Answer:
37.5 N Hard
Explanation:
Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.
Using the expression for hook's law,
F = ke.............. Equation 1
F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.
Given: k = 750 N/m, e = 5.0 cm = 0.05 m
Substitute into equation 1
F = 750(0.05)
F = 37.5 N
Hence the athlete is pushing 37.5 N hard
Answer: At 34°c
Explanation:
Using The Arrhenius Equation:
k = Ae − Ea/RT
k represents rate constant
A represents frequency factor and is constant
R represents gas constant which is = 8.31J/K/mol
Ea represents the activation energy
T represents the absolute temperature.
By taking the natural log of both sides,
ln k = ln A- Ea/RT
Reactions at temperatures T1 and T2 can be written as;
ln k1= ln A− Ea/RT1
ln k2= ln A− Ea/RT2
Therefore,
ln(k1/k2) = −Ea/RT1 + Ea/RT2
Since k2=2k1 this becomes:
ln(1/2) = Ea/R*[1/T2 − 1/T1]
Theefore,
-0.693 = 37.2 x 10^3/8.31 * [ 1/T2 - 1/293]
1/T2 - 1/293 = -1.55 x 10^-4
1/T2 = -1.55 x 10^-4 + 34.13x 10^-4
1/T2 = 32.58 x 10^-4
Therefore T2 = 307K
T2 = 307 - 273 = 34 °c
Answer:
The box displacement after 6 seconds is 66 meters.
Explanation:
Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object (
), in meters, can be determined by the following expression:
(1)
Where:
- Initial velocity, in meters per second.
- Time, in seconds.
- Acceleration, in meters per square second.
If we know that 
, 
and 
, then the box displacement after 6 seconds is:
The box displacement after 6 seconds is 66 meters.
Answer:
piece of submerged iron
Explanation:
The buoyant force is given by:
where
is the density of the liquid (in this case, water)
is the submerged part of the object
g is the acceleration due to gravity
In this problem, the liquid used in the two examples is the same, water, so is the same; and g is the same as well. Therefore, the only difference between the two examples is .
For the piece of wood, we are told that the object is floating, so only a fraction of its total volume of 25.0 cm3 will be submerged: therefore the submerged volume will be less than 25.0 cm3. On the contrary, the problem tells us that the piece of iron is submerged, so the submerged volume in this case is 25.0 cm3. Therefore, the submerged volume for the iron is greater than for the piece of wood, so the buoyant force of the piece of iron is greater.
