By Newton's second law,
<em>n</em> + (-<em>w</em>) = 0
<em>p</em> + (-<em>f</em> ) = (20 kg) (2 m/s²)
where <em>n</em> is the magnitude of the normal force, <em>w</em> is the weight of the box, <em>p</em> is the magnitude of the applied force (<em>p</em> for <u>p</u>ush or <u>p</u>ull), and <em>f</em> is the magnitude of the friction force.
Calculate the weight of the box:
<em>w</em> = (20 kg) (9.80 m/s²) = 196 N
Then
<em>n</em> = <em>w</em> = 196 N
and
<em>f</em> = <em>µ</em> <em>n</em> = 0.5 (196 N) = 98 N
Now solve for <em>p</em> :
<em>p</em> - 98 N = 40 N
<em>p</em> = 138 N
The answer is commensalism because commensalism is a relationship where an organism is benefitted and the other is neither benefitted nor harmed. The barnacle is being benefited and the whale is not being benefited or harmed.
Answer:
d = 0m
v_avg = 2.5 m/s
Explanation:
A) The distance traveled by the swimmer was:
However, the displacement is zero because the final position of the swimmer is the same as the initial position.
displacement = 0m
The average velocity was:
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
A) La distancia recorrida por el nadador fue:
Sin embargo, el desplazamiento es cero porque la posición final del nadador es la misma que la posición inicial.
desplazamiento = 0m
La velocidad promedio fue:
Answer:
A) E = 4.96 x 10³ eV
B) E = 4.19 x 10⁴ eV
C) E = 3.73 x 10⁹ eV
Explanation:
A)
For photon energy is given as:
where,
E = energy of photon = ?
h = 6.625 x 10⁻³⁴ J.s
λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m
Therefore,
<u>E = 4.96 x 10³ eV</u>
<u></u>
B)
The energy of a particle at rest is given as:
where,
E = Energy of electron = ?
m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
<u>E = 4.19 x 10⁴ eV</u>
<u></u>
C)
The energy of a particle at rest is given as:
where,
E = Energy of alpha particle = ?
m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg
c = speed of light = 3 x 10⁸ m/s
Therefore,
<u>E = 3.73 x 10⁹ eV</u>
Answer:
V(t) = (q0/C) * e^(−t/RC
)
Explanation:
If there were a battery in the circuit with EMF E , the equation for V(t) would be V(t)=E−(RC)(dV(t)/dt) . This differential equation is no longer homogeneous in V(t) (homogeneous means that if you multiply any solution by a constant it is still a solution). However, it can be solved simply by the substitution Vb(t)=V(t)−E . The effect of this substitution is to eliminate the E term and yield an equation for Vb(t) that is identical to the equation you solved for V(t) . If a battery is added, the initial condition is usually that the capacitor has zero charge at time t=0 . The solution under these conditions will look like V(t)=E(1−e−t/(RC)) . This solution implies that the voltage across the capacitor is zero at time t=0 (since the capacitor was uncharged then) and rises asymptotically to E (with the result that current essentially stops flowing through the circuit).