1. A car moving to the right at 30 m/s, slows down 5 m/s every second until it comes to a stop.

Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc

romanna [79]

Answer:

a) The speed of the electron as it reaches the inner conductor is closest to:

v = 1.45 × 10⁷m/s

b) The electric field magnitude between the cylinders is

E = 10,000V/m

Explanation:

given

inner radius of the cylinder r₁ = 20mm = 0.02m

outter radius of the cylinder r₂ = 80mm = 0.08m

potential difference V= 600V

mass of electron = 9.1×10⁻³¹kg

charge on electron = 1.6×10⁻¹⁹C

calculating the work done in bringing electron at inner conductor is

$W =\frac{1}{2}mv^{2}$

note:

$V = \frac{W}{q}$

∴W = (ΔV)q

(ΔV)q = $\frac{1}{2}mv^{2}$

(600)1.6×10⁻¹⁹ = ¹/₂ × 9.1×10⁻³¹ × v²

v² ≈ 2.11 × 10¹⁴

v = 1.45 × 10⁷m/s

According to the energy conservation law, the total energy of an isolated system is always constant.

The energy of an isolated system can neither be created nor be destroyed, it can only convert one form to another form.

∴ the maximum electric field

E = ΔV/d

E = 600/d

where d is the distance between the two points

where d = 0.06m

E = 600/0.06

E = 10,000V/m

Note: the electric field due to the potential difference between to points depends upon the potential difference V and the distance between both points d.

5 0

3 years ago

A 10 m long steel beam is accidentally dropped by a construction crane from a height of 4.89 m. The horizontal component of the

Otrada [13]

Answer:

e = 1.21 mV

Explanation:

given,

length of rod = 10 m

height of drop = 4.89 m

Earth’s magnetic field = 12.4 µT

acceleration of gravity = 9.8 m/s²

velocity of the beam

$v = \sqrt{2gh}$

$v = \sqrt{2\times 9.8 \times 4.89}$

v = 9.79 m/s

emf of the beam

e = B l v

e = 12.4 x 10⁻⁶ x 9.79 x 10

e = 1.21 x 10⁻³ V

e = 1.21 mV

4 0

3 years ago

What happens to the velocity of an object when balanced force acts on it?

Pani-rosa [81]

When the entire group of forces acting on an object is balanced,
the individual forces all add up to zero, and the effect on the object
is the same as it would be if there were no force at all acting on it.

No force on it means no acceleration ... no change in speed or
direction of motion.

No change in speed or direction of motion means constant velocity.

8 0

4 years ago

A luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 25 degrees above the horizontal by a force F of magnitude 145 N

Nonamiya [84]

Answer:

A) 667 J

B) 381.4 J

C) 0 J

D) 245.4 J

E) 40.2J

F) 2 m/s

Explanation:

Let g = 9.81 m/s2

A) The work done on the suitcase is the product of the force applied and the distance travelled:

w = Fs = 145 * 4.6 = 667 J

B) The work done by gravitational force the dot product between the gravity vector and the distance vector

$W_g = \vec{P}\vec{s} = mgs sin\alpha = 20*9.81*4.6*sin25^o = 381.4 J$

C) As the normal force vector is perpendicular to the distance vector, the work done by the normal force is 0

D) The work done on the suitcase by friction force is the product of the force applied and the distance travelled, whereas friction force is the product of normal force and coefficient

$W_f = F_fs = \mu N s = \mu s mgcos\alpha = 0.3* 4.6 * 20*9.81*cos25^o = 245.4 J$

E) The total workdone on the suite case would be the pulling work subtracted by gravity work and friction work

$W = w – W_g – W_f = 667 – 381.4 – 245.4 = 40.2 J$

F) As the suit case has 0 kinetic and potential energy at the bottom, and the total work done is converted to kinetic energy at 4.6 m along the ramp, we can conclude that:

$E_k = W = 40.2 j$