Ask question

Ask question

All categories

3 years ago

13

If a golf ball and a bowling ball are moving and both have the same genetic energy if they move at the same speed which one has

more kinetic energy

1 answer:

Anastaziya [24]3 years ago

5 0

If they both have the same kinetic energy then they both have the same kinetic energy. otherwise it would be the bowling ball due to factors like weight

You might be interested in

Which planetery body has the greast gravitational pull?

Nataly [62]

Answer should be the earth

7 0

3 years ago

A student swings a container of water in a vertical circle of radius 1.0 m. Calculate the minimum speed of the container so that

12345 [234]

Answer:

Explanation:

The centripetal acceleration requirement must equal gravity at the top of the circle

mg = mv²/R

v = √Rg

v = √(1.0(9.8))

v = 3.1304951...

v = 3.1 m/s

6 0

3 years ago

A spherical shell is rolling without slipping at constant speed on a level floor. What percentage of the shell's total kinetic e

IgorC [24]

Answer:

41.667 per cent of the total kinetic energy is translational kinetic energy.

Explanation:

As the spherical shell is rolling without slipping at constant speed, the system can be considered as conservative due to the absence of non-conservative forces (i.e. drag, friction) and energy equation can be expressed only by the Principle of Energy Conservation, whose total energy is equal to the sum of rotational and translational kinetic energies. That is to say:

$E = K_{t} + K_{r}$

Where:

$E$ - Total energy, measured in joules.

$K_{r}$ - Rotational kinetic energy, measured in joules.

$K_{t}$ - Translational kinetic energy, measured in joules.

The spherical shell can be considered as a rigid body, since there is no information of any deformation due to the motion. Then, rotational and translational components of kinetic energy are described by the following equations:

Rotational kinetic energy

$K_{r} = \frac{1}{2}\cdot I_{g}\cdot \omega^{2}$

Translational kinetic energy

$K_{t} = \frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2}$

Where:

$I_{g}$ - Moment of inertia of the spherical shell with respect to its center of mass, measured in $kg\cdot m^{2}$ .

$\omega$ - Angular speed of the spherical shell, measured in radians per second.

$R$ - Radius of the spherical shell, measured in meters.

After replacing each component and simplifying algebraically, the total energy of the spherical shell is equal to:

$E = \frac{1}{2}\cdot (I_{g} + m\cdot R^{2})\cdot \omega^{2}$

In addition, the moment of inertia of a spherical shell is equal to:

$I_{g} = \frac{2}{3}\cdot m\cdot R^{2}$

Then, total energy is reduced to this expression:

$E = \frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2}$

The fraction of the total kinetic energy that is translational in percentage is given by the following expression:

$\%K_{t} = \frac{K_{t}}{E}\times 100\,\%$

$\%K_{t} = \frac{\frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2} }{\frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2} } \times 100\,\%$

$\%K_{t} = \frac{5}{12}\times 100\,\%$

$\%K_{t} = 41.667\,\%$

41.667 per cent of the total kinetic energy is translational kinetic energy.

7 0

3 years ago

Two similar sized stones one heavy and one light dropped from the same height into a pond explain why the impact of the heavy st

Zigmanuir [339]

In energy point of view, the larger stone had more potential energy before dropping. impacting the water, the larger one, having more kinetic energy which changed from potential energy, tranfered energy to the water and formed wave. the amplitude of the wave indicate the energy of the wave. more energy more amplitude.

5 0

3 years ago

A car of 900 kg mass is moving at the velocity of 60 km/hr. It is brought into rest at 50 meter distance by applying a brake. No

kozerog [31]

Answer: $-2502N$

Explanation:

$(V_2)^2=(V_1)^2+2ad$

where;

$V_2$ = final velocity = 0

$V_1$ = initial velocity = 60 km/h = 16.67 m/s

$a$ = acceleration

$d$ = distance

First all of, because acceleration is given in m/s and not km/h, you need to convert 60km/h to m/s. Our conversion factors here are 1km = 1000m and 1h = 3600s

$60km/h(\frac{1000m}{1km} )(\frac{1h}{3600s} )=16.67m/s$

Solve for a;

$(V_2)^2=(V_1)^2+2ad$

Begin by subtracting $(V_1)^2$

$(V_2)^2-(V_1)^2=2ad$

Divide by 2d

$\frac{(V_2)^2-(V_1)^2}{2d} =a$

Now plug in your values:

$a=\frac{(0)^2-(16.67 m/s)^2}{2(50m)}$

$a=\frac{0-277.89m^2/s^2}{100m}$

$a=-2.78m/s$

If you're wondering why I calculated acceleration first is because in order to find force, we need 2 things: mass and acceleration.

$F=ma$

m = mass = 900kg

a = acceleration = -2.78m/s

$F=(900kg)(-2.78m/s)\\F=-2502N$

It's negative because the force has to be applied in the opposite direction that the car is moving.

8 0

3 years ago