Answer:
If a car skids 66 ft on wet concrete, it will move at 243 ft/s when the brake is applied.
Explanation:
To determine how fast the car was moving, after skidding, the formula below is used:
V = √32*fd
V is the car's speed (ft/s)
d is skid length (ft) = 66 ft
f is the coefficient of friction determined by the material the car was skidding on.
Coefficient of friction for wet concrete is 0.65
V = √32*fd
V = √32 *0.65* 66
V = 242.679 ft/s ≅ 243 ft/s (nearest whole number)
If a car skids 66 ft on wet concrete, it will move at 243 ft/s when the brake is applied.
Threshhold frequency is defined as the minimum light frequency necessary to start photo-electric emission from the surface. This frequency is just effecient enough to emit electrons without any additional force or energy. this minimum frequency creates the necessary result that is sought by the person. The photo electric emission for certain metals can occir with threshhold frequency. The importance of this threshhold frequency lies in the fact that this has been used to create numerous instruments like amplifiers for use of people.
<span>since sound travels using mechanical waves and needs a material medium to propagate and since mechanical waves spread through vibrations ...and since hard materials have their atoms packed closely....they need to vibrate with a smaller amplitude to pass on the wave....thus sound travels faster in a denser medium than a less dense one.</span>
To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as 
, the relation with the final volume as
Initial temperature = 
Let T be the temperature after expanding by the formula of volume expansion
we have,
Where 
is the volume coefficient of copper 
Therefore the temperature is 53.06°C
Answer:
100 Joule
Explanation:
Amount of heat in agiven body is given by Q = m•C•ΔT
where m is the mass of the body
c is the specific heat capacity of body. It is the amount of heat stored in 1 unit weight of body which raises raises the temperature of body by 1 unit of temperature.
ΔT is the change in the temperature of body
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coming back to problem
m = 5g
C = 2J/gC
since, it is given that temperature of body increases by 10 degrees, thus
ΔT = 10 degrees
Using the formula for heat as given
Q = m•C•ΔT
Q = 5* 2 * 10 Joule= 100 Joule
Thus, 100 joule heat must be added to a 5g substance with a specific heat of 2 J/gC to raise its temperature go up by 10 degrees.
