If Zack has a mass of 80 kg calculate his weight on Earth. Please show work, thank you!

Gekata [30.6K]

WAIT.... What are the items?

8 0

3 years ago

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.220 T. If the kinetic energy of the electr

Gwar [14]

Answer:

$971605.66\ \text{m/s}$

$25.1\ \mu\text{m}$

Explanation:

m = Mass of electron = $9.11\times 10^{-31}\ \text{kg}$

B = Magnetic field = 0.22 T

K = Kinetic energy of electron = $4.3\times 10^{-19}\ \text{J}$

q = Charge = $1.6\times 10^{-19}\ \text{C}$

v = Velocity of electron

r = Radius of curved path

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 4.3\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=971605.66\ \text{m/s}$

The speed of the electron is $971605.66\ \text{m/s}$

The force balance of the system is given by

$qvB=\dfrac{mv^2}{r}\\\Rightarrow r=\dfrac{mv}{qB}\\\Rightarrow r=\dfrac{9.11\times 10^{-31}\times 971605.66}{1.6\times 10^{-19}\times 0.22}\\\Rightarrow r=0.0000251=25.1\ \mu\text{m}$

The radius of the curved path is $25.1\ \mu\text{m}$

6 0

3 years ago

According to ohm's law if you don't change the value of the resistor & you double the voltage in a circuit the amount of cur

Nana76 [90]

They double!! Hope im not too late!!

3 0

3 years ago

Two small spheres, each carrying a net positive charge, are separated by 0.400 m. You have been asked to perform measurements th

neonofarm [45]

Answer:

a) q₁ = 15. 28 10⁻⁶C, b) q₂ = 5.64 10⁻⁶ C

Explanation:

For this exercise we use Newton's second law where force is Coulomb's electric force

Case 1. Distance (x₁ = 0.200 m) from the third sphere

F₁ = F₁₃ - F₂₃

F₁ = k q₁q₃ / x₁² - k q₂ q₃ / (0.4 - x₁)²

F₁ = k q₃ (q₁ / x₁² - q₂ / (0.4- x₁)²

Case2 Distance (x₂ = 0.6 m) from the third sphere

F₂ = F₁₃ + F₂₃

F₂ = k q₁q₃ / x₂² + k q₂q₃ / (0.4- x₂)²

F₂ = k q₃ (q₁ / x₂² + q₂ / (0.4-x₂)²

The distance is between the spheres, in the annex you can see the configuration of the charge and forces

Let's replace the values

F₁ = 8.99 10⁹ 3.00 10⁻⁶⁶ (q₁ / 0.2² - q₂ / (0.4-0.2)²

F₂ = 8.99 10⁹ 3.00 10⁻⁶ (q₁ / 0.6² + q₂ / (0.4-0.6)²

6.50 = 674. 25 10³ (q₁ –q₂)

3.50 = 26.97 10³ (q₁ / 0.36 + q₂ / 0.04)

We have a system of two equations with two unknowns, let's solve it. Let's clear q1 in the first and substitute in the second

q₁ = q₂ + 6.50 / 674 10³

3.50 / 26.97 10³ = (q₂ + 9.64 10⁻⁶) /0.36 + q₂ / 0.04

1.2978 10⁻⁴ = q₂ / 0.36 + q₂ / 0.04 + 26.77 10⁻⁶

q₂ (1 / 0.36 + 1 / 0.04) = 129.78 10⁻⁶ + 26.77 10⁻⁶

q₂ 27,777 = 156,557 10⁻⁶

q₂ = 156.557 10-6 /27.777

q₂ = 5.636 10⁻⁶ C

We look for q1 in the other equation

q₁ = q₂ + 6.50 / 674 10³

q₁ = 5.636 10⁻⁶ + 9.6439 10⁻⁶

q₁ = 15. 28 10⁻⁶C

4 0

3 years ago

Liz has designed a sailboat. She wants to build a prototype of the boat. Liz has $12,000 to spend on the boat. $7,000 of that mo

Naddika [18.5K]

Answer:

A. Marine plywood only.

Explanation:

3 0

3 years ago

Read 2 more answers