A block of mass m=3.0 kg is pushed down a 50° incline surface with a force of magnitude F=40 N.

An alpha particle can be produced in certain radioactive decays of nuclei and consists of two protons and two neutrons. The part

Varvara68 [4.7K]

Answer:

Explanation:

charge, q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

mass, m = 4 u = 4 x 1.661 x 10^-27 kg = 6.644 x 10^-27 kg

Radius, r = 4.5 cm = 0.045 m

Magnetic field, B = 1.20 T

(a) Let the speed is v.

$v=\frac{Bqr}{m}$

$v=\frac{1.20\times 3.2\times 10^{-19}\times 0.045}{6.644\times 10^{-27}}$

v = 2.6 x 10^6 m/s

(b) Let T be the period of revolution

$T=\frac{2\pi r}{v}$

$T=\frac{2\times 3.14\times 0.045}{2.6\times 10^{6}}$

T = 1.09 x 10^-7 s

(c) The formula for the kinetic energy is

$K=\frac{B^{2}\times q^{2}\times r^{2}}{2m}$

$K=\frac{\left ( 1.20\times 3.2 \times 10^{-19}\times 0.045 \right )^{2}}{2\times 6.644\times 10^{-27}}$

K = 2.25 x 10^-14 J

(d) Let the potential difference is V.

K = qV

$V = \frac{K}{q}$

$V= \frac{2.25\times 10^-14}{3.2\times 10^{-19}}$

V = 70312.5 V

5 0

3 years ago

Un cuerpo se lanza verticalmente hacia arriba con una velocidad de 13 m/s. ¿Cuánto tiempo tarda en alcanzar la altura máxima? a)

Elena L [17]

Answer:

D. 1.33 segundos.

Explanation:

El cuerpo es experimenta un movimiento en caída libre al modificarse su velocidad por efecto de la gravitación terrestre. Este cuerpo alcanza instantáneamente el reposo cuando se encuentra a su altura máxima, el tiempo puede obtenerse sabiendo la aceleración y las velocidades incial y final a partir de la siguiente ecuación cinemática:

$v = v_{o}+g\cdot t$

Donde:

$v$ - Velocidad final del cuerpo, medida en metros por segundo.

$v_{o}$ - Velocidad inicial del cuerpo, medida en metros por segundo.

$g$ - Aceleración gravitacional, medida en metros por segundo al cuadrado.

$t$ - Tiempo, medido en segundos.

Ahora se despeja el tiempo:

$t = \frac{v-v_{o}}{g}$

Si $v_{o} = 13\,\frac{m}{s}$ , $v=0\,\frac{m}{s}$ y $g = -9.807\,\frac{m}{s^{2}}$ , entonces:

$t = \frac{0\,\frac{m}{s}-13\,\frac{m}{s}}{-9.807\,\frac{m}{s^{2}} }$

$t = 1.326\,s$

Por ende, la respuesta correcta es D.

6 0

3 years ago

Plants depend on the atmosphere for the largest source of

Sonja [21]

The correct answer that would best complete the given statement above would be CARBON DIOXIDE (CO2). Plants <span>depend on the atmosphere for the largest source of carbon dioxide. Carbon dioxide is considered as an important trace gas in the Earth's atmosphere which constitutes about 0.04% of it. Hope this answer helps. </span>

6 0

3 years ago

Someone help me solve this please and explain

vfiekz [6]

Answer:

i think you minus The weights I think

8 0

3 years ago

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

kherson [118]

Explanation:

The Coulomb's law states that the magnitude of each of the electric forces between two point-at-rest charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=\frac{kq_1q_2}{d^2}$

In this case we have an electron (-e) and a proton (e), so:

$F=-\frac{ke^2}{d^2}\\F=-\frac{8.99*10^9\frac{N\cdot m^2}{s^2}(1.6*10^{-19}C)^2}{(933*10^{-9}m)^2}\\F=-2.64*10^{-16}N$

In this case, the electric force is negative, therefore, the force is repulsive and its magnitude is:

$F=2.64*10^{-16}N$

3 0

3 years ago