Shale, sandstone, and limestone are the most commoc types of sedimentary rocks. They are formed by the most common mineral that is found on or near the surface of the Earth

4 0

3 years ago

What is the value of acceleration due to gravity at heght h=4re

Norma-Jean [14]

Answer: Formula for Acceleration Due to Gravity

These two laws lead to the most useful form of the formula for calculating acceleration due to gravity: g = G*M/R^2, where g is the acceleration due to gravity, G is the universal gravitational constant, M is mass, and R is distance.please mark as brainliest

Explanation:

5 0

3 years ago

You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resul

photoshop1234 [79]

1) First of all, let's find the resistance of the wire by using Ohm's law:
$V=IR$
where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have:
$R= \frac{V}{I}= \frac{5.70 V}{17.6 A}=0.32 \Omega$

2) Now that we have the value of the resistance, we can find the resistivity of the wire $\rho$ by using the following relationship:
$\rho = \frac{RA}{L}$
Where A is the cross-sectional area of the wire and L its length.
We already have its length $L=2.90 m$ , while we need to calculate the area A starting from the radius:
$A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2$

And now we can find the resistivity:
$\rho = \frac{RA}{L}= \frac{(0.32 \Omega)(1.34 \cdot 10^{-6}m^2)}{2.90m}= 1.48 \cdot 10^{-7}\Omega \cdot m$

7 0

3 years ago

At some distance from a point charge, the electric potential is 635.0 V and the magnitude of the electric field is 189.0 N/C. Fi

Nataliya [291]

Answer:

The distance from the charge is 3.35 m.

Explanation:

Given that,

Electric potential, V = 635 V

Magnitude of electric field, E = 189 N/C

We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :

$E=\dfrac{V}{d}$

d is the distance from charge

$d=\dfrac{V}{E}\\\\d=\dfrac{635}{189}\\\\d=3.35\ m$

So, the distance from the charge is 3.35 m. Hence, this is the required solution.

8 0

3 years ago