The frictional force is 218.6 N
Explanation:
The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.
There are two forces acting along this direction:
- The component of the weight parallel to the incline, downward along the plane, of magnitude
where
m = 46 kg is the mass
is the acceleration of gravity
is the angle of the incline
- The (static) frictional force, acting upward, of magnitude 
Since the block is in equilibrium, we can write
And substituting, we find the force of friction:
Learn more about frictional force along an inclined plane:
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What property/properties of water allowed the water to stick to the penny?
Solution's:
Explanation:
By these properties, the water molecules gets attracted to the other molecules. Hence, this makes the water get stick to the penny.
You see its position before and then its position after.
The final velocity is 3.47 m/s east
Explanation:
We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the two players before and after the collision must be conserved:
where:
is the mass of the first player
is the initial velocity of the first player (we take east as positive direction)
is the mass of the second player
is the initial velocity of the second player
is their final combined velocity after the collision
Re-arranging the equation and substituting the values, we find:
So, their velocity afterwards is 3.47 m/s east.
Learn more about momentum:
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From conservation of momentum, the ram force can be calculated similarly to rocket thrust:
F = d(mv)/dt = vdm/dt.
<span>In other words, the force needed to decelerate the wind equals the force that would be needed to produce it.
</span><span> v = 120/3.6 = 33.33 m/s
</span><span> dm/dt = v*area*density
</span> dm/dt = (33.33)*((45)*(75))*(1.3)
dm/dt = <span>
146235.375 </span><span>kg/s
</span><span> F = v^2*area*density
</span> F = (33.33)^2*((45)*(75))*(1.3) = <span>
<span>4874025 </span></span><span>N
</span> This differs by a factor of 2 from Bernoulli's equation, which relates velocity and pressure difference in reference not to a head-on collision of the fluid with a surface but to a fluid moving tangentially to the surface. Also, a typical mass-based drag equation, like Bernoulli's equation, has a coefficient of 1/2; however, it refers to a body moving through a fluid, where the fluid encountered by the body is not stopped relative to the body (i.e., brought up to its speed) like is the case in this problem.
