The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V
<u>Explanation:</u>
Given data,
E= 3 ×10 ⁶ Δx=0.06/100
We have to find the minimum potential difference
E= -ΔV/Δx
ΔV=- E × Δx
ΔV =-3 ×10 ⁶ . 0.06/100
ΔV=-1800 V
The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V
Wildlife researcher starts from a and then reaches b, he turns towards north 40 degree to move towards c.
Total displacement is ac
Total horizontal displacement = 4+4 cos40 =7.06 km
Total vertical displacement = 4 sin40 =2.57 km
Total displacement 
= 7.51 km
Answer:
Speed of block after the bullet emerges = 1.5 m/s
Explanation:
Here momentum is conserved.
Initial momentum = Final momentum.
Mass of bullet = 10 g = 0.01 kg
Initial Velocity of bullet = 300 m/s
Mass of block = 1 kg
Initial Velocity of block = 0 m/s
Final Velocity of bullet = 50% of initial velocity. = 150 m/s
We need to find final velocity of block. Let it be v
We have
Initial momentum = 0.01 x 300 + 1 x 0 = 3 kg m/s
Final momentum = 0.01 x 150 + 1 x v = 1.5 + v
Equating
3 = 1.5 + v
v = 1.5 m/s
Speed of block after the bullet emerges = 1.5 m/s
Answer:
A. a = 814.815 ft/s^2.
B. Distance, S = 320 ft.
Explanation:
Equations of motion
i. vf = vi + a*t
ii. S = vi*t + 1/2*(a*t)
iii. vf^2 = vi^2 + 2a*S
Given:
vi = 0 ft/s
vf = 1000 miles/hr
t = 1.8 s
g = 32 ft/ s2
Converting miles/hr to ft/s,
1 mile = 5280 ft
Also, 1 hr = 60 mins * 60s
= 3600 s
Therefore, 1000 miles/hr * 5280 ft/1 mile * 1 hr/3600 s
= 1466.7 ft/s.
A.
Using the i. Equation of motion,
1466.7 = 0 + a*1.8
a = 814.815 ft/s^2
Comparing a to g,
a = (814.815/32) * g
= 25.46 g
B.
Using the ii. Equation of motion,
S = 1/2 * (814.815) * (1.8)^2
= 1320 ft.
