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2 years ago

Which change is an example of transforming potential energy to kinetic energy?.

Physics

1 answer:

labwork [276]2 years ago

5 0

Answer:

Kinetic energy is energy an object has because of its motion. A ball held in the air, for example, has gravitational potential energy. If released, as the ball moves faster and faster toward the ground, the force of gravity will transfer the potential energy to kinetic energy.

Explanation:

there hope this helps

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3 years ago

Suppose an element had an atomic number of 52. How many protons would that element have?

frutty [35]

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3 years ago

Read 2 more answers

A ball is dropped and falls with an acceleration of 9.8 m/s2 downward. It hits the ground with a velocity of 49 m/s downward. Ho

il63 [147K]

Hey!

NOTE-:

u= initial velocity
v= final velocity
g= acceleration due to gravity
t= time

u= 0
v= 49 m/s
t=?
g= 9.8 m/s^2

Using first equation of motion -

v-u=at
49-0= 9.8×t
49 = 9.8t
49/9.8= t
t= 5 second

Hope it helps...!!!

6 0

4 years ago

Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of

viktelen [127]

Answer:

$W=2.76\times 10^{-6}\ J$

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

$U=\dfrac{Q^2}{2C}$

Put all the values,

$U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J$

So, the work done is $2.76\times 10^{-6}\ J$ .

8 0

3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$