True or false? rocks have organic materials and fossil remains in them

A material that provides resistance to the flow of electric current is called a(an)

stira [4]

An insulator is a material that provides resistance to the flow of electric current. Insulators include rubber, plastic, glass since they do not allow electricity to flow through them.

8 0

3 years ago

Read 2 more answers

If the kinetic and potential energy in a system are equal, then the potential energy increases. What happens as a resul

Kobotan [32]

-- If the system is 'closed', then nothing ... including energy ... can get in or out, and the total energy inside has to be constant.

If half of the energy in the system starts out as potential energy and the rest starts out as kinetic, and then the potential energy increases, there's only one place the increase could have come from ... it could only have been converted from kinetic energy. So the kinetic energy in the system must decrease.

In fact, this isn't even a "result". The kinetic energy has to decrease before the potential energy can increase, because that's where the increase has to come from.

If the system is 'open', then energy can come in and go out. If the potential energy inside suddenly increases, we don't know where it came from, so we can't say anything about what happens to the system.

7 0

2 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

4 0

3 years ago

A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plan

pshichka [43]

Answer:

a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N

Explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight

sin θ = Wₓ / W

cos θ = W_y / W

Wₓ = W sin θ

W_y = W cos θ

Wₓ = 1200 sin 30 = 600 N

W_y = 1200 cos 30 = 1039.23 N

Y axis

N- W_y = 0

N = W_y = 1039.23 N

Remember that the friction force always opposes the movement

a) in this case, the system will begin to move upwards, which is why friction is static

P -Wₓ -fr = 0

P = Wₓ + fr

as the system is moving the friction coefficient is dynamic

fr = μ N

fr = 0.20 1039.23

fr = 207.85 N

we substitute

P = 600+ 207.85

P = 807.85 N

b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static

P + fr -Wx = 0

fr = μ N

fr = 0.20 1039.23

fr = 207.85 N

we substitute

P = Wₓ -fr

P = 600 - 207,846

P = 392.15 N

c) as the movement is continuous, the friction coefficient is dynamic

P - Wₓ + fr = 0

P = Wₓ - fr

fr = 0.15 1039.23

fr = 155.88 N

P = 600 - 155.88

P = 444.12 N

6 0

2 years ago

The Hubble telescope’s orbit is 5.6 × 105 meters above Earth’s surface. The telescope has a mass of 1.1 × 104 kilograms. Earth e

djverab [1.8K]

The gravitational field is the Force divided by the mass

Call g the gravitational fiel, F the force exerted by the earth and m the mass of the telescope.

g = F / m

g=9.1x10^4 N / 1.1 x 10^4 kg = 8.27 N/kg

Note that the unit N/kg is equivalent to m/s^2

8 0

3 years ago