}^{0.5}(30.77-38.46 s) d s$

$\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}$

$\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right]$

$\left[\frac{v^{2}}{2}\right]=[15.385-4.807]$

$\left[\frac{v^{2}}{2}\right]=10.578$

$v^{2}=10.578 \times 2$

$v^{2}=21.15$

$v = \sqrt{21.15}$

$v = 4.6 m/s^2$

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3 years ago

Examine the reactants of the incomplete double displacement reaction.

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<h3>Answer:</h3>

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<h3>Explanation:</h3>

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In this case, silver nitrate and sodium chloride exchange anions and cations to form silver chloride and sodium nitrate.

Therefore, the complete reaction is given by;

AgNO3 + NaCl → AgCl + NaNO₃

But since silver chloride is a precipitate, the reaction may also be an example of a precipitation reaction.

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3 years ago

Goshia [24]

Answer:

the anwer is true , because production will decrease

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3 years ago

If the frequency of this beam is increased while the intensity is held constant, does the number of electrons ejected per second

Ivan

Answer:

if the intensity of photons is constant then number of ejected electrons will remain same

Explanation:

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Here the energy of photons is used to eject out the electrons from metal surface and to give the kinetic energy to the ejected electrons

so we have

$h\nu = W + KE$

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KE = kinetic energy of ejected electrons

now if we increase the frequency of the photons that incident on the metal surface then in that case the incident energy will increase

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