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Anni [7]
3 years ago
12

A 0.54 kg particle has a speed of 5.0 m/s at point A and kinetic energy of 7.5 J at point B. What is

Physics
1 answer:
Wewaii [24]3 years ago
4 0

<u>Answers</u>

(a)  6.75 Joules.

(b)  5.27 m/s

(c) 0.75 Joules


<u>Explanation</u>

Kinetic energy is the energy possessed by a body in motion.

(a) its kinetic energy at A?

K.E = 1/2 mv²

       = 1/2 ×  0.54 × 5²

       = 6.75 Joules.

(b) its speed at point B?

K.E = 1/2 mv²

7.5 = 1/2 × 0.54 × V²

V² = 7.5 ÷ 0.27

     = 27.77778

V = √27.77778

   = 5.27 m/s

(c) the total work done on the particle as it moves from A to B?

Work done = 7.5 - 6.75

                 = 0.75 Joules

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8 0
1 year ago
An unstrained horizontal spring has a length of 0.36 m and a spring constant of 320 N/m. Two small charged objects are attached
lakkis [162]

Answer:

1.been both -ve charged or both +be charged particles

2. 3.52mC

Explanation:

For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.

Now the Force sustain by the extended string is

F = Ke;

Where K is the force constant of the string, 320 N/m

e is the extension,0.033 m

F = 320 × 0.033 =10.56N

2.But according to columns law of charge;

F = kQ1 Q2

But Q1=Q2{ since the charge are of the same magnitude}.

Hence F = KQ^2

Where K is columns constant =9×10^9F/m

Hence Q=√F/K

Q= √10.56/9×10^9

=3.52×10^-3C

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8 0
3 years ago
An electron is projected with an initial speed v0 = 1.10 x 10⁶ m/s into the uniform field between the parallel plates. The dista
ArbitrLikvidat [17]

Answer:

a) E=364N/C

b) No

Explanation:

A) Because the electron is affected by an acceleration force in this case by the electric field, we can use the formulas of 2-dimension movement.

We will assume the electron missed the upper plate, so we need to calculate the time to travel all the way through the plate, that is:

x=v_x*t

where:\\x=distance\\v=speed\\t=time

so:

t=\frac{x}{v_x}=\frac{0.02m}{1.6\cdot 10^6m/s}\\\\t=1.25\cdot10^{-8}

the electron experiences an accelerated motion in the vertical direction, so we can obtain the acceleration of the electron:

y=\frac{1}{2}.a.t^2\\\\where:\\y=vertical\_distance\\a=acceleration\\t=time

so:

a=\frac{2.y}{t^2}\\\\a=\frac{2*(\frac{0.01}{2}m)}{(1.25\cdot10^{-8}s)^2}\\\\a=6.4\cdot10^{13} m/s^2

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F=m.a=E.q\\so\\E=\frac{m.a}{q}

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Now we can calculate the electric field:

E=\frac{9.1\cdot10^{-31}kg\cdot6.4\cdot10^{13}m/s^2}{1.6\cdot10^{-19}C}\\\\E=364N/C

B) Because the proton has the same charge but positive it will go in the opposite direction, so because we assume the electron didn't touch the plate, the proton won't.

5 0
3 years ago
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|W|=|F|\times distance

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