-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless. I guess that's what "smooth" means.
-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.
-- Your force is 6.4 N.
Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
</em> That's about <em>2.634 m/s²</em> <em>
</em>(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)
-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is
Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = <em>2.45 N</em>.
That's the force that's accelerating the little block, so that must be the tension
in the string.
ANSWER:
F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.
STEP-BY-STEP EXPLANATION:
F(h) is Horizontal Force = 200 N
V is Speed = 2.4 m/s
The total weight increase by 42%
coefficient of rolling friction decrease by 19%
Since the velocity is constant so acceleration is zero; a=0
Now the horizontal force required to move the pickup is equal to the frictional force.
F(h) = F(f)
F(h) = mg* u
m is mass
g is gravitational acceleration = 9.8 m/s^2
200 = mg*u
Since weight increases by 42% and friction coefficient decreases by 19%
New weight = 1+0.42 = 1.42 = (1.42*m*g)
New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u
F(h) = (0.81μ) (1.42 m g)
= (0.81) (1.42) (μ m g)
= (0.81) (1.42) (200)
= 230 N
Answer:
Average accelation = -4V
Explanation:
V=0 m/s (because the frog stopped)
V0 = V (average velocity)
t= 0,25 s
So;
Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:
<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
![r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%20r_%7B2%7D%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7BT_%7B1%7D%7D%7BT_%7B2%7D%7D%29%5E%7B2%7D%7D%20)
![r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%203.84%5Ccdot%2010%5E%7B5%7D%20km%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7B1%20d%7D%7B0.07481%20y%20%5Ccdot%20%5Cfrac%7B365%20d%7D%7B1%20y%7D%7D%29%5E%7B2%7D%7D%20)
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!
With time, momentum increases as it builds speed assuming their is nothing in the way to stop it. Based on the graph, you can see that example being displayed as the line on the graph gets higher
