Answer:
76 V/m
Explanation:
= Electric potential at initial location = 119 V
= Electric potential at final location = 43 V
= distance between initial and final location = 1 m
= magnitude of electric field
Magnitude of electric field is given as
= 76 V/m
Answer:
u₂ = 3.7 m/s
Explanation:
Here, we use the law of conservation of momentum, as follows:
where,
m₁ = mass of the car = 1250 kg
m₂ = mass of the truck = 2020 kg
u₁ = initial speed of the car before collision = 17.4 m/s
u₂ = initial speed of the tuck before collision = ?
v₁ = final speed of the car after collision = 6.7 m/s
v₂ = final speed of the truck after collision = 10.3 m/s
Therefore,
<u>u₂ = 3.7 m/s</u>
Answer:
his acceleration rate is -0.00186 m/s²
Explanation:
Given;
initial position of the car, x₀ = 100 miles = 160, 900 m ( 1 mile = 1609 m)
time of motion, t₀ = 60 minutes = 60 mins x 60 s = 3,600 s
final position of the car, x₁ = 150 miles = 241,350 m
time of motion, t₁ = 100 minutes = 100 mins x 60 s = 6,000 s
The initial velocity is calculated as;
u = 160, 900 m / 3,600 s
u = 44.694 m/s
The final velocity is calculated as;
v = 241,350 m / 6,000 s
v = 40.225 m/s
The acceleration is calculated as;
Therefore, his acceleration rate is -0.00186 m/s²