A cat is running at 24 m/s. It then accelerates at 7 m/s2. How long will it take the cat to reach a speed of 49 m/s?
1 answer:
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Answer:
life (N) of the specimen is 117000 cycles
Explanation:
given data
ultimate strength Su = 120 kpsi
stress amplitude σa = 70 kpsi
solution
we first calculate the endurance limit of specimen Se i.e
Se = 0.5× Su .............1
Se = 0.5 × 120
Se = 60 kpsi
and we know strength of friction f = 0.82
and we take endurance limit Se is = 60 kpsi
so here coefficient value (a) will be
a = ......................1
put here value and we get
a =
a = 161.4 kpsi
so coefficient value (b) will be
b =
b =
b = −0.0716
so here number of cycle N will be
N =
put here value and we get
N =
N = 117000
so life (N) of the specimen is 117000 cycles
Answer:
Q = 7272 Kilojoules.
Explanation:
<u>Given the following data;</u>
Mass = 2.0*101kg = 202kg
Initial temperature, T1 = 10°C
Final temperature, T2 = 90°C
We know that the specific heat capacity of iron = 450J/kg°C
*To find the quantity of heat*
Heat capacity is given by the formula;
Where;
- Q represents the heat capacity or quantity of heat.
- m represents the mass of an object.
- c represents the specific heat capacity of water.
- dt represents the change in temperature.
dt = T2 - T1
dt = 90 - 10
dt = 80°C
Substituting the values into the equation, we have;
Q = 7272KJ or 7272000 Joules.