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4 years ago

A concave mirror is used as a reflector ina torch light?

Physics

1 answer:

4vir4ik [10]4 years ago

8 0

Answer:

A concave mirror is used as a torch reflector. ... When a light bulb is placed at the focus of a concave mirror reflector, the diverging light rays of the bulb are collected by the reflector. These rays are then reflected to produce a strong, parallel-sided beam of light.

Explanation:

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What is the formula of kinetics?

snow_lady [41]

Answer:

KE = 1/2 mv v^{2}

Explanation:

Sorry if this didn't helped you.

4 0

3 years ago

A 1000-kg car traveling north at 15 m/s collides with a2000-kg truck traveling east at 10 m/s. The occupants, wearing seat belts

Aleksandr-060686 [28]

Answer:

8.33 m/s, 36.87° North of East

Explanation:

$m_n$ = Mass of car = 1000 kg

$v_n$ = Velocity of car = 15 m/s

$m_e$ = Mass of truck = 2000 kg

$v_e$ = Velocity of truck = 10 m/s

M = Combined mass = 1000+2000 = 3000 kg

Momentum

$p_n=m_nv_n\\\Rightarrow p_n=1000\times 15\\\Rightarrow p_n=15000\ kgm/s$

Momentum of car traveling East is 15000 kgm/s

$p_e=m_ev_e\\\Rightarrow p_n=2000\times 10\\\Rightarrow p_n=20000\ kgm/s$

Momentum of truck traveling North is 20000 kgm/s

Angle

$\theta=tan^{-1}\frac{p_n}{p_e}\\\Rightarrow \theta=tan^{-1}\frac{15000}{20000}\\\Rightarrow \theta=36.87^{\circ}$

As the two vehicles are vectors, the resultant velocity is

$(Mv)^2=p_n^2+p_e^2\\\Rightarrow v=\sqrt{\frac{p_n^2+p_e^2}{M^2}}\\\Rightarrow v=\sqrt{\frac{15000^2+20000^2}{3000^2}}\\\Rightarrow v=8.33\ m/s$

Velocity of the two vehicles when they are locked together is 8.33 m/s and direction is 36.87° North of East

5 0

3 years ago

a 10kg cement block horizontally at 6 m/s plows into a bank of sand and comes to a stop in 2.0 m/s. What is the average impact f

Gnesinka [82]

I assume the block plows into the bank of sand with a velocity of 6 m/s and comes to a stop in 2 s.

$\bar Ft = \Delta mv \\ \\ \bar F = \frac{\Delta mv }{t} \\ \\ \bar F \ avarage \ Force \\ m \ momentum \\ v \ velocity \\ t \ time$

7 0

4 years ago

A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby

8090 [49]

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, $O=(0\ nm,0\ nm)$

Position of desired magnetic field, $D\equiv(1\ nm,8\ nm)$

Magnitude of desired magnetic field, $E=0\ N.C^{-1}$

Let q be the positive charge magnitude placed at origin.

<u>We know the distance between the two Cartesian points is given as:</u>

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

<u>For the electric field effect to be zero at point D we need equal and opposite field at the point.</u>

$\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }$

$\therefore (1-0)^2+(8-0)^2=r^2$

$r^2=65\ nm$

$r=\sqrt{65}$

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

$x=2\times 1=2\ nm$