All categories

3 years ago

The beginning of the Phanerozoic is marked by what occurrence

Physics

1 answer:

Sergio039 [100]3 years ago

3 0

The beginning of the Phanerozoic is marked by the development of hard body parts, such as shells and bones.

You might be interested in

An airplanes lands with a velocity of +75 m/s and comes to a stop in 20 seconds. What is the acceleration of the airplane while

erma4kov [3.2K]

Answer:

The acceleration of the car, a = -3.75 m/s²

Explanation:

Given data,

The initial velocity of the airplane, u = 75 m/s

The final velocity of the plane, v = 0 m/s

The time period of motion, t = 20 s

Using the I equations of motion

v = u + at

a = (v - u) / t

= (0 - 75) / 20

= -3.75 m/s²

The negative sign indicates that the plane is decelerating

Hence, the acceleration of the car, a = -3.75 m/s²

7 0

3 years ago

Lionel makes a graphic organizer to compare the electric field around a positive charge with the electric field around a negativ

krek1111 [17]

X- points away from the charge
y- points in the direction of the force on the positive charge
z- points toward the charge

6 0

3 years ago

Read 2 more answers

How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume

Svetach [21]

Answer:

$9.56\cdot 10^{-7} C$

Explanation:

A parallel-plate capacitors consist of two parallel plates charged with opposite charge.

Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.

The electric field between two infinite sheets with opposite charge is:

$E=\frac{\sigma}{\epsilon_0}$

where

$\sigma=\frac{Q}{A}$ is the surface charge density, where

Q is the charge on the plate

A is the area of the plate

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

In this problem:

- The side of one plate is

L = 19 cm = 0.19 m

So the area is

$A=L^2=(0.19)^2=0.036m^2$

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

$E=3\cdot 10^6 N/C$

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

$E=\frac{Q}{A\epsilon_0}\\Q=EA\epsilon_0 = (3\cdot 10^6)(0.036)(8.85\cdot 10^{-12})=9.56\cdot 10^{-7} C$

7 0

3 years ago

Two masses of size m and 4m are connected by a massless thread and are strung over a frictionless pulley of radius R and moment-

tekilochka [14]

Answer:

Total kinetic energy of entire system is 3 mgl

Explanation:

Given two masses: m and 4m.

Since the pulley is frictionless and the thread is massless, the energy here is linked to the two masses.

Total kinetic energy of entire system = decrease in gravitational potential energy of the system.

Therefore, we have :

ΔKE = Δp

ΔKE = 4mgl - mgl

= 3 mgl

Total kinetic energy of entire system is 3 mgl

6 0

3 years ago

A flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the

Ostrovityanka [42]

Answer:

The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

Explanation:

Given;

number of turns of the flat circular loop, N = 18 turns

radius of the loop, R = 15.0 cm = 0.15 m

current through the wire, I = 0.51 A

The magnetic field through the center of the loop is given by;

$B = \frac{N\mu_o I}{2R}$

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

$B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T$

Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

6 0

3 years ago