Answer:
A) 37 m
Explanation:
The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:
(1)
where
v = 0 m/s is the final velocity of the car
u = 24 m/s is the initial velocity
a is the acceleration
d is the length of the skid
We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:
where
m = 1000 kg is the mass of the car
is the coefficient of friction
a is the deceleration of the car
g = 9.8 m/s^2 is the acceleration due to gravity
The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:
And we can substitute it into eq.(1) to find d:
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Answer:
The resulting net force on the airplane would be 250N.
Explanation:
It would be 250N because to find the amount of Newtons you would have to do 450 minus 200. 450 minus 200 equals 250. You can check by adding 200 plus 250 and it equals 450 the which is the "thrust".
Answer:
11.) g = 3.695 m/s^2
12.) g = 8.879 m/s^2
13.) E = 8127 N/C
Explanation:
11.) Given that the
Mercury mass M = 3.3 × 10^23kg
Radius r = 2.44 ×10^6 m
Gravitational constant G = 6.67408 × 10^-11 m3kg-1 s^-2
Gravitational field strength g can be calculated by using the formula below
g = GM/r^2
Substitutes all the parameters into the formula
g = (6.67408 × 10^-11 × 3.3 × 10^23)/(2.44×10^6)^2
g = 2.2×10^13/5.954×10^12
g = 3.695 m/s^2
12.) Given that the
Venus mass M = 4.87×10^24kg
Radius r = 6.05 × 10^6 m
Using the same formula for gravitational field strength g
g = GM/R2
Substitute all the parameters into the formula
g = (6.67408 × 10^-11 × 4.87×10^24)/(6.05×10^6)^2
g = 3.25×10^14/3.66×10^13
g = 8.879 m/s^2
13.) Given that the
Charge = 2.26 nC = 2.26×10^-9
Distance d = 0.05m
Electric field strength E can be calculated by using the formula below
E = Kq/d^2
Where
K = electrostatic constant 8.99 × 10^9 Nm2/C2
Substitutes all the parameters into the formula
E = (8.99 × 10^9 × 2.26×10^-9)/0.05^2
E = 20.3174/2.5×10^-3
E = 8126.96 N/C
Hi there!
Normal rain has a pH between 5.0 and 5.5, which is slightly acidic on the pH scale. Acid rain on the other hand has an average pH of 4.0, much more acidic than your average rain.
Therefore, the most likely pH of acid rain would be 4.0.
