The solution would
be like this for this specific problem:
<span>
F=−</span>k∗x∗<span>q∗</span>Q<span>/(</span>+)<span>F−≈</span><span><span><span>k∗x∗<span>q∗</span>Q</span><span>/R3</span></span>[(</span>1−<span><span>3/2</span><span><span>*x2</span><span>/R3</span></span>]
</span><span>F=−</span><span><span>k∗x∗<span>q∗/</span>Q</span><span>R<span>3
</span></span></span><span>F=</span><span>ma
</span>−<span><span><span><span>k∗<span>q∗</span>Q</span><span>/R3</span></span>*</span>x</span>=<span>ma
</span>−k∗x=m∗<span>a
a</span>==<span><span><span>ω2</span>x
</span>ω</span><span>=(</span>k/<span>m<span>)<span><span>1/</span><span>2
</span></span></span></span>ω<span>=(</span><span>kqQ</span>/<span><span>R3</span><span>)<span><span>1/</span>2
</span></span></span>
<span>I am hoping that
this answer has satisfied your query and it will be able to help you in your
endeavor, and if you would like, feel free to ask another question.</span>
S=d/t
30/0.75=40
Answer: 40km/h
Answer:
14.0 cm
Explanation:
Draw a free body diagram of the block. There are three forces: weight force mg pulling down, elastic force k∆L pulling down, and buoyancy ρVg pushing up.
Sum of forces in the y direction:
∑F = ma
ρVg − mg − k∆L = 0
(1000 kg/m³) (4.63 kg / 648 kg/m³) (9.8 m/s²) − (4.63 kg) (9.8 m/s²) − (176 N/m) ∆L = 0
∆L = 0.140 m
∆L = 14.0 cm
Answer:
Final velocity, v = 25.3 m/s
Explanation:
Initial velocity of a locomotive, u = 19 m/s
Acceleration of the locomotive, a = 0.8 m/s²
Length of station, d = 175 m
We need to find its final velocity (v) when the nose leaves the station. It can be calculated using the third law of motion :
v = 25.31 m/s
v = 25.3 m/s
When the nose leaves the station, it will move with a velocity of 25.3 m/s. Hence, this is the required solution.