Answer:
The shortest braking distance is 35.8 m
Explanation:
To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down
On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis
Y axis
N- W = 0
N = W = mg
X axis
-Fr = m a
-μ N = m a
-μ mg = ma
a = μ g
a = - 0.32 9.8
a = - 3.14 m/s²
We calculate the distance using the kinematics equations
Vf² = Vo² + 2 a x
x = (Vf² - Vo²) / 2 a
When the train stops the speed is zero (Vf = 0)
Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s
x = ( 0 - 15²) / 2 (-3.14)
x= 35.8 m
The shortest braking distance is 35.8 m
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13 g —> 0.013 kg
KE = 1/2(m)(v)^2
KE = 1/2(0.013)(8.5)^2
KE = 0.47 J
Answer:
7.65x10^3 m/s
Explanation:
The computation of the satellite's orbital speed is shown below:
Given that
Earth mass, M_e = 5.97 × 10^24 kg
Gravitational constant, G = 6.67 × 10^-11 N·m^2/kg
Orbital radius, r = 6.80 × 10^6m
Based on the above information
the satellite's orbital speed is
V_o = √GM_e ÷ √r
= √6.67 × 10^-11 × 5.97 × 10^24 ÷ √6.80 × 10^6
= 7.65x10^3 m/s
Answer:
<h2>A.
6pF</h2>
Explanation:
If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown

Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;

C₄ ≈ 6pF
Hence the value of the X capacitor is approximately 6pF
When you are talking about calculating the kinetic energy of an object, the formula is as stated:
Ek =1/2mv²
Where m is mass and v is velocity. Sub all those numbers in and you'll get 25J of kinetic energy.