Answer:
Explanation:
Speed of car =22.5miles/hr
U=22.5miles/hour
Applied brake and come to rest
Final velocity, =0
t, =2sec
Given that,
Speed=distance /time
Then,
Distance, =speed, ×time
Converting mile/hour to m/s
Given that
Use: 1 mile= 1600 m, 1 h= 3600s
22.5miles/hour × 1600m/mile × 1hour/3600s
Therefore, 22.5mile/hour=10m/s
Using speed =10m/s
Distance =speed ×time
Distance=10×2
Distance, =20m
The distance travelled before coming to rest is 20m.
To claculate the gravitational attraction between two bodies with mass 1(m1) and mass 2 (m2) you need to use the equation:
F= G ((m1*m2)/r^2)
Where;
G is the gravitational constant (6.67E-11 m^3 s-2 Kg-1) and
r is the distance between the two objects.
Answer: C and D
Explanation: I’m pretty sure it’s C and D I got the same question but cold water let me know please, thanks!
Answer:
P V = n R T ideal gas equation
P2 V2 / P1 V1 = T2 / T1
V2 / V1 = T2 / T1 * P1 / P2 = T2 P1 / (T1 P2)
V2 / V1 = (1.17 T1) / T1 * (P1 / .22 P1) assuming absolute temp as 1.17 P1
V2 / V1 = 1.17 / .22 = 5.32
V = 4/3 pi R^3 = 4/3 pi (D/2)^3 = 4/3 pi D^3 / 8 = pi D^3 / 6
V2 / V1 = D2^3 / D1^3
D2 = (V2 / V1 * D1^3)^1/3
D2 = 5.32^1/3 * D = 1.75 D (D1 = D)
Answer:
4) Driving in a straight line at 60 miles per hour
Explanation:
1) Driving 60 miles per hour around a curve
2) Going from 0 to 60 miles per hour in 10 seconds
3) Slamming on the brakes to come to a stop at a stop sign
4) Driving in a straight line at 60 miles per hour
1) The speed is constant here, but in circular motion you have an acceleration that is v^2/r, where v is the speed and r the radius
2) You are accelerating from 0 to 60
3) You are desaccelerating
4) constant speed , no acceleration