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oee [108]
4 years ago
6

Explain how momentum is transferred when a golfer hits a ball with a golf club.

Physics
1 answer:
melomori [17]4 years ago
7 0
Momentum is transferred through the convert wich in this case is the golf club
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50. A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 335 kV. The secondar
Sedaia [141]

Answer:

Explanation:

a )  The transformer steps  up  the voltage from 12000 V  to 335000 V . Voltage in primary is 12000 V and in the secondary it is 335000 V in old transformer

If n₁ be no of turns in primary coil and n₂ be no of turns in secondary coils

the formula is

n₂ / n₁ = voltage in secondry / voltage in primary

n₂ / n₁ = 335000 / 12000

ratio of turns  in old transformer

= 27.9

ratio of turns  in new transformer

n₃ / n₁ = 750 / 12 ( n₃ is no of turns in the  secondary of new transformer )

= 62.5

T he ratio of turns in the new secondary compared with the old secondary

n₃ / n₂ = 62.5 / 27.9

= 2.24

b ) Current in secondary / current in primary

= turns in primary / turns in secondary

current output ratio of old

= Current in secondary / current in primary

= n₁ / n₂

= 12 / 335

= .03582

current output ratio of new

= Current in secondary / current in primary

= n₁ / n₃

= 12 / 750  

= .016

The ratio of new current output to old output (at 335 kV) for the same power

= .016 / .03582

= .4466

c ) power loss in new

=  (current in secondary )² x resistance of secondary

=( .016 x current in primary )² x R

= 2.56 X 10⁻⁴ X ( current in primary )² x R

power loss in old  

=  (current in secondary )² x resistance of secondary

=( .03582 x current in primary )² x R

= 12.83 X 10⁻⁴ X ( current in primary )² x R

ratio of new line power loss to old

= 2.56 / 12.83

= .199

7 0
3 years ago
A 170 g air-track glider is attached to a spring. The glider is pushed in 11.2 cm against the spring, then released. A student w
4vir4ik [10]

Answer:k = 10.83 N/m²

Explanation: The angular frequency (ω), spring constant (k) and mass is related by the formulae below

ω = √k/m

But ω = 2πf, where f = frequency.

f = number of oscillations /time taken

Number of oscillations = 14, time taken = 11s

f = 14/11 = 1.27Hz.

ω = 2×22/7×1.27

ω = 7.98 rad/s.

By substituting this parameters into ω = √k/m

Where ω = 7.98rad/s, m = 170g = 170/1000 = 0.17kg.

7.98 = √k/0.17

By squaring both sides

(7.98)² = k/ 0.17

k = (7.98)² × 0.17

k = 10.83 N/m²

7 0
4 years ago
An object with a mass of 1500kg accelerates 10.0m/s when unknown force is applied to it. What is the amount of force?
Hatshy [7]

Answer:

<h3>The answer is 15000 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 1500 × 10

We have the final answer as

<h3>15,000 N</h3>

Hope this helps you

5 0
3 years ago
Read 2 more answers
The planets move around the stars in different orbits .which planet will have the shortest of revolution (year)?
Ivahew [28]

4 because it's closest to the star

3 0
3 years ago
Add a vector whose magnitude is 13 with angle 27 degrees to one whose magnitude is 11 with angle 45 degrees? Put the length firs
mafiozo [28]

Answer:

Magnitude of the vector is 23.75\ \text{units} and the direction is 35.23^{\circ}

Explanation:

Magnitude of first vector = |A| = 13\ \text{units}

Angle = \theta_1=27^{\circ}

Magnitude of second vector = |B| = 11\ \text{units}

Angle = \theta_2=45^{\circ}

x component of first vector

A_{x}=|A|\cos\theta_1\\\Rightarrow A_x=13\cos27^{\circ}\\\Rightarrow A_x=11.6\ \text{units}

y component of first vector

A_{y}=|A|\sin\theta_1\\\Rightarrow A_y=13\sin27^{\circ}\\\Rightarrow A_y=5.9\ \text{units}

x component of second vector

B_{x}=|B|\cos\theta_2\\\Rightarrow B_x=11\cos45^{\circ}\\\Rightarrow B_x=7.8\ \text{units}

y component of first vector

B_{y}=|B|\sin\theta_2\\\Rightarrow B_y=11\sin45^{\circ}\\\Rightarrow A_y=7.8\ \text{units}

Adding the magnitudes

C_x=A_x+B_x=11.6+7.8\\\Rightarrow C_x=19.4\ \text{units}

C_y=A_y+B_y=5.9+7.8\\\Rightarrow C_y=13.7\ \text{units}

Magnitude of the sum of the vectors would be

|C|=\sqrt{C_x^2+C_y^2}\\\Rightarrow |C|=\sqrt{19.4^2+13.7^2}=23.75\ \text{units}

The direction would be

\theta=\tan^{-1}\dfrac{C_y}{C_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{13.7}{19.4}\\\Rightarrow \theta=35.23^{\circ}

The magnitude of the vector is 23.75\ \text{units} and the direction is 35.23^{\circ}

4 0
3 years ago
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