The magnitude of the component of the box’s weight
perpendicular to the incline can be olve using the formula:
F = wcos(a)
Where F is the box’s weight perpendicular to the incline
W is the weight of the box
A is the angle of the incline
F = (46)cos(25)
F = 42 N
Answer: 70.5 km/h
Justification:
The question is not clearly stated but it seems you are asking for the x - component of the velocity of the helicopter.
You can find the x and y - components of the velocity using the trigonometric ratios sine and cosine.
The sine ratio relates the y-component and the velocity by:
sin(angle) = y-component of velocity / velocity
The cosine ratio related the x-component and the velocity by:
cos(angle) = x-component of velocity / velocity.
Since you have the angle and the velocity and are asked by the x-component of the velocity, you need to use the cosine ratio:
cos(35°)= x-component / 86.0 km/h
=> x -component = 86.0 km/h * cos(35°) = 70.5 km/h
Answer:
Please find the answer in the explanation
Explanation:
Given that A 1.0 g plastic bead, with a charge of -6.0 nC, is suspended between the two plates by the force of the electric field between them.
Since it is suspended, it must have been repelled by the bottom negative plate and trying to be attracted to the top plate.
We can therefore conclude that the upper plate, is positively charged
B.) The charge on the positive plate of parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates must be less than 6.0 nC
<h2>The hiker will go up to 850 m on the hill</h2>
Explanation:
The total energy gained by the hiker = 140 x 4186 J
This energy is consumed in the potential energy acquired , while climbing up the hill.
The potential energy P.E = mass of hiker x acceleration due to gravity x height
Thus
140 x 4186 = 69 x 10 x h
or h = 
= 850 m
If the 20% of the total energy is used
the height h₀ = 
= 170 m
Answer:
h = v₀² / 2g
, h = k/4g x²
Explanation:
In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches
Starting point. Lower compressed spring
Em₀ = K = ½ m v²
Final point. Highest on the path
= U = mg h
As or no friction the energy is conserved
Em₀ = Em_{f}
½ m v₀²² = m g h
h = v₀² / 2g
We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse
½ k x² = 2 g h
h = k/4g x²
