Answer:
Explanation:
AVerage acceleration is the cjange in velocity with time
a = v-u/t
v is the final velocity = 48m/s
u is the initial velocity = 40m/s
t is the time = 6.5s
a = 48-40/6.5
a = 8/6.5
a = 1.23m/s²
Hence the magnitude of the car’s average acceleration during this period is 1.23m/s²
Answer:
Explanation:
Given
Initially Reading on the odometer is 
Final reading on the odometer is 
Time taken is 
average velocity 
Thus the average velocity of mail truck is
Answer:
330.5 m
Explanation:
In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .
The maximum height will be calculated as;
where ∝ is the angle of launch = 30°
vi= initial launch velocity = 40 m/s
g= 10 m/s²
h= 40²*sin²40° / 2*10
h={1600*0.4132 }/ 20
h= 661.1/2 = 330.5 m
Answer:
La longitud del camino recorrido es de 25.9 [m]
Explanation:
Se reemplaza el valor de tiempo en segundos en la ecuación dada de desplazamiento
x=10+20*(3) - 4.9*(3)^2
x= 25.9 [metros]
Explanation:
Heat flow = conductivity × area × change in temperature / thickness
q' = kAΔT/t
13.3 W = k (1.56 m²) (7.8°C) / (0.0234 m)
k = 0.0256 W/m/°C
Heat lost by water = heat gained by ice
-mCΔT = mL + mCΔT
-(1000 g) (1 cal/g/°C) (12°C − 37°C) = m (79.7 cal/g) + m (1 cal/g/°C) (12°C − 0°C)
25,000 cal = (91.7 cal/g) m
m = 272.6 g
