Ask question

Ask question

All categories

2 years ago

12

A ball drops some distance and gains 30 J of kinetic energy. How much gravitational potential energy did the ball start with? Do

not ignore the effects of air resistance.

1 answer:

Dovator [93]2 years ago

4 0

Answer:

The gravitational potential energy of the ball is more than 30 J.

Explanation:

Given that,

Gains kinetic energy = 30 J

We need to calculate the gravitational potential energy

According to law of conservation of energy

If there is no air resistance then the potential energy of the ball fully converted into kinetic energy.

If the air resistance is considered then the potential energy used to do work against air resistance.

$\Delta U\geq \Delta K.E$

According to question,

A ball drops some distance and gains 30 J of kinetic energy and considers air resistance then 30 J of potential energy can be loosed by the ball.

Hence, The gravitational potential energy of the ball is more than 30 J.

You might be interested in

James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m

Over [174]

Answer:

$4.1\cdot 10^8 N$

Explanation:

First of all, we need to find the pressure exerted on the sphere, which is given by:

$p=p_0 + \rho g h$

where

$p_0 =1.01\cdot 10^5 Pa$ is the atmospheric pressure

$\rho = 1000 kg/m^3$ is the water density

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

$h=11,000 m$ is the depth

Substituting,

$p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa$

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m

So the total area of the sphere is

$A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2$

And so, the inward force exerted on it is

$F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N$

8 0

2 years ago

Read 2 more answers

In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.

denis23 [38]

The magnetic force exerted by a field E to a charge q is given by F=Eq. In this case, F=4.30*10^4*(6.80mu C). 1mu C=10^-6C, so F=4.30*6.80=10^-2=0.29N. The direction is in the x direction, the direction that the field is applied because the charge is positive.

5 0

2 years ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$

the vertical component of the velocity is

$v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.$

which gives a speed $v$ of

$v = \sqrt{v_x^2+v_y^2}$

$\boxed{v =33.3m/s.}$

4 0

2 years ago

Two resistors ( 3 ohms & 6 ohms) in a series circuit with a power supply = 12 volts. The current through resistor 6 ohms is

Scorpion4ik [409]

In a series circuit . . .

-- The total resistance is the sum of the individual resistors.

-- The current is the same at every point in the circuit.

The total resistance in this circuit is (3Ω + 6Ω ) = 9Ω

The current at every point is (V/R) = (12v / 9Ω ) = <em>1.33 A</em> .

Pick choice<em> (a)</em>.

6 0

1 year ago

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a

Lapatulllka [165]

Answer:

Amplitude, A = 0.049 meters

Explanation:

Given that,

A harmonic wave travels in the positive x direction at 6 m/s along a taught string. A fixed point on the string oscillates as a function of time according to the equation :

$y = 0.049 \cos(7t)$ .......(1)

The general equation of a wave is given by :

$y=A\cos(\omega t)$ .......(2)

A is amplitude of wave

On comparing equation (1) and (2) we get :

A = 0.049 meters

So, the amplitude of the wave is 0.049 meters.

3 0

2 years ago