A 5.0\, \text {kg}5.0kg5, point, 0, start text, k, g, end text box is at rest on a table. The static friction coefficient \mu_sμ
2 answers:
Answer:
25 N
Explanation:
When a force is applied to an object, for it to move it has to overcome the frictional force. The frictional force is perpendicular to the surface on which it acts. For an object at rest, it is acted upon by static friction, this friction must be overcome before the object can move, while a moving object is resisted by kinetic friction that prevents it from moving.
Since the box is at rest, it is acted upon by a static friction.
mass = 5 kg, acceleration due to gravity = 10 m/s², static friction coefficient = 0.5
Weight of the box = mass * acceleration due to gravity = 5 * 10 = 50 N
The applied horizontal force (F) is:
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Answer:
Q = 1.2*10⁻¹² C
Explanation:
- For any capacitor, by definition the capacitance C is equal to the relationship between the charge on one of the conductors and the potential difference between them, as follows:
- For the special case of a parallel plate capacitor, just by application of Gauss' law to a rectangular surface half out of the outer surface, and half inside it, it can be showed that the value of the capacitance C is a parameter defined only by geometric constants, as follows:
- So, due to the left sides in (1) and (2) are equal each other, right sides must be equal too.
- Replacing ε₀, εr (dielectric constant), A, d and V by their values, we can solve for Q, as follows:
Answer: The length of the shadow on the wall is decreasing by 0.6m/s
Explanation:
the specified moment in the problem, the man is standing at point D with his head at point E.
At that moment, his shadow on the wall is y=BC.
The two right triangles ΔABC and ΔADE are similar triangles. As such, their corresponding sides have equal ratios:
ADAB=DEBC
8/12=2/y,∴y=3 meters
If we consider the distance of the man from the building as x then the distance from the spotlight to the man is 12−x.
(12−x) /12=2/y
1− (1 /12x )=2 × 1/y
Let's take derivatives of both sides:
−1 / 12dx = −2 × 1 / y^2 dy
Let's divide both sides by dt:
−1/12⋅dx/dt=−2/y^2⋅dy/dt
At the specified moment:
dxdt=1.6 m/s
y=3
Let's plug them in:
−1/121.6) = - 2/9 × dy/dt
dy/dt = 1.6/12 ÷ 2/9
dy/dt = 1.6/12 × 9/2
dy/dt = 14.4/24 = 0.6m/s
