<em>Since the wagon is being pulled down hill with a constant velocity, all the forces of the wagon would be (C) increasing.</em>
<em>You are correct! **</em>
Answer:
Loss, 
Explanation:
Given that,
Mass of particle 1, 
Mass of particle 2, 
Speed of particle 1, 
Speed of particle 2, 
To find,
The magnitude of the loss in kinetic energy after the collision.
Solve,
Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.
Applying the conservation of momentum to find the speed of two particles after the collision.



V = 6.71 m/s
Initial kinetic energy before the collision,



Final kinetic energy after the collision,



Lost in kinetic energy,



Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.
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Answer:
change in relative vorticity 0.0590
Explanation:
Given data
pressure = 1000 hPa
temperature lapse rate q1 = 3.1◦C per 50 hPa
pressure = 850 hPa
temperature lapse rate q2= -0.61◦C per 50 hPa
to find out
change in relative vorticity
solution
we will apply here formula that is
N = (g / potential temperature ) × (potential vertical temperature) × exp^1/2 ............................1
here we know g = 9.8 m/s
and q1 = potential temperature=3.3 degree celsius
potential vertical temperature gradient = 3.1 - 0.61 / 1000 -850
potential vertical temperature gradient = 0.0166 degree celsius/hpa
so
N = 9.8 / 2.75 × 0.0166 × exp^1/2
N = 0.0590
Answer:
The value of A is 1.5m/s^2 and B is 0.5m/s^³
Explanation:
The mass of the rocket = 2540 kg.
Given velocity, v(t)=At + Bt^2
Given t =0
a= 1.50 m/s^2
Now, velocity V(t) = A*t + B*t²
If, V(0) = 0, V(1) = 2
a(t) = dV/dt = A+2B × t
a(0) = 1.5m/s^²
1.5m/s^² = A + 2B × 0
A = 1.5m/s^2
now,
V(1) = 2 = A× 1 + B× 1^²
1.5× 1 +B× 1 = 2m/s
B = 2-1.5
B = 0.5m/s^³
Now Check V(t) = A× t + B × t^²
So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s
Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)