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Crank
1 year ago
14

A piece of copper has a volume of 100.0 cm³. If the mass is 890 g, what is the density of copper

Physics
1 answer:
Hoochie [10]1 year ago
3 0

Answer:

Explanation:

mass = m =890 g       volume = v =100cm³              

Density = m/v =890g / 100cm³ =8.9g/cm³

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Combine these three velocity vectors into a resultant: 3.0 m/s north, 4.0 m/s east 1.0 m/s west. Identify the resultant vector
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Answer:

The resultant vector is 1 m/s

Explanation:

The resultant vector is 1 m/s west based on triangle law of vector addition, when two sides of a triangle is represented by two vectors, the resultant vector is the third side of the triangle.

5 0
3 years ago
The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole huma
gayaneshka [121]

Explanation:

Below is an attachment containing the solution.

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3 years ago
Two equally charged, 1.00 g spheres are placed with 2.00 cm between their centers. when released, each begins to accelerate at 2
Leya [2.2K]
1) Force = m*a = 1.00 g * (1kg / 1000 g) * 225 m/s^2 = 0.225 N

2) Charge

Force = K (charge)^2 /(distance)^2 => charge = √ [Force * distance^2 / k]

k = 9.00 * 10^9 N*m^2 / C^2

charge = √ [0.225 N * (0.02 m)^2 / 9.00* 10^9 N*m^2 / C^2 ]

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3 0
3 years ago
The length of a rectangle is increasing at a rate of 4 cm/s and its width is increasing at a rate of 6 cm/s. When the length is
Fudgin [204]

Answer:

24cm/s

Explanation:

A=L*w

A'=L'*w'

L=13

w=5

L'=4

w'=6

A=?

A'=?

A=L*w

A=13*5

A=65

A'=L'*w'

A'=4*6

A'=24

*the given lengths are just to throw you off*

3 0
2 years ago
Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin
Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

\mu_o=4\pi \times 10^{-7}\ T-m/A

Resistivity, \rho=1.72\times 10^{-8}\ \Omega-m

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

R=\rho \dfrac{l}{A}

R=\rho \dfrac{l}{\pi r^2}

r=\sqrt{\dfrac{\rho l}{R\pi}}

r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}

r = 0.00199 m

or

r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m

The magnetic field on a current carrying wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

I=\dfrac{2\pi rB}{\mu_o}

I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0
2 years ago
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