A circuit has a resistance of 2.5 Ω and is powered by a 12.0 V battery.

Can it is possible to complete half syllabus of physics class 11 in 8 days?

castortr0y [4]

Answer:

In the human thoughts, there is possible and impossible. In God's world everything is possible. If you keep thinking this you'll never start to study. Just get up, open the book and start to study like you do. Believe in yourself!

3 0

3 years ago

What will the reading of the voltmeter be at the instant the switch returns to position a if the inertia of the d'Arsonval movem

Step2247 [10]

Answer:

hello your question is incomplete attached below is the complete question

answer :

20.16 v

Explanation:

The reading of the voltmeter at the instant the switch returns to position a

L = 5H

i ( current through inductor ) = 1/L ∫ V(t) d(t) + Vo

= 1/5 ∫ 3*10^-3 d(t) + 0 = 0.6 * 10^-3 t

iL ( 1.6 s ) = 0.6 * 10^-3 * 1.6 = 0.96 mA

Rm ( resistance ) = 21 * 1000 = 21 kΩ

The reading of the voltmeter ( V )

V = IR

= 0.96 mA * 21 k Ω = 20.16 v

3 0

3 years ago

a ball is projected with a certain angle with initial velocity u. it covers horizontal range R. With what initial velocity it mu

Elena L [17]

Answer:

1.5 u

Explanation:

The range equation is:

R = u² sin(2θ) / g

When u = v, R = 2.25 R.

2.25 R = v² sin(2θ) / g

2.25 u² sin(2θ) / g = v² sin(2θ) / g

2.25 u² = v²

1.5 u = v

8 0

3 years ago

Consider an electron that is 10-10 m from an alpha particle (9 = 3.2 x 10-19 C). (Enter the magnitudes.) (a) What is the electri

Studentka2010 [4]

Answer:

a) $E=2.88*10^{11}N/C$

b) $E=1.44*10^{11}N/C$

c) $F=4.61*10^{-8}N$

Explanation:

We use the definition of a electric field produced by a point charge:

$E=k*q/r^2$

a)Electric Field due to the alpha particle:

$E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(10^{-10})^2=2.88*10^{11}N/C$

b)Electric Field due to electron:

$E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(10^{-10})^2=1.44*10^{11}N/C$

c)Electric Force on the alpha particle, on the electron:

The alpha particle and electron feel the same force but with opposite direction:

$F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(10^{-10})^2=4.61*10^{-8}N$

4 0

3 years ago

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

$L$ = length of the meter stick = 1 m

$m$ = mass of the meter stick

$w$ = angular speed of the meter stick as it hits the floor

$v$ = speed of the other end of the stick

we know that, linear speed and angular speed are related as

$v = r w\\w = \frac{v}{r}$

$h$ = height of center of mass of meter stick above the floor = $\frac{L}{2} = \frac{1}{2} = 0.5 m$

$I$ = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

$I = \frac{mL^{2} }{3}$

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

$(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}$

7 0

3 years ago