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<span>KE = 1/2mv^2 = (1/2)(2000)(2^2) = 4000 J This must equal the net work acting on the car. W=Fd The net force is 1140-950= 190N. so, d=W/F = 4000/190 = 21.05 m</span>
Answer:
Correct answer is A.
The higher the enzyme, the higher the Vmax
Explanation:
Although, in the absence of enzyme, the rate of a reaction(Vmax) increase linearly with substrate concentration. The reaction rate is given as dp/dt.
The rate of a reaction involving enzyme also increases.
At low enzyme concentrations or high substrate concentrations, all of the available enzyme active sites could be occupied with substrates. Therefore, increasing the substrate concentration further will not change the rate of diffusion. In other words, there is some maximum reaction rate (Vmax) when all enzyme active sites are occupied. The reaction rate will increase with increasing substrate concentration, but must asymptotically approach the saturation rate, Vmax. Vmax is directly proportional to the total enzyme concentration, E
Answer:
28.5 m/s
Explanation:
There are 2 different velocities: the train velocity and yours velocity. They're in opposite directions, so one is positive and other is negative. Take the forward direction as the positive:
V = +30 -1.5
V = 28.5 m/s
Answer:
Vx = 15V
Explanation:
To find the speed of the third car you take into account that the distance that they travel is the same. Only time is different in one second. Mathematically you have:

you equal the first and second equation obtain the time t:

then, you can equal the third and first equation:

hence, Vx = 15V
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TRANSLATION:
Para encontrar la velocidad del tercer automóvil, tenga en cuenta que la distancia que recorren es la misma. Solo el tiempo es diferente en un segundo. Matemáticamente tienes:
iguala la primera y segunda ecuación obtiene el tiempo t:
entonces, puedes igualar la tercera y primera ecuación:
por lo tanto, Vx = 15V
Answer:
B) A body at rest will stay at rest
Explanation:
A body in motion will stay in motion unless acted on by another force.