Answer:
Explanation:
a). Find the graph attached for the motion.
b). If a shopper walk 5.4 m westwards then 7.8 m eastwards,
Distance traveled by the shopper = Distance traveled in eastwards + Distance traveled westwards
= 5.4 + 7.8
= 13.2 m
c). Displacement of the shopper = Distance walked westwards - Distance traveled eastwards
= 5.4 - 7.8
= -2.4 m
Therefore, magnitude of the displacement of the shopper is = 2.4 m
And the direction of the displacement is eastwards.
F equals 3N with respect to the circle's center, moving in the same direction as the centripetal acceleration.
<h3>How much centripetal force is there in a centrifuge?</h3>
Centripetal force is the force that pushes an item in the direction of its center of curvature. It is fundamental to how a centrifuge operates.
<h3>On a roller coaster, what is centripetal force?</h3>
An item travelling in a circle is pushed inward toward what is known as the center of rotation, which is essentially what a roller coaster accomplishes when it travels through a loop. The force that maintains an object moving along a curved route is this pull toward the center, or centripetal force.
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Explanation:
F = Gm1m2/r^2
kya nikalna hai bhai isme
Answer:
The effective spring constant of the firing mechanism is 1808N/m.
Explanation:
First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

(This is correct because the horizontal motion has acceleration zero). Then:

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

Then, plugging in the given values, we obtain:

Finally, the effective spring constant of the firing mechanism is 1808N/m.
Answer:
Electric potential = 0.00054 V
Explanation:
We are given;
Charge; q = 3 pC = 3 × 10^(-12) C
Radius; r = 2 cm = 0.02 m
Formula for the electric potential of this surface will be;
V = kqr
Where;
K is a constant = 9 × 10^(9) N⋅m²/C².
Thus;
V = 9 × 10^(9) × 3 × 10^(-12) × 0.02
V = 0.00054 V