A cat runs in a straight line. Which of the following statements about the cat's motion must be true?

A 6.3 μC electric charge is placed in an Electric Field with a magnitude of 5.0 x 105 charge due to the Electric Field? N/C. Wha

EastWind [94]

Answer:

F = 3.15 N

Explanation:

Given electric charge, q = 6.3 μC

The magnitude of electric field, $E=5\times 10^5\ N/C$

We need to find the electric force on the charge due to the electric field. The electric force is given by :

F = qE

Putting all the values,

$F=6.3\times 10^{-6}\times 5\times 10^5\\\\F=3.15\ N$

So, the required force on the charge is 3.15 N.

7 0

3 years ago

The table lists the speed of light in four different materials at the same temperature. Through which material would light trave

bija089 [108]

I would say diamond because of how durable of a material it is. The light also travels the slowest for m/s.

7 0

2 years ago

Read 2 more answers

The uniform crate has a mass of 150 kg. The coefficient of static friction between the crate and the floor is μs = 0.2. The coef

Tatiana [17]

Answer:

The man will be able to move the crate.

Explanation:

$\mu_s$ = Coefficient of static friction between the crate and the floor = 0.2

$\mu'_s$ = Coefficient of static friction between the man's shoes and the floor = 0.4

$m_c$ = Mass of crate = 150 kg

$m_p$ = Mass of man = 85 kg

g = Acceleration due to gravity = 9.81 m/s²

Horizontal force in order to move the crate is given by

$F_h=\mu_sm_cg\\\Rightarrow F_h=0.2\times 150\times 9.81\\\Rightarrow F_h=294.3\ N$

Maximum force that the man can apply

$F_m=\mu'_sm_pg\\\Rightarrow F_m=0.4\times 85\times 9.81\\\Rightarrow F_m=333.54\ N$

Here it can be seen that $F_m>F_h$ .

So, the man will be able to move the crate.

3 0

3 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

balandron [24]

Answer with Explanation:

We are given that

Radius of solid core wire=r=2.28 mm= $2.28\times 10^{-3} m$

$1mm=10^{-3} m$

Radius of each strand of thin wire=r'=0.456 mm= $0.456\times 10^{-3} m$

Current density of each wire= $J=3750 A/m^2$

a.Area = $\pi r^2$

Where $\pi=3.14$

Using the formula

Cross section area of copper wire has solid core = $3.14\times (2.28\times 10^{-3})^2=16.3\times 10^{-6} m^2$

Current density = $J=\frac{I}{A}$

Using the formula

$3750=\frac{I}{16.3\times 10^{-6}}$

$I=3750\times 16.3\times 10^{-6}=0.061 A$

Total number of strands=19

Area of strand wire= $A'=19\times 3.14\times (0.456\times 10^{-3})^2=12.4\times 10^{-6} m^2$

$J'=\frac{I'}{A'}$

$3750=\frac{I'}{19\times 3.14(0.456\times 10^{-3})^2}$

$I'=3750\times 19\times 3.14(0.456\times 10^{-3})^2$

$I'=0.047 A$

b.Resistivity of copper wire= $\rho=1.69\times 10^{-8}\Omega-m$

Length of each wire =6.25 m

Resistance, R= $\frac{\rho l}{A}$

Using the formula

Resistance of solid core wire= $R=\frac{1.69\times 10^{-8}\times 6.25}{16.3\times 10^{-6}}=6.5\times 10^{-3}\Omega$

Resistance of strand wire= $R'=\frac{1.69\times 10^{-8}\times 6.25}{12.4\times 10^{-6}}=8.5\times 10^{-3}\Omega$

7 0

3 years ago

The greatest ocean depths on the earth are found in the marianas trench near the philippines. calculate the pressure due to the

tensa zangetsu [6.8K]

First, let us derive our working equation. We all know that pressure is the force exerted on an area of space. In equation, that would be: P = F/A. From Newton's Law of Second Motion, force is equal to the product of mass and gravity: F = mg. So, we can substitute F to the first equation so that it becomes, P = mg/A. Now, pressure can also be determined as the force exerted by a fluid on an area. This fluid can be measure in terms of volume. Relating volume and mass, we use the parameter of density: ρ = m/V. Simplifying further in terms of height, Volume is the product of the cross-sectional area and the height. So, V = A*h. The working equation will then be derived to be:

P = ρgh

This type of pressure is called the hydrostatic pressure, the pressure exerted by the fluid over a known height. Next, we find the literature data of the density of seawater. From studies, seawater has a density ranging from 1,020 to 1,030 kg/m³. Let's just use 1,020 kg/m³. Substituting the values and making sure that the units are consistent:

P = (1,020 kg/m³)(9.81 m/s²)(11 km)*(1,000 m/1km)
P = 110,068,200 Pa or 110.07 MPa

6 0

3 years ago