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1 year ago

Whats the usefulness or inconvenience of frictional force in a soccer player running with soccer boots (studs underneath)

Physics

1 answer:

elena55 [62]1 year ago

4 0

Answer:

They're are useful in providing larger frictional force

Explanation:

The football shoes have spikes or studs because the studs or spikes provides larger frictional force than normal shoes while running on the grass.

The studs helps to prevent a player from slipping on the grass and help to run faster and change direction quickly without slipping.

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How does the constitution protect the rights of individuals against vivo

balandron [24]

The constitution protects the rights of people because in the constitution is the bill of rights. Which are admendments, which contain all or our rights.

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3 years ago

A proton has been accelerated from rest through a potential difference of -1350 V. What is the proton's kinetic energy, in elect

Luba_88 [7]

Answer:

1 eV = 1.60 * 10^-19 J work done in accelerating electron thru 1 V

KE (total energy) = 1350 ^ 1 eV (note proton goes from + to -)

KE = 1.60 * 10^-19 * 1350 = 2.16 * 10^-16 Joules

1/2 m v^2 = KE = 2.16 * 10^-16 J

v^2 = 4.32 * 10E-16 / 1.67 * 10-27 = 2.59 * 10^11

v = 5.09 * 10^5 m/s

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3 years ago

At sunset, the sun appears reddish. What is MOST LIKELY the reason for this phenomenon?

Alexeev081 [22]

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3 years ago

Read 2 more answers

A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height

rewona [7]

Answer:

$W=17085KJ$

Explanation:

From the question we are told that:

Height $H=16m$

Radius $R=3$

Height of water $H_w=9m$

Gravity $g=9.8m/s$

Density of water $\rho=1000kg/m^3$

Generally the equation for Volume of water is mathematically given by

$dv=\pi*r^2dy$

$dv=\frac{\piR^2}{H^2}(H-y)^2dy$

Where

y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

$dw=(pdv)g (H-y)$

Substituting dv

$dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)$

$dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy$

Therefore

$W=\int dw$

$W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy$

$W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)$

$W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0$

$W=3420.84*0.25[2401-65536]$

$W=17084965.5J$

$W=17085KJ$

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3 years ago

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sattari [20]

I can't make sense of this question. Julie's throwing the ball, so it's leaving her rather than arriving at her ???

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