Answer:
B. Is its acceleration constant
Explanation:
Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. ... An object undergoing uniform circular motion is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction.
It would be 12W because: 6v is half of 12v so half of 24w would be 12w
This question is based on the fundamental assumption of vector direction.
A vector is a physical quantity which has magnitude as well direction for its complete specification.
The magnitude of a physical quantity is simply a numerical number .Hence it can not be negative.
A negative vector is a vector which comes into existence when it is opposite to our assumed direction with respect to any other vector. For instance, the vector is taken positive if it is along + X axis and negative if it is along - X axis.
As per the first option it is given that a vector is negative if its magnitude is greater than 1. It is not correct as magnitude play no role in it.
The second option tells that the magnitude of the vector is less than 1. Magnitude can not be negative. So this is also wrong.
Third one tells that a vector is negative if its displacement is along north. It does not give any detail information about the negativity of a vector.
In a general sense we assume that vertically downward motion is negative and vertically upward is positive. In case of a falling object the motion is vertically downward. So the velocity of that object is negative .
So last option is partially correct as the vector can be negative depending on our choice of co-ordinate system.
Answer:
1. -8.20 m/s²
2. 73.4 m
3. 19.4 m
Explanation:
1. Apply Newton's second law to the car in the y direction.
∑F = ma
N − mg = 0
N = mg
Apply Newton's second law to the car in the x direction.
∑F = ma
-F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
Given μ = 0.837:
a = -(9.8 m/s²) (0.837)
a = -8.20 m/s²
2. Given:
v₀ = 34.7 m/s
v = 0 m/s
a = -8.20 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx
Δx = 73.4 m
3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.
d = v₀t
d = (34.7 m/s) (0.56 s)
d = 19.4 m