speed of the car is given on the road
reaction time is given as
now we can find the distance that it will move during this reaction time
now the deceleration of the car is 10 m/s^2
so the distance that it will move before stop is given by
so the total distance that it require to stop is given as
while the deer is standing at distance 38 m
so the car will stop 2.8 m before the position of deer
Answer:
give me brainliest first and ill give u the correct anwers
Explanation:
The salesman is telling you the average magnitude of the car's acceleration.
| Acceleration | = (change in speed) / (time for the change)
| Acceleration | = (60 mi/hr) / (6 sec)
| Acceleration | = 10 miles/hr-sec
That would be 36,000 miles per hour squared,
or 0.0028 mile per second squared.
<em>500 sec</em>
<em>8 min 20 sec</em>
<em>Hi there !</em>
<em />
<em>8 m ................ 1 s </em>
<em>4000 m ........ x s</em>
<em>x = 4000m×1s/8m = 500 sec = 8 min 20 sec</em>
<em />
<em>Good luck ! </em>
Answer:
0.265
Explanation:
Draw a free body diagram. There are four forces:
Normal force Fn pushing up.
Weight force mg pulling down.
Tension force T at an angle θ.
Friction force Fn μ pushing left.
Sum the forces in the y direction:
∑F = ma
Fn + T sin θ − mg = 0
Fn = mg − T sin θ
Sum the forces in the x direction:
∑F = ma
T cos θ − Fn μ = 0
Fn μ = T cos θ
μ = T cos θ / Fn
μ = T cos θ / (mg − T sin θ)
Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:
μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)
μ = 0.265
