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3 years ago

A small plastic bead has been charged to -15nC.

Physics

1 answer:

marishachu [46]3 years ago

8 0

Answer:

Explanation:

q = - 15 n C = - 15 x 10^-9 c

(a) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on proton, F = charge on proton x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (1.67 x 10^-27)

a = 1.6 x 10^14 m/s²

(B) the direction of acceleration is towards the bead, as the force is attractive.

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on electron, F = charge on electron x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (9.1 x 10^-31)

a = 3 x 10^17 m/s²

(D) the direction of acceleration is away from the bead, as the force is repulsive.

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An object of mass 2kg raised to a height 10m possess potential energy of 200J. What is the kinetic energy and potential energy a

Shtirlitz [24]

Explanation:

${\bold{\sf{\underline{Understanding \: the \: concept}}}}$

✠ This question says that there is an object and its mass is 2 kg ; it's raised to a height 10 m and possess potential energy of 200 J. Now this question ask us to find the kinetic energy and the potential energy at a height 4 metre.

$\bold{↬{ }}$ ${\bold{\sf{\underline{Given \: that}}}}$

✰ Mass = 2 kilograms

✰ Raised height = 10 metres

✰ Posses potential energy = 200 Joules

$\bold{↬{ }}$ ${\bold{\sf{\underline{To \: find}}}}$

✰ Kinetic energy at a height 4 metre

✰ Potential energy at a height 4 metre

${\bold{\sf{\underline{Solution}}}}$

✰ Kinetic energy at a height 4 metre = 120 J

✰ Potential energy at a height 4 metre = 80 J

${\bold{\sf{\underline{Using \: concepts}}}}$

✰ Potential energy formula.

${\bold{\sf{\underline{Using \: formula}}}}$

✰ Potential energy = mgh

${\bold{\sf{\underline{We \: also \: write \: these \: as}}}}$

✰ Potential energy as P.E

✰ Mass as m

✰ Joules as J

✰ Height as h

✰ Raised height as g

${\bold{\sf{\underline{Full \: solution}}}}$

<h3>✠ Let us find the Potential energy.</h3>

↦ Potential energy = mgh

↦ Potential energy = 2 × 10 × 4

↦ Potential energy = 20 × 4

↦ Potential energy = 80 J

<h3>✠ Now according to the question let us find the kinetic energy</h3>

↦ Kinetic energy = Posses potential energy - Finded potential energy

↦ Kinetic energy = 200 J - 80 J

↦ Kinetic energy = 120 Joules

4 0

3 years ago

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4.Quick reactions help you defeat opponents.

5.Sports may be competitive but they teach you valuable life lessons.

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3 years ago

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After fertilization in the fallopian tube, how many days will the zygote travel before arriving at uterus?.

natta225 [31]

After fertilization in the fallopian tube, It will take 6-12 days for the zygote to travel before arriving at the uterus.

<h3>What is a zygote?</h3>

A zygote is, generally speaking, a cell created by the fusion of two gametes; the growing person is created from such a cell.

It takes the zygote around 6–12 days following fertilization in the fallopian tube for the fertilized egg to travel to the uterus and attach to the uterus, a process known as implantation.

Hence It will take 6-12 days for the zygote to travel before arriving at the uterus.

To learn more about the zygote refer;

brainly.com/question/465851

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6 0

2 years ago

4. A force of 5 N gives a mass m,, an acceleration of 10 m's, and a mass

Art [367]

Answer:

For mass m 1 newton 2nd law

F=m 1 a 1

5=m 1 ×10

m 1 = 2 1 kg

For mass m 2

F=m 2 a 2

5=m 2 ×20

m 2 = 4 1 kg

if tied together

Total mass =m 1 +m 2 = 1/2 +1/4=3/4kg

Now

F=M T Q T

a T = 5/m T = 5×4/3 = 3 /20m/s^2

Explanation:

8 0

3 years ago

When a 2.60-kg object is hung vertically on a certain light spring described by Hooke's law, the spring stretches 2.86 cm. (a) W

Goryan [66]

Answer:

a) k = 891.82 N/m

b) e = 0.0143 m = 1.43 cm

c) W = 5.02 J

Explanation:

Step 1: Data given

Mass = 2.60 kg

the spring stretches 2.86 cm = 0.0286

Step 2: What is the force constant of the spring?

Force constant, k = force applied / extension produced

k = (2.60kg * 9.81N/kg) / 0.0326 m

k = 891.82 N/m

b) If the 2.60-kg object is removed, how far will the spring stretch if a 1.30-kg block is hung on it

Extension = F/k = (1.30 kg * 9.81) / 891.82 = 0.0143 m = 1.43 cm

Half the mass means half the extension

c) How much work must an external agent do to stretch the same spring 7.50 cm from its unstretched position?

W = average force used * distance

W = 1/2 * k*e * e = 1/2 k*e²

W = 1/2 * 891.82 * (0.075)² = W = 5.02 J

5 0

3 years ago