Answer:
The height at which the object is moved is 10 meters.
Explanation:
Given that,
Force acting on the object, W = F = 490 N
The gravitational potential energy, P = 4900 J
We need to find the height at which the object is moved. We know that the gravitational potential energy is possessed due to its position. It is given by :
So, the height at which the object is moved is 10 meters. Hence, this is the required solution.
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
1 pound ≈ 0.4536 kg
170 pounds ≈ 170 * 0.4536 kg
≈ 77.112 kg
Answer:
a) 37.70 m/s
b)710.6 m/s²
Explanation:
Given that ;
Mass of object = 2 kg
Radius of the motion = 2m
Frequency of motion = 3 rev/s
The formula to apply is;
v= 2πrf where v is linear speed
v = 2×π×2×3 =12π = 37.70 m/s
Centripetal acceleration is given as;
a= 4×π²×r×f²
a= 4×π²×2×3²
a=710.6 m/s²
