Answer:
v2 = 0.6 m/s in opposite direction
Explanation:
The computation of the speed of the raft is as follows
The center of system mass would be remains at the same place also it has no external force applied
Therefore
V = 0
So,
M1.V1= - (M2.V2)
where,
M1 = 60 kg, V1 = 1.5 m/s
M2 = 150 kg
Now
60 × 1.5 = - 150 × v2
v2 = -0.6 m/s
v2 = 0.6 m/s in opposite direction
Answer:
I. Aceleración, a = -2 m/s²
II. Distancia, S = 100 metros
Explanation:
<u>Dados los siguientes datos;</u>
Velocidad inicial = 72 km / h
Tiempo = 10 segundos
Velocidad final = 0 m/s
<u><em>Conversión:</em></u>
72 km/h a metros por segundo = 72 * 1000/3600 = 72000/3600 = 20 m/s
I. Para encontrar la aceleración, usaríamos la primera ecuación de movimiento;
Dónde;
- V es la velocidad final.
- U es la velocidad inicial.
- a es la aceleración.
- t es el tiempo medido en segundos.
Sustituyendo en la fórmula, tenemos;
<em>Aceleración, a = -2 m/s²</em>
<em>Nota: el signo negativo indica desaceleración o retraso.</em>
II. To find the acceleration, we would use the third equation of motion;
Dónde;
- V es la velocidad final.
- U es la velocidad inicial.
- a es la aceleración.
- S es la distancia.
Sustituyendo en la fórmula, tenemos;
<em>Distancia, S = 100 metros</em>
Answer:
C
Explanation:
The Coulomb's Law gives us the Electric Field strength of charge as follows:
E = k* (Q) / R^2
Where,
k : Coulomb constant 8.99 * 10^9 N /C
Q: Charge generating Electric Field strength
R: Distance from source to any point
In our case the value of E = 4.106 N /C is experienced by charge +3 C. This charge is replaced by an equal magnitude of charge but with negative sign -3 C. Scrutinizing on the above formula we can see that all quantities remain same hence, the magnitude of field strength remains same.
However, we also know a positive magnitude is directed away from the source charge while a negative magnitude is directed towards the source. We can conclude that only the direction of electric field reverses while magnitude remains same.
Answer:
<em>The horizontal velocity vector of the canonball does not change at all, but is constant throughout the flight.</em>
Explanation:
First, I'll assume this is a projectile simulation, since no simulation is shown here. That been the case, in a projectile flight, there is only a vertical component force (gravity) acting on the body, and no horizontal component force on the body. The effect of this on the canonball is that the vertical velocity component on the canonball goes from maximum to zero at a deceleration of 9.81 m/s^2, in the first half of the flight. And then zero to maximum at an acceleration of 9.81 m/s^2 for the second half of the flight before hitting the ground. <em>Since there is no force acting on the horizontal velocity vector of the canonball, there will be no acceleration or deceleration of the horizontal velocity component of the canonball. This means that the horizontal velocity component of the canonball is constant throughout the flight</em>
The molecules of ice stick together in the process of cohesion. They are tightly packed so there isn't much room to move. Liquid water is a looser hold. The molecules can go past one another, and they will take the shape of whatever container they occupy. Water vapor is loosely contained, and it will will fill whatever container it is kept in, and it will take its shape, too.
