Answer:
The mass of the child + skateboard is 50 kg
Explanation:
In this problem, we can apply Newton's second law:
F = ma
where
F is the net force on a system
m is the mass of the system
a is the acceleration of the system
In this problem, our system is the child + the skateboard. The net force on them is
F = 75 N
and their acceleration is
So we can re-arrange the equation above to find their combined mass:
1.velocity and acceleration
2.
3.inertia
4.
5.speed
Answer:
a.) The main scale reading is 10.2cm
b.) Division 7 = 0.07
c.) 10.27 cm
d.) 10.31 cm
e.) 10.24 cm
Explanation:
The figure depicts a vernier caliper readings
a.) The main scale reading is 10.2 cm
The reading before the vernier scale
b.) Division 7 = 0.07
the point where the main scale and vernier scale meet
c.) The observed readings is
10.2 + 0.07 = 10.27 cm
d.) If the instrument has a positive zero error of 4 division
correct reading = 10.27 + 0.04 = 10.31cm
e.) If the instrument has a negative zero error of 3 division
correct reading = 10.27 - 0.03 = 10.24cm
Answer:
I_weight = M L²
this value is much larger and with it it is easier to restore balance.I
Explanation:
When man walks a tightrope, he carries a linear velocity, this velocity is related to the angular velocity by
v = w r
For man to maintain equilibrium needs the total moment to be zero
∑τ = I α
S τ = 0
The forces on the home are the weight of the masses, the weight of the man and the support on the rope, the latter two are zero taque the distance to the center of rotation is zero.
Therefore the moment of the masses and the open is the one that must be zero.
If the man carries only the bar, we could approximate it by two open one on each side of the axis of rotation formed by the free of the rope
I = ⅓ m L² / 4
As the length of half the length of the bar and the mass of the bar is small, this moment is small, therefore at the moment if there is some imbalance it is difficult to recover.
If, in addition to the opening, each of them carries a specific weight, the moment of inertia of this weight is
I_weight = M L²
this value is much larger and with it it is easier to restore balance.
Answer:
because speed is the modulus of velocity which is a vector
the velocity to be zero it must be a round trip
Explanation:
This is because speed is the modulus of velocity which is a vector.
For the velocity to be zero it must be a round trip, therefore the resulting vector zero
On the other hand, the speed of the module is the same in both directions
