The final speed of C is equal to that of B.
The travel time of A is less than that of B.
The final speed of B is equal to that of A.
The travel time of C is greater than that of B.
The final speed of A is equal to that of C.
The travel time of C is greater than that of A.
<h3>What is speed?</h3>
The speed of any moving object is the ratio of the distance covered and the time taken to cover that distance.
Consider the three inclines shown in the figure (attached below). The vertical line is an incline with a 90° angle. All three inclines are frictionless. Three small identical objects are released from the top of the inclines.
When the objects are at the same height, they travel with the same height. The travel time is directly proportional to the incline.
Thus, all the statements are completed.
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Answer:
The correct answer is the IONIC bond.
Explanation:
To solve the problem it is necessary to take into account the concepts of the kinetic equations for the description of the torque at the rate of force and distance.
By definition the torque is given by,

where,


For the problem in question the mass of the trophy is 1.64Kg and the distance of the tropeo to the board (the shoulder) is 0.655m
PART A) For part A, the torque with the given mass and the stipulated torque in the horizontal plane must be calculated as well,

For Newton's second law



PART B) For part B there is an angle of 26 degrees with respect to the horizontal, therefore to know the net torque it is necessary to know the horizontal component to the formed angle, that is,




Answer:Correct answer: 15.85 kg·m/s
Explanation:
A 30 kg gun is standing on a frictionless sur-face. The gun fires a 50 g bullet with a muzzlevelocity of 317 m/s.The positive direction is that of the bullet.Calculate the momentum of the bullet im-mediately after the gun was fired.
Answer:
a) v = 88.54 m/s
b) vf = 26.4 m/s
Explanation:
Given that;
m = 1400.0 kg
a)
by using the energy conservation
loss in potential energy is equal to gain in kinetic energy
mg × ( 3200-2800) = 1/2 ×m×v²
so
1400 × 9.8 × 400 = 0.5 × 1400 × v²
5488000 = 700v²
v² = 5488000 / 700
v² = 7840
v = √7840
v = 88.54 m/s
b)
Work done by all forces is equal to change in KE
W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)
we substitute
1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf² -0 )
488000 = 700 vf²
vf² = 488000 / 700
vf² = 697.1428
vf = √697.1428
vf = 26.4 m/s