Calculate the gravitational potential energy of 3kg book sitting on shelf 4m high

A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?

garri49 [273]

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

$U_i +K_i = U_f + K_f$

where:

$U_i$ is the initial potential energy, at the bottom

$K_i$ is the initial kinetic energy, at the bottom

$U_f$ is the final potential energy, at the top

$K_f$ is the final kinetic energy, at the top

We can rewrite the equation as:

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m is the mass of the hopper

$g=9.8 m/s^2$ is the acceleration of gravity

$h_i = 0$ is the initial height

u is the launch speed of the hopper

$h_f = 1.20 m$ is the maximum altitude reached by the hopper

v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

$u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s$

Learn more about kinetic energy and potential energy:

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#LearnwithBrainly

4 0

3 years ago

What's the Coulomb's law?

Ulleksa [173]

In layman's term: like charges don't attract while opposite charges doelectrostatic forces between point A( which is charged) and point B (which is also charged) are proportional to the charge of point A and point B. there is also something else about this law that I don't quite remember.

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Here is the formula:

F = k x Q1 x Q2/d^2

What the formula means:

F=force between charges

Q1 and Q2= amount of charge

d=distance between these two charges

k= Coulombs constant (proportionally constant)

________________________________________________

I think that about covers it and hopefully this helped.

4 0

3 years ago

Read 2 more answers

Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25

mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp = 1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0

3 years ago

Which is the resulting direction of a surface wave

Sliva [168]

In a surface wave (like water) the water goes up and down, but the wave travels across (parallel to) the surface.

3 0

3 years ago

A farsighted boy has a near point at 2.3 m and requires eyeglasses to correct his vision. Corrective lenses are available in inc

tino4ka555 [31]

Answer:

P = 3.5 D

Explanation:

As we know that convex lens is to be used to make the near point of eye to be correct

So we will have

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

here we have

$d_i = 2.3 m = 230 cm$

$d_o = 25 cm$

now plug in all values into the formula

$-\frac{1}{230} + \frac{1}{25} = \frac{1}{f}$

$f = 28 cm$

now for power of lens

$P = \frac{1}{f}$

$P = \frac{1}{0.28} = 3.5 D$

so the power in dioptre is

P = 3.5 D

5 0

3 years ago