Answer:
a) 72.54° and It will take the woman 33.4m
Explanation:
The woman runs with a constant speed V1 = 6m/s
V2 = 20m/s
V1 = V2 cos θ
Cos θ= V1/V2= 6/20
Cos θ= 0.3
Cos^-1 0.3=72.54°
Using Range formular for projectile
R= (V2 Cos θ)/g (V2 Sin θ)^2 +sqrt(V2 Sin θ)^2 + 2gh)
R= (20cos72.54)(2Sin72.54+sqrt(20Sin72.54)^2 + 2×9.8×45
R=33.4m
b )
see the attached file
Explanation:
✠ This question says that there is an object and its mass is 2 kg ; it's raised to a height 10 m and possess potential energy of 200 J. Now this question ask us to find the kinetic energy and the potential energy at a height 4 metre.
✰ Mass = 2 kilograms
✰ Raised height = 10 metres
✰ Posses potential energy = 200 Joules
✰ Kinetic energy at a height 4 metre
✰ Potential energy at a height 4 metre
✰ Kinetic energy at a height 4 metre = 120 J
✰ Potential energy at a height 4 metre = 80 J
✰ Potential energy formula.
✰ Potential energy = mgh
✰ Potential energy as P.E
✰ Mass as m
✰ Joules as J
✰ Height as h
✰ Raised height as g
↦ Potential energy = mgh
↦ Potential energy = 2 × 10 × 4
↦ Potential energy = 20 × 4
↦ Potential energy = 80 J
<h3>✠ Now according to the question let us find the kinetic energy</h3>↦ Kinetic energy = Posses potential energy - Finded potential energy
↦ Kinetic energy = 200 J - 80 J
↦ Kinetic energy = 120 Joules
Answer:
The induced emf 1.43 s after the circuit is closed is 4.19 V
Explanation:
The current equation in LR circuit is :
.....(1)
Here I is current, V is source voltage, R is resistance, L is inductance and t is time.
The induced emf is determine by the equation :
Differentiating equation (1) with respect to time and put in above equation.
Substitute 6.05 volts for V, 0.655 Ω for R, 2.55 H for L and 1.43 s for t in the above equation.