Static frictional force = ƒs = (Cs) • (Fɴ)
2.26 = (Cs) • m • g
2.26 = (Cs) • (1.85) • (9.8)
Cs = 0.125
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
1.49 = (Cκ) • m • g
1.49 = (Cκ) • (1.85) • (9.8)
Cκ = 0.0822
Use the equation for the acceleration
A = final velocity - initial velocity divided by time final - time initial
A= 54 - 32 / 8 - 0
A= 22 / 8
A= 2.75 m/s^2
Hope this helps!
Answer:
Quick question do you mean what are some safety rules
Explanation:
Crosswalk, Stop sign,
Answer:
-8.4°C
Explanation:
From the principle of heat capacity.
The heat sustain by an object is given as;
H = m× c× (T2-T1)
Where H is heat transferred
m is mass of substance
T2-T1 is the temperature change from starting to final temperature T2.
c- is the specific heat capacity of ice .
Note : specific heat capacity is an intrinsic capacity of a substance which is the energy substained on a unit mass of a substance on a unit temperature change.
Hence ; 35= 1× c× ( T2-(-25))
35= c× ( T2+25)
35 =2.108×( T2+25)
( T2+25)= 35/2.108= 16.60°{ approximated to 2 decimal place}
T2= 16.60-25= -8.40°C
C, specific heat capacity of ice is =2.108 kJ/kgK{you can google that}
Answer:
71.85 m/s
Explanation:
Given the following :
Length of skid marks left by jaguar (s) = 290 m
Skidding Acceleration (a) = - 8.90m/s²
Final velocity of jaguar (v) = 0
Speed of Jaguar before it Began to skid =?
Hence, initial speed of jaguar could be obtained using the formula :
v² = u² + 2as
Where
v = final speed of jaguar ; u = initial speed of jaguar(before it Began to skid) ; a = acceleration of jaguar ; s = distance /length of skid marks left by jaguar
0² = u² + (2 × (-8.90) × 290)
0 = u² + (-5,162)
u² = 5162
Take the square root of both sides
u = √5162
u = 71.847 m/s
u = 71.85m/s