A lightweight ball is thrown horizontally. What forces act on the ball just after being released from the thrower's hand
1 answer:
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Answer:
The magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.
Explanation:
Given;
length of the straight wire, L = 0.56 m
conventional current, I = 0.4 A
distance of magnetic field from the wire, r = 2.6 cm = 0.026 m
To determine magnitude of magnetic field made by current in the wire, we will apply Bio-Savart Law;
Therefore, the magnitude of the magnetic field made by current in the wire is 3.064 x 10⁻⁶ T.
<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>
Explanation:
a) Given S(t) = 76 + 128t − 16t²
s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft
s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft
Displacement in 4 seconds = 332 - 76 = 256 ft
Time = 4 - 0 = 4 s
Average velocity in first 4 seconds is 64 ft/s upward
a) Given S(t) = 76 + 128t − 16t²
s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft
s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft
Displacement in 4 seconds = 78 - 332 = -254 ft
Time = 4 - 0 = 4 s
Average velocity in second 4 seconds is 63.5 ft/s downward