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zvonat [6]
3 years ago
12

A 60 mAs results in an exposure of 85 mGya, with all factors remaining the same, what would the new exposure be if 120 mAs is us

ed?
Physics
1 answer:
Cloud [144]3 years ago
8 0

Answer: d₂ = 170 mGya

Explanation:

the relationship between absonbed 'd' and exposure 'E' is given as;

D(Gv) = F . x (AS/xB)

F is a conversion coefficient depending on medium

so we can simply write

d₁/d₂ = x₁/x₂

Given that;

our x₁ = 60 mAs, x₂ = 120 mAs,  d₁ = 85 mGya, d₂ = ?

from the given formula,

d₂ = (x₂d₁ / x₁)

now we substitute

d₂ = (120 × 85) / 60

d₂ = 170 mGya

∴ if 120 mAa is used,  the new exposure will be 170 mGya

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Thermal conductivity of a material is given as 129Btuft–¹ h–¹°F–¹.Calculate this thermal conductivity in Jm–¹s–¹°C–¹(Given: 1Btu
ss7ja [257]

Answer:

223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

Explanation:

The thermal conductivity of an object is defined as the measure or the ability of the object to transfer heat or conduct heat through its body.

In the context, the thermal conductivity of the material is given as

$=129 \text{ Btu ft}^{-1}\text{h}^{-1}^\circ\text{F}^{-1}$

And it is given that :

1 Btu = 1055 J

1 ft = 0.3048 m

$1^\circ F = \frac{5}{9}^\circ C$

We know that 1 h = 3600 s

So the thermal conductivity of the material in $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$ is  :

Thermal conductivity :

$=\frac{129 \text{ Btu}}{1 \text{ ft }\times \text{1 h}\times 1^\circ\text{F}}$

$=\frac{129 \times 1055 \text{ J}}{0.3048 \text{ m} \ \times 3600 \text{ s}\ \times \frac{5}{9}^\circ \text{C}}$

=  223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

3 0
3 years ago
Resistors and reactors, for use over 600 volts, shall not be installed in close enough proximity to combustible materials to con
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Resistors and reactors, for use over 600 volts, shall not be installed in close enough proximity to combustible materials to constitute a fire hazard and shall have a clearance of not less than<u> 300 mm </u>from combustible materials.

Explanation:

  • The hazards associated with high power industrial resistors are primarily due to their open construction, which is necessary for cooling.
  • The exposed conductors which make up the resistors can be not only a shock hazard but also a thermal burn hazard.
  • When a resistor fails, it either goes open or the resistance increases. When the resistance increases, it can burn the board, or burn itself up.
  • Avoid touching non-flammable resistors in operation; the surface temperature ranges from approximately 350 °C to 400°C when utilized at the full rated value. Maintaining a surface temperature of 200°C or less will extend resistors service life.
  • Do not apply power to a circuit while measuring resistance. When you are finished using an ohmmeter, switch it to the OFF position if one is provided and remove the leads from the meter.
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4 0
3 years ago
What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =
iren2701 [21]
1) The electric potential at a distance r from a single point charge is given by
V(r) = k  \frac{q}{r}
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.

The charge in this problem is
q=2.2 \mu C =2.2 \cdot 10^{-6} C
So the potential at distance r=6.3 m is
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2 C^{-2})  \frac{2.2 \cdot 10^{-6} C}{6.3 m}=3139 V

2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2C^{-2})  \frac{2.2 \cdot 10^{-6} C}{99 m}=198 V
3 0
3 years ago
Read 2 more answers
A man pushing a crate of mass
marin [14]

(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.

(b) The net work done on the crate while it is on the rough surface is 23.7 J.

(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.

<h3>Magnitude of net force on the crate</h3>

F(net) = F - μFf

F(net) = 280 - 0.351(92 x 9.8)

F(net) = -36.46 N

<h3>Net work done on the crate</h3>

W = F(net) x L

W = -36.46 x 0.65

W = - 23.7 J

<h3>Acceleration of the crate</h3>

a = F(net)/m

a = -36.46/92

a = - 0.396 m/s²

<h3>Speed of the crate</h3>

v² = u² + 2as

v² = 0.845² + 2(-0.396)(0.65)

v² = 0.199

v = √0.199

v = 0.45 m/s

Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.

The net work done on the crate while it is on the rough surface is 23.7 J.

The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.

Learn more about work done here: brainly.com/question/8119756

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7 0
2 years ago
Difference between acceleration and decceleration​
mario62 [17]

acceleration is considered to describe an increase or positive change of speed or velocity But deceleation is considered to describe a decrease or negative change of speed or velocity

8 0
3 years ago
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