Find the velocity of the student using KE= 1/2 mv^2. A 50-kilogram student is running and has 225 joules of kinetic energy. The
1 answer:
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To solve this, we use the Wien's Displacement Law as shown in the attached picture. First, convert the temperature to Kelvin.
C to F:
C = (F - 32)*5/9
C = (325 - 32)*5/9 = 162.78 °C
C to K:
K = C + 273
K = 162.78 + 273 = 435.78 K
λmax = 2898/435.78 = <em>6</em><em>.65 μm</em>
C to F:
C = (F - 32)*5/9
C = (325 - 32)*5/9 = 162.78 °C
C to K:
K = C + 273
K = 162.78 + 273 = 435.78 K
λmax = 2898/435.78 = <em>6</em><em>.65 μm</em>
Answer:
Magnitude of static friction force is 70 sin40° = 44.99 N.
No, it is not necessary that it is maximum static friction.
Normal force is equal to 70 cos40° = 53.62 N.
Explanation:
We apply newton law of moton equation along the plane and perpendicular to plane;
Along the plane,
70 sin 40° = ---------------(1)
70 cos 40° = N --------------(2)
= μN -----------------(3)
So, it depends on the value of μ that the friction is maximum or not .