Answer:
The coefficient of friction causes the force on the object to be less than its initial reading on the spring scale.
Explanation:
Since the block reads 24.5 N before the block starts to move, this is its weight. Now, when the block starts to move at a constant velocity, it experiences a frictional force which is equal to the force with which the student pulls.
Now, since the velocity is constant so, there is no acceleration and thus, the net force is zero.
Let F = force applied and f = frictional force = μN = μW where μ = coefficient of friction and N = normal force. The normal force also equals the weight of the object W.
Now, since F - f = ma and a = 0 where a = acceleration and m = mass of block,
F - f = m(0) = 0
F - f = 0
F = f
Since the force applied equals the frictional force, we have that
F = μW and F = 23.7 N and W = 24.5 N
So, 23.7 N = μ(24.5 N)
μ = 23.7 N/24.5 N
μ = 0.97
Since μ = 0.97 < 1, the coefficient of friction causes the force on the object to be less than its initial reading on the spring scale.
Answer:
Explanation:
The way to show a cubed substance is either like this³ or like this x^3. The small three is found at the bottom toolbar at the bottom of the question space marked by the Ω symbol.
100 mmHg
Givens
V1 = 20 cm^3
V2 = 80 cm^3
P1 = 400 mmHg
P2 = ?
Formula
V1 * P1 = V2 * P2
Solution
20 * 400 = 80 * P2 Divide by 80
20 * 400/80 = P2
P2 = 8000 / 80
P2 = 100 mmHg
Se necesita 105 j de trabajo para mover 2,7 c de carga de la placa negativa a la placa positiva de un condensador de placa paralela. Qué diferencia de tensión existe entre las placas
Answer:
3430000 J
Explanation:
The formula for potential energy is PE=mgh.
M being the mass, g being the force of gravity, and h being the height.
First thing you want to do is convert 250 kg to g (grams).
From there you get 25000g and you have to multiply that by 14m and 9.8m/s^2 (the force of gravity is constant, at least on earth).
Answer:
Explanation:
a ) Between r = 0 and r = r₁
Electric field will be zero . It is so because no charge lies in between r = 0 and r = r₁ .
b ) From r = r₁ to r = r₂
At distance r , charge contained in the sphere of radius r
volume charge density x 4/3 π r³
q = Q x r³ / R³
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q x r³ / ε₀R³
E= Q x r / (4πε₀R³)
E ∝ r .
c )
Outside of r = r₂
charge contained in the sphere of radius r = Q
Applying Gauss's law
4πr² E = q / ε₀
4πr² E = Q / ε₀
E = Q / 4πε₀r²
E ∝ 1 / r² .
