Answer:
13.33 K
Explanation:
Given that,
Heat absorbed, Q = 30 J
Mass of nail, m = 5 g = 0.005 kg
The specific heat capacity of the nail is 450 J/kg-°C.
We need to find the increase in the temperature during the process. The heat absorbed in a process is as follows:
So, the increase in temperature is 13.33 K.
Answer:
this measurement if feet is: 2.624672 ft
Explanation:
Notice that 80 cm can be expressed as 0.8 meters, and In order to convert from meters to feet, one needs to multiply the meter measurement times 3.28084. Therefore:
0.80 m can be written in feet as: 0.80 * 3.28084 feet = 2.624672 feet
Answer:
387 volts
Explanation:
Ohm's law is used to relate voltage, current and resistance.
The formula is as follows:V = I * R
where:
V is the applied voltage (measured in volts)
I is the current flowing (measured in amperes)
R is the resistance (measured in ohm)
In the given, we have:
current (I) = 9 amperes
resistance (R) = 43 ohm
Substitute with the givens in the above formula to get the voltage as follows:
V = 9 * 43
V = 387 volts
Hope this helps :)
When it comes to optics, Snell's law is the basic formula to be used. If you notice, when light hits the water, the light does not travel in the same direction. After, it hits the water, it changes in angle. Light becomes refracted. This is observed when your hands tend to become bigger if you place it underwater. The formula for Snell's Law is
n₁ sin θ₁ = n₂sin θ₂, where n is the index of refraction. This depends on the type of medium. For example, for air, n=1. The parameters θ₁ is the angle of incidence, and θ₂ is the angle of refraction. Critical angle is the incident angle needed so that the refract angle is 90°. So, modifying the equation:
n₁ sin θcrit = n₂sin 90°, since sin 90°=1,
sin θcrit = n₂/n₁
θcrit = sin ⁻¹ (n₂/n₁)
Since liquid comes first before glass, n₁=1.75 and n₂=1.52. Substituting,
θcrit = sin ⁻¹ (1.52/1.75)
θcrit = 60.29°
Answer:
Exposure time limitation, shielding and distance.
Explanation:
- Limitation of exposure time, since the dose received is directly proportional to the exposure time, so that, at a shorter time, lower dose. For this reason, planning is suggested, to reduce time.
-
Use of shields. This allows a reduction in the dose received by the technician when filtered by the barrier (screen). There are two types of shields or screens, the primary barriers (attenuate the radiation of the primary beam) and the secondary barriers (avoid diffuse radiation).
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Distance to the radioactive source. The dose received is inversely proportional to the square of the distance to the radioactive source. Therefore, if the distance is doubled, the dose received will decrease by a quarter. Reason for this, it is advisable to use devices or remote controls whenever possible.
