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Zepler [3.9K]
1 year ago
11

Tamsin strikes a golf ball so that it flies off with a speed of 60 m/s. The golf ball has a mass of 45 g. Calculate the kinetic

energy of the golf ball.
Physics
1 answer:
hram777 [196]1 year ago
7 0

81 joule is the kinetic energy of the golf ball.

K.E= mv²/2

K.E=0.045×60×60÷2

K.E=81 joule

Kinetic energy is a particular kind of power that is present in moving particles or objects. An object gains kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force. Kinetic energy is a property of motion that depends on the mass and speed of an object or particle. Motion is any combination of vibration, axis rotation, translation, and movement (along a path from one location to another). A body's translational kinetic energy, which is determined by multiplying its mass, m, by the square of its speed, v, is equal to 1/2mv2.

To know more about  kinetic energy visit : brainly.com/question/14838079

#SPJ4

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The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

Explanation:

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First charge q_{1}= 2.9\mu C

Second chargeq_{2}= 2.9\mu C

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

E_{1}=\dfrac{kq}{r'^2}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}

E_{1}=52200=52.2\times10^{3}\ N/C

For E₂,

Using formula of electric field

E_{1}=\dfrac{kq}{r^2}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}

E_{2}=104400=104.4\times10^{3}\ N/C

We need to calculate the horizontal electric field

E_{x}=E_{1}\cos\theta

E_{x}=52.2\times10^{3}\times\cos45

E_{x}=36910.97=36.9\times10^{3}\ N/C

We need to calculate the vertical electric field

E_{y}=E_{2}+E_{1}\sin\theta

E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45

E_{y}=141310.97=141.3\times10^{3}\ N/C

We need to calculate the net electric field

E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

E_{net}=146038.69\ N/C

E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

\theta=75.36^{\circ}

Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

4 0
3 years ago
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