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.
This is false. I hope this helps.
The diesel engine has no ignition system. (C)
It depends on the compression of the fuel/air mixture to ignite the mixture.
This may be a big part of the reason that Diesel engines always
sound to me as if they are falling apart.
Answer:
a). V = 3.13*10⁶ m/s
b). T = 1.19*10^-7s
c). K.E = 2.04*10⁵
d). V = 1.02*10⁵V
Explanation:
q = +2e
M = 4.0u
r = 5.94cm = 0.0594m
B = 1.10T
1u = 1.67 * 10^-27kg
M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg
a). Centripetal force = magnetic force
Mv / r = qB
V = qBr / m
V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27
V = 2.09088 * 10^-20 / 6.68 * 10^-27
V = 3.13*10⁶ m/s
b). Period of revolution.
T = 2Πr / v
T = (2*π*0.0594) / 3.13*10⁶
T = 1.19*10⁻⁷s
c). kinetic energy = ½mv²
K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²
K.E = 3.27*10^-14J
1ev = 1.60*10^-19J
xeV = 3.27*10^-14J
X = 2.04*10⁵eV
K.E = 2.04*10⁵eV
d). K.E = qV
V = K / q
V = 2.04*10⁵ / (2eV).....2e-
V = 1.02*10⁵V
The distance covered is 115 m
Explanation:
The motion of Ileana is a uniformly accelerated motion (constant acceleration), therefore we can use the following suvat equation:
where
s is the distance covered
u is the initiaal velocity
v is the final velocity
t is the time elapsed
In this problem, we have:
u = 4.20 m/s
v = 5.00 m/s
t = 25.0 s
Therefore, we can re-arrange the equation to find the distance covered:
Learn more about accelerated motion:
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