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3 years ago

8

An athlete whirls a 8.00 kg hammer m from the axis of rotation in a horizontal circle, as shown in the figure below . The hammer

makes one revolution in s. What is the centripetal force on the hammer ?

1 answer:

Zigmanuir [339]3 years ago

4 0

Answer:

$mv \div r \\ mass \times velocity \div radius$

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Give an example of a system whose mass is not constant.

Sloan [31]

A spinning top is the answer

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3 years ago

A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

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Explanation:

Given: $\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}$

$\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}$

$\:\:\:\:\:= 6×10^{14}\:\text{Hz}$

The work function $\phi$ is then

$\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})$

$\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}$

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2 years ago

Two light bulbs are installed in two sockets, the sockets are connected in series, and power is then applied to the combination

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Answer:

False

Explanation:

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This is because the removal of one of the bulbs would interrupt the flow of current though the entire circuit.

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Well the Answer is True.

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