In objects motion will not change is the force is acting on the object are blank

Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe

matrenka [14]

(a) does an electric field exert a force on a stationary charged object?
Yes. The force exerted by an electric field of intensity E on an object with charge q is
 $F=qE$
As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with charge q and speed v is
 $F=qvB \sin \theta$
where $\theta$ is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

(c) does an electric field exert a force on a moving charged object?
Yes, The intensity of the electric force is still
 $F=qE$
as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

(d) does a magnetic field do so?
Yes. As we said in point b, the magnetic force is
$F=qvB \sin \theta$
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

(f) does a magnetic field do so?
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
 $F=qE$
So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
 $F=qvB \sin \theta$
is different from zero because v is different from zero.

6 0

3 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

Ipatiy [6.2K]

k = $\dfrac{ (\dfrac{h}{ \lambda} )^{2} }{2m}$

k = (6.626×10-¹⁹/590 × 10-⁹ )^{2} /2 × 1.673 × 10-²⁷

k = (1.12 × 10-³⁰)^2/3.346×10-²⁷

k = 1.25 × 10-⁶⁰ /3.346×10-²⁷

k = 0

ldk why, my answer is coming this :(

4 0

3 years ago

Consider a highway composed of concrete. If the road is 4.5 km long, 30 m wide, and 0.70 m thick, what is its mass?

Aloiza [94]

Answer:

Explanation:

We need to assume that the density of the concrete is about 2350 Kg/m^3. And using the dimensions of the highway we can calculate the volume of the highway.

$vh=4500 * 30 * 0,70\\vh=94500 m^3\\den=m/v\\m=den*v\\m=2350*94500\\\\m=222075 ton$

5 0

3 years ago

Calcite (CaCO_3) is a crystal with abnormally large birefringence. The index of refraction for light with electric field paralle

katrin2010 [14]

Answer:

(a) 42.28°

(b) 37.08°

Explanation:

From the principle of refraction of light, when light wave travels from one medium to another medium, we have:

$\frac{n_{b} }{n_{a} }$ = sinθ $_{a}$ /sinθ $_{b}$

In the given problem, we are given the refractive indices of light which are parallel and perpendicular to the axis of the optical lens as 1.4864 and 1.6584 respectively.

For critical angle θ $_{a}$ = θ $_{c}$ , θ $_{b}$ = 90°; $n_{b} = 1$

(a) $n_{a} = 1.4864$

$\frac{1 }{1.4864 }$ = sinθ $_{c}$ /sin90°

0.6728 = sinθ $_{c}θ[tex]_{c} = sin^(-1) 0.6728 = 42.28°(b) [tex]n_{a} = 1.6584$

$\frac{1 }{1.6584}$ = sinθ $_{c}$ /sin90°

0.60299 = sinθ[tex]_{c}

θ[tex]_{c} = sin^(-1) 0.60299 = 37.08°

7 0

4 years ago

mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan

Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, $v_e$ , for the planet is $v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }$

Where;

$v_e$ = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, $M_E$

r = The radius of the planet = 3 × The radius of Earth, $R_E$

The escape velocity for Earth, $v_e_E$ , is therefore given as follows;

$v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }$

$\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } = \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E$

$v_e$ = 16 × $v_e_E$

Given that the escape velocity for Earth, $v_e_E$ ≈ 11,186 m/s, we have;

The escape velocity on the planet = $v_e$ ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

3 0

3 years ago