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2 years ago

A set of data is collected for object in an inelastic collision, as recorded in the table.

Physics

1 answer:

TiliK225 [7]2 years ago

5 0

Answer:

To identify the momentum of object 1, you must multiply mass (m) and velocity(v) to find momentum.

Object 1 has momentum of 8 kg. m/s before collision.

Object 1 has momentum of 0 kg. m/s before collision.

The combined mass after the collision had a total momentum of 8 kg. m/s.

Explanation:

Momentum of the object is given by,

Momentum = mass × velocity

For object 1:

Momentum = mass × velocity

Momentum = 2 × 4

Momentum = 8 kg. m/s

For object 2:

Momentum = mass × velocity

Momentum = 6 × 0

Momentum = 0 kg. m/s

For object 1 + object 2:

Momentum = mass × velocity

Momentum = 8 × 1

Momentum = 8 kg. m/s

To identify the momentum of object 1, you must multiply mass (m) and velocity(v) to find momentum.

Object 1 has momentum of 8 kg. m/s before collision.

Object 1 has momentum of 0 kg. m/s before collision.

The combined mass after the collision had a total momentum of 8 kg. m/s.

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A densly wound cylindrical coil has 210 turns per meter, a 5 cm radius, and carries 38 mA. What is the magnitude of the uniform

kaheart [24]

Answer:

(a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

Explanation:

Given that,

Number of turns = 210

Radius = 5 cm

Current = 38 mA

Current in the wire = 500 mA

We need to calculate the magnetic field inside a coil

Using formula of magnetic field

$B=\mu_{0} ni$

Put the value into the formula

$B_{c}=4\pi\times10^{-7}\times210\times38\times10^{-3}$

$B_{c}=0.0000100\ T$

$B_{c}=10.02\times10^{-6}\ T$

Now a straight wire is inserted into the coil that carries 500 mA. It travels down the central axis of the coil

Distance $d=\dfrac{5}{2}$

$d=2.5\ cm$

We need to calculate the magnetic field from the straight wire

Using formula of magnetic field

$B_{w}=\dfrac{\mu_{0}I}{2\pi d}$

Put the value into the formula

$B_{w}=\dfrac{4\pi\times10^{-7}\times500\times10^{-3}}{2\times\pi\times2.5\times10^{-2}}$

$B_{w}=4.0\times10^{-6}\ T$

This field is perpendicular to the wire.

The magnitude of magnetic field is

$B=\sqrt{B_{c}^2+B_{w}^2}$

$B=\sqrt{(10.02\times10^{-6})^2+(4.0\times10^{-6})^2}$

$B=0.00001078\ T$

$B=10.78\times10^{-6}\ T$

$B=10.78\ \mu\ T$

Hence, (a). The magnitude of the uniform magnetic field is 10.02 μT

(b). The the magnitude of the total magnetic field is 10.78 μT.

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1 year ago

A CCD has a greatest possible pixel value of 4095. what is the bit level of this CCD?

Shtirlitz [24]

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2 years ago

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tiny-mole [99]

Explanation:

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2 years ago

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Korvikt [17]

Answer:

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