Answer:
the volume decreases at the rate of 500cm³ in 1 min
Explanation:
given
v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min
PV=C
vΔp + pΔv = 0
differentiate with respect to time
v(Δp/t) + p(Δv/t) = 0
(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0
40000 + 80kPa(Δv/t) = 0
Δv/t = -40000/80
= -500cm³/min
the volume decreases at the rate of 500cm³ in 1 min
Answer:
6.5454 m
Explanation:
Let the distance from the wire carrying 3 A current is x
Then the distance from the the carrying current 8 A is 24-x
We know that magnetic field due to long wire is given by
It is given that magnetic field is zero at some distance so

Here
So 
Answer: 80m
Explanation:
Distance of balloon to the ground is 3150m
Let the distance of Menin's pocket to the ground be x
Let the distance between Menin's pocket to the balloon be y
Hence, x=3150-y------1
Using the equation of motion,
V^2= U^s + 2gs--------2
U= initial speed is 0m/s
g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s
40m/s is contant since U (the coin is at rest is 0) hence V =40m/s
Slotting our values into equation 2
40^2= 0^2 + 2 * 10* (3150-y)
1600 = 0 + 63000 - 20y
1600 - 63000 = - 20y
-61400 = - 20y minus cancel out minus on both sides of the equation
61400 = 20y
Hence y = 61400/20
3070m
Hence, recall equation 1
x = 3150 - 3070
80m
I hope this solve the problem.
Answer:
La velocidad media es 5
, que equivale a 1.389 
Explanation:
La velocidad es una magnitud física que expresa la relación entre el espacio recorrido por un objeto y el tiempo empleado para ello.
La velocidad media relaciona el cambio de la posición con el tiempo empleado en efectuar dicho cambio. Por lo que se calcula como la distancia recorrida por un objeto dividido por el tiempo transcurrido:

En este caso:
- distancia= 10 km= 10,000 m (siendo 1 km= 1,000 m)
- tiempo= 2 h= 7,200 s (siendo 1 h= 3,600 s)
Entonces, reemplazando en la definición de velocidad media:

Resolviendo se obtiene:

<u><em>La velocidad media es 5 </em></u>
<u><em>, que equivale a 1.389 </em></u>
<u><em></em></u>