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Irina-Kira [14]
3 years ago
13

When Woods hits a 0.04593 kg golf ball, the golf ball is usually traveling around 281 kilometers per hour. What average force do

es he exert on a golf ball if the club head and golf ball are in contact for 0.030 seconds?
Physics
1 answer:
Semenov [28]3 years ago
7 0

Answer:

128.9 N

Explanation:

The force exerted on the golf ball is equal to the rate of change of momentum of the ball, so we can write:

F=\frac{\Delta p}{\Delta t}

where

F is the force

\Delta p is the change in momentum

\Delta t=0.030 s is the time interval

The change in momentum can be written as

\Delta p = m(v-u)

where

m = 0.04593 kg is the mass of the ball

u = 0 is the initial velocity of the ball

v=281 km/h =78.1 m/s is the final velocity of the ball

Substituting into the original equation, we find the force exerted on the golf ball:

F=\frac{m(v-u)}{\Delta t}=\frac{(0.04593)(78.1-0)}{0.030}=128.9 N

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Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet
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Answer:

(a).The maximum current in the wire is 4.217\times10^{5}\ A.

(b). The electric field in the wire is 11.2\times10^{5}\ N/C.

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

Explanation:

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Diameter of metal plates = 10 cm

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Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

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Put the value into the formula

A=\pi\times(5\times10^{-2})^2

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We need to calculate the capacitor

Using formula of capacitor

C=\dfrac{\epsilon_{0}A}{d}

Put the value into the formula

C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}

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Using formula of resistivity

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Put the value into the formula

R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}

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We need to calculate the voltage

Using formula of charge

q=CV

V=\dfrac{q}{C}

Put the value into the formula

V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}

V=1.801\times10^{3}\ V

(a). We need to calculate the current

Using formula of current

I=\dfrac{V}{R}

I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}

I=421779.85\ A

I=4.217\times10^{5}\ A

(b). We need to calculate the electric field

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}

E=11.2\times10^{5}\ N/C

The electric field in the wire is 11.2\times10^{5}\ N/C.

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

E=\dfrac{1}{2}CV^2

Put the value into the formula

E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2

E=1.126\times10^{-5}\ J

The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

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(b). The electric field in the wire is 11.2\times10^{5}\ N/C.

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

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