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3 years ago

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A 0.280 m radius, 475 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a

uniform magnetic field. (This is 60 rev/s.) Find the magnetic field strength needed to induce an average emf of 10,000 V.

1 answer:

KonstantinChe [14]3 years ago

8 0

Hope it helps u mate. Plz follow me .

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The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.

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Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let $\rho_{m}$ be the density of metal and $\rho_{w}$ be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

$W =0.10400\ N$

$mg=0.10400$

$\rho V g=0.10400$

$Vg=\dfrac{0.10400}{\rho_{m}}$ .....(I)

The weight of metal in water is

Using buoyancy force

$F_{b}=0.10400-0.08400$

$F_{b}=0.02\ N$

We know that,

$F_{b}=\rho_{w} V g$ ....(I)

Put the value of $F_{b}$ in equation (I)

$\rho_{w} Vg=0.02$

Put the value of Vg in equation (II)

$\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02$

$1000\times\dfrac{0.10400}{0.02}=\rho_{m}$

$\rho_{m}=5200\ kg/m^3$

Hence, The density of the metal is 5200 kg/m³.

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Explanation:

This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:

$V_{f}=V_{o}+a.t$ (1)

Where:

$V_{f}$ is the final velocity of the supply bag

$V_{o}=0$ is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)

$a=g=-9.8m/s^{2}$ is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)

$t=3.2s$ is the time

Knowing this, let's solve (1):

$V_{f}=0+(-9.8m/s^{2})(3.2s)$ (2)

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