tectonic plates rubbing against each other.
Answer:
v= 13 m/s
Explanation:
Velocity is defined as the derivative of displacement with respect to time
v= ds/dt
Known data
s(t) = 5t + 2t² : distance that the ball has rolled after t seconds
vi= 5 m/s : initial velocity
t= 2 s
Problem develoment
s(t) = 5t + 2t²
v= ds/dt= 5 + 4t : velocity of the ball in function of the time
We replace t =2 s in the equation of velocity
v= 5 + 4(2)
v= 13 m/s : velocity after 2 seconds
Answer:
Explanation:
Let t represent the time for Tina to catch David.
Hence, considering the equation of linear motion S = ut + 1/2at^2..... 1
For David u = 28.0 m/s where 'a' is set to nought
S = ut
S = 28t.......2
For Tina consider equation 1
Where acceleration = 2.90m/s^2 and u is set at nought
S = 1/2×2.90 m/s×t^2.......3
Equate 2 and 3
28t = 1.45t^2
Divide through by t
28 = 1.45t
t = 28/1.45
t = 19.31seconds
Now put the value of t into equation 3
S = 1/2×2.90 m/s×t^2.......3
= 1.45×20×20
= 580m
Tina must have driven 580meters before passing David
Considering the equation of linear motion : V^2 = U^2+2as
Where u is set at nought
V^2 = 2as
V^2 = 2×2.9×580
V^2 = 3364
V = √3364
V = 58m/s
Her speed will be 58m/s
Correct question is;
A thermal tap used in a certain apparatus consists of a silica rod which fits tightly inside an aluminium tube whose internal diameter is 8mm at 0°C.When the temperature is raised ,the fits is no longer exact. Calculate what change in temperature is necessary to produce a channel whose cross-sectional is equal to that of the tube of 1mm. (linear expansivity of silica = 8 × 10^(-6) /K and linear expansivity of aluminium = 26 × 10^(-6) /K).
Answer:
ΔT = 268.67K
Explanation:
We are given;
d1 = 8mm
d2 = 1mm
At standard temperature and pressure conditions, the temperature is 273K.
Thus; Initial temperature; T1 = 273K,
Using the combined gas law, we have;
P1×V1/T1 = P2×V2/T2
The pressure is constant and so P1 = P2. They will cancel out in the combined gas law to give:
V1/T1 = V2/T2
Now, volume of the tube is given by the formula;V = Area × height = Ah
Thus;
V1 = (πd1²/4)h
V2 = (π(d2)²/4)h
Thus;
(πd1²/4)h/T1 = (π(d2)²/4)h/T2
π, h and 4 will cancel out to give;
d1²/T1 = (d2)²/T2
T2 = ((d2)² × T1)/d1²
T2 = (1² × T1)/8²
T2 = 273/64
T2 = 4.23K
Therefore, Change in temperature is; ΔT = T2 - T1
ΔT = 273 - 4.23
ΔT = 268.67K
Thus, the temperature decreased to 268.67K
Respuesta:
0,0560 cal / gºC.
Explicación:
Cantidad de calor; (Q)
Q = mcΔt; Δt = t2 - t1
m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura
c de agua = 1 cal / gºC
c de aluminio = 0,22 cal / gºC
QTotal = Q de agua + Q de aluminio
Q de agua = 450 * 1 * (26 - 23) = 1350 cal
Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal
QTotal = 1350 + 39,6 = 1389,6 cal
Calor perdido = calor ganado
QTotal = calor perdido
- 1389,6 = 335,2 * c * (26 - 100)
-1389,6 = −24804,8 * c
c = 1389,6 / 24804,8
c = 0,056021 cal / gºC.
Capacidad calorífica específica de la plata = 0,0560 cal / gºC.
