An egg is thrown horizontally off the top row bleachers at the Brickyard,

El valor más preciso de la masa de un electron es 9.11*10^-11kg ¿ Cuánto aumentaría la masa de un cuerpo que se carga con -1 c d

chubhunter [2.5K]

Answer:

$m = 569.375\times 10^{6}\,kg$

Explanation:

Un electrón tiene una carga negativa de $1.6\times 10^{-19}\,C$ y la masa total es igual al producto del número de electrones y esa carga unitaria. El número de electrones se obtiene al dividir la carga total por la carga unitaria. (An electron has a negative charge of $1.6\times 10^{-19}\,C$ and the total mass is equal to the product of the qunatity of electrons and such unit charge. The quantity of electrons is found by diving the total charge by the unit charge):

$m = \frac{Q}{q} \cdot m_{e}$

$m = \left(\frac{1\,C}{1.6\times 10^{-19}\,C} \right)\cdot (9.11\times 10^{-11}\,kg)$

$m = 569.375\times 10^{6}\,kg$

7 0

3 years ago

A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$

This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.

3 0

3 years ago

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total

Lynna [10]

Explanation:

Given Data

Total mass=93.5 kg

Rock mass=0.310 kg

Initially wagon speed=0.540 m/s

rock speed=16.5 m/s

To Find

The speed of the wagon

Solution

As the wagon rolls, momentum is given as

P=mv

where

m is mass

v is speed

put the values

P=93.5kg × 0.540 m/s

P =50.49 kg×m/s

Now we have to find the momentum of rock

momentum of rock = mv

momentum of rock = (0.310kg)×(16.5 m/s)

momentum of rock =5.115 kg×m/s

From the conservation of momentum we can find the wagons momentum So

wagon momentum=50.49 -5.115 = 45.375 kg×m/s

Speed of wagon = wagon momentum/(total mass-rock mass)

Speed of wagon=45.375/(93.5-0.310)

Speed of wagon= 0.487 m/s

Throwing rock backward,

momentum of wagon = 50.49+5.115 = 55.605 kg×m/s

Speed of wagon = wagon momentum/(total mass-rock mass)

speed of wagon = 55.605 kg×m/s/(93.5kg-0.310kg)

speed of wagon= 0.5967 m/s

7 0

2 years ago

Sound waves have relatively long wavelengths. We can hear people around the corner before we can see them. Which wave behavior d

pogonyaev

I'd say diffraction since sound waves can bend around objects like corners. Let's say you're in the hallway and you can hear sound coming from a door. The sound waves diffract around the door and spread out into the hallway, making it possible for you to hear.
Also, you can hear it before you see it because light waves are shorter than sound waves and hardly diffract around doors.

6 0

3 years ago

Meg goes swimming on a hot afternoon. When she comes out of the pool, her foot senses that the prevement is unbearably hot. Supp

mart [117]

1.Record her observation with the time it was hot.
2. Gather info about the pavement and its surroundings. Find out what it's made of and what its temp. is at different times of the day.
3. Come up with a hypothesis about why it is hot.
4. Design an experiment to test the hypothesis. If she thinks the Sun is responsible (which she should b smart enough to know), keep it covered during the day time and check it's temp.
5. Come up with a conclusion. If her hypothesis is not supported, design a new experiment or gather more info.

7 0

3 years ago