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koban [17]
3 years ago
5

a particle with charge Q is on the y axis a distance a from the origin and a particle with charge qi is on the x axis at a dista

nce d from the origin. The value of d for which the x component of the force on the second particle is the greatest is
Physics
1 answer:
vova2212 [387]3 years ago
3 0

Answer:

Explanation:

Force on Q due to q₁

= k Q.q₁ / r² , r is distance between two.

X component

= k Q.q₁ / r²  x cos θ ,  θ is angle r makes with x - axis.

= k Q.q₁ / r²  x cos θ

r = a / sinθ

X component

= k Q.q₁ sin²θ . cosθ/ a²  

= k Q.q₁ / 2 a² x sin2θ . sinθ

for maximum value of it ,

sin2θ = 1  , θ = 45°

a / d = tan45

a = d .

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3 years ago
A fireworks shell is accelerated from rest to a velocity of 68.0 m/s over a distance of 0.230 m.
Ugo [173]

Answer:

(A) 10052.2 m/s²

(B)  0.00678 seconds

Explanation:

From the question,

(A) Applying

V² = U²+2as..................... Equation 1

Where V = Final velocity, U = Initial velocity, a = acceleration due to gravity, s = distance.

make a the subject of the  equation

a = (V²-U²)/2s........................ Equation 2

Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m

Substitute these values into equation 2

a = (68²-0²)/(2×0.230)

a = 10052.2 m/s²

(B) Using,

a = (V-U)/t......................... Equation 3

Where t= time.

make t the subject of the equation

t = (V-U)/a......................... Equation 4

Given: V = 68 m/s, U = 0 m/s, a = 10052.2

Substitute into equation 4

t = (68-0)/10052.2

t = 0.00678 seconds

5 0
3 years ago
A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05
7nadin3 [17]

Answer:

W_{drag} = 4.223\,J

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}

The work done on the ball due to drag is:

W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})

W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})

W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]

W_{drag} = 4.223\,J

7 0
3 years ago
Read 2 more answers
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