When pushing on a wall while wearing skates, what provides the force that causes you to accelerate away from the wall?

Which one of the following statements is true concerning the magnitude of the electric field at a point in space? It is a measur

IceJOKER [234]

Answer:

It is a measure of the electric force per unit charge on a test charge.

Explanation:

The magnitude of the electric field is defined as the force per charge on the test charge.

Since we define electric field as the force per charge, it will have the units of force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).

5 0

2 years ago

Maple syrup, which comes from the sap of maple trees, contains water and natural sugars. It's a clear, brown liquid and the suga

Amanda [17]

The answer would be colloid

7 0

2 years ago

The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these

Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

$r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\$

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

$r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\$

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

$F = \frac{kq_{1}q_{2}}{r^2}\\$

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

$F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\$

5 0

2 years ago

The space shuttle travels at a speed of about 7.6times10^3 m/s. The blink of an astronaut's eye lasts about 110 ms. How many foo

sveta [45]

Answer:

It covers distance of 9.15 football fields in the said time.

Explanation:

We know that

$Distance=Speed\times Time$

Thus distance covered in blinking of eye =

$Distance=7.6\times 10^{3}m/s\times 110\times 10^{-3}s\\\\Distance=836 meters$

Thus no of football fields= $\frac{936}{91.4}=9.15Fields$

7 0

2 years ago

Which of the following is a theory stating that one plate is forced beneath another plate?

morpeh [17]

Answer:

theroy of plate tectonics

7 0

2 years ago