An object kept in front of lens, which is thick at the edges, produces a ________

How far apart would you have to place the poles of a 1. 5 v battery to achieve the same electric field?

Zarrin [17]

To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m

The potential difference is related to the electric field by:

∆V=Ed

where,

∆V is the potential difference

E is the electric field

d is the distance

what is potential difference?

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

We want to know the distance the detectors have to be placed in order to achieve an electric field of

E=1v/cm=100v/cm

when connected to a battery with potential difference

∆v=1.5v

Solving the equation,we find

$d = \frac{ \:Δv}{e}$

$= \frac{1.5v}{100v/m}$

$= 1.5 \times 10 {}^{ - 2} m$

learn more about potential difference from here: brainly.com/question/28166044

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6 0

8 months ago

One of the element carbon combines with one of the element oxygen to form one of the compound carbon dioxide.

saul85 [17]

False. C + O --> CO not CO2. Carbonmonoxide

3 0

3 years ago

Read 2 more answers

To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl

Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁ + 2.43 x 0 = v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂² = 0.07032

v₂² = 0.07032 / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

5 0

2 years ago

The temperatures (in degrees Fahrenheit) in Long Island recorded by the weather bureau over a week were 42, 49, 53, 55, 50, 47,

lora16 [44]

Answer:

Inter Quartile Range

Explanation:

Quartile is a positional statistical average, which divided the data into 4 equal halves.

Q1 (Lower Quartile) has 25% data below it, 75% above it. Q3 (Upper Quartile) has 75% data below it, 25% above it.

Interquartile range is the measure used to calculate how far the lower & upper quartiles are.

7 0

3 years ago

A laboratory technician drops a 72.0 g sample of unknown solid material, at a temperature of 80.0°C, into a calorimeter. The cal

Natalija [7]

Answer : The specific heat of unknown sample is, $8748.78J/kg^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-[q_2+q_3]$

$m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]$

where,

$c_1$ = specific heat of unknown sample = ?

$c_2$ = specific heat of water = $4186J/kg^oC$

$c_3$ = specific heat of copper = $390J/kg^oC$

$m_1$ = mass of unknown sample = 72.0 g = 0.072 kg

$m_2$ = mass of water = 203 g = 0.203 kg

$m_2$ = mass of copper = 187 g = 0.187 kg

$T_f$ = final temperature of calorimeter = $39.4^oC$

$T_1$ = initial temperature of unknown sample = $80.0^oC$

$T_2$ = initial temperature of water and copper = $11.0^oC$

Now put all the given values in the above formula, we get

$0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]$

$c_1=8748.78J/kg^oC$

Therefore, the specific heat of unknown sample is, $8748.78J/kg^oC$

7 0

3 years ago

An object kept in front of lens, which is thick at the edges, produces a _________,