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LuckyWell [14K]

2 years ago

5

An object of mass 0.5 kg is swung in uniform circular motion. The radius is 2 meters, and the force exerted is 4 N. Calculate th

e magnitude of the velocity.

1 answer:

Katarina [22]2 years ago

8 0

Using
F= mv²/r
4 = 0.5×v² / 2
8 /0.5 = v²
v²=16
v= √16
v= 4 ms-¹

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A system of 4 electrons, 18 protons, and 4 neutrons has a net charge of?

slamgirl [31]

Answer:

+ 14

Explanation:

18 protons make a positive 18 charge (+18)

4 electrons make a negative 4 charge (-4)

both combined give + 18 - 4 = + 14

The four neutrons don't carry net charge, so the addition of the electrons doesn' affect the net charge found above which still gives + 14.

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2 years ago

A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and at its highest point is 1

SCORPION-xisa [38]

Answer:

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b) 52.38 m/s

c) -75.58 m

Explanation:

See attachment for calculation

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2 years ago

In an extrasolar planetary system containing a single planet, the parent star is measured to move about its center of mass every

Mademuasel [1]

Answer:

Orbital Time Period is 24 years

Explanation:

This can be explained by the definition of time period.

Time period can be defined as the time taken by an object to complete one cycle, here, time taken to complete one revolution.

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Therefore, the time taken by the parent star to move about its mass center is the orbital time period that is 24 years.

4 0

2 years ago

A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul

GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that for spherical shell and pure rolling conditions

$v = R \omega$

$I = \frac{2}{3}mR^2$

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh$

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$h = \frac{5v^2}{6g}$

$h = \frac{5(8^2)}{6(9.81)} = 5.44 m$

Part b)

If ball is not rolling and just sliding over the hill then in that case

$\frac{1}{2}mv^2 = mgh$

$h = \frac{v^2}{2g}$

$h = \frac{8^2}{2(9.81)} = 3.16 m$

3 0

2 years ago

A car traveling at 30 m/s speeds up to 35 m/s over a period of 5 seconds. What is the acceleration of the car?

Delvig [45]

Answer:

u =30 m/s

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t = 5 secs

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a = (v- u)/t

a = (35-30)/5

a = 5/5

a = 1 m/s^2

PLS MARK BRAINLIEST

8 0

2 years ago