Have All existing elements have been discovered

A 39.5 kg child stands at the center of a 125 kg playground merry-go-round which rotates at 3.10 rad/s. If the child moves to th

Neporo4naja [7]

Using the law of conservation of angular momentum, we have

I1 w1 = I2 w2 

ie., m1r^2/2 x w1 = ( m1r^2/2 + m2r^2 ) w2 

ie., new angular velocity w2 = m1 w1 / ( m1+ 2m2) = 125 x 3.1 / ( 125 + 2 x39.5 ) 

= 1.8995 = 1.9 rad /sec ( nearly )

3 0

3 years ago

A force of 8 N accelerates by 4 m/s^2. What would be the amount of force needed to give a final acceleration of 5.3 m/s^2

Elis [28]

Answer:

10.6 N

Explanation:

F = ma

8 = m * 4

m = 2 kg

F = 2 * 5.3

F = 10.6 N

4 0

3 years ago

A. Calculate the electric potential energy stored in a capacitor that stores <img src="https://tex.z-dn.net/?f=3.40%20x%2010%5E%

sertanlavr [38]

Part a)

As we know that energy stored inside the capacitor is given as

$U = \frac{1}{2}CV^2$

for a given capacitor we know

$Q = CV$

Now we can use it in above equation to find the energy

$U = \frac{1}{2}QV$

$U = \frac{1}{2}(3.4\times 10^{-6})(24)$

$U = 40.8\times 10^{-6} J$

PART b)

If two negative charges are hold near to each other and then released

Then due to mutual repulsion they start moving away from each other

Due to mutual repulsion as the two charges moving away the electrostatic potential energy of two charges will convert into kinetic energy of the two charges.

So here as they move apart kinetic energy will increase while potential energy will decrease

Part c)

As we know that capacitance is given as

$C = \frac{Q}{V}$

here we know that

$Q = 3\times 10^{-10}C$

$V = 35 volts$

$C = \frac{3\times 10^{-10}}{35}$

$C = 8.6 \times 10^{-12} F$

5 0

3 years ago

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A baby carriage is sitting at the top of a hill that is 21 m high. The carriage with the baby has a mass of 4kg. The carriage ha

Ira Lisetskai [31]

Answer:

E=252J

Explanation:

The total mechanical energy of an object or system is given by:

E mech=K+U

Where K is the kinetic energy of the object and U is the potential energy of the object. The carriage, sitting motionless at the top of the hill, has only potential energy in the form of gravitational potential energy.

Gravitational potential energy is given by:

Ug=mgh

Where m is the mass of the object, g is the gravitational acceleration constant, and h is the height of the object above some specific reference point, in this case the ground 21 m below.

The weight of a stationary object at the surface of the earth is equal to the force of gravity acting on the object.

W=→Fg=mg

We are given that the carriage weighs 12 N, therefore mg=12N.

Ug=12N⋅21m

⇒Ug=252Nm=252J

Hope it helped, God bless you!

5 0

3 years ago

Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts

Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground = 3.13 m/s

4 0

3 years ago