All categories

3 years ago

A vector must always have both size and ..

Physics

1 answer:

DochEvi [55]3 years ago

5 0

Both magnitude and DIRECTION
For example,
• 12m East
• -2 miles
•9 meter north
• 8 miles up

You might be interested in

A tray filled with ice is removed from the freezer. After a short period of time, the ice begins to melt.

Sav [38]

A: decreases in specific heat capacity

3 0

3 years ago

A small metal sphere has a mass of 0.19 gg and a charge of -23.0 nCnC. It is 10.0 cmcm directly above an identical sphere that h

RoseWind [281]

Answer:

a = -7.29 m / s²

Explanation:

For this exercise we must use Newton's second law,

F -W = m a

Force is electrical force

F = k q₁ q₂ / r²

k q₁ q₂ / r² -mg = m a

indicate that the charge of the two spheres is equal

q₁ = q₂ = q

a = (k q² / r² - m g) / m

a = k q² / m r² - g

Let's reduce the magnitudes to the SI system

m = 0.19 g (1kg / 1000 g) = 1.9 10⁻⁴ kg

q1 = q2 = q = -23.0 nC (1C / 10⁹ nC) = -23.0 10⁻⁹ C

r = 10.0 cm (1m / 100cm) = 0.1000 m

let's calculate

a = 9 10⁹ (23.0 10⁻⁹)² / (0.1000² 1.9 10⁻⁴) - 9.8

a = -7.29 m / s²

The negative sign indicates that the direction of this acceleration is downward

6 0

3 years ago

During which phase of the moon may a solar eclipse occur?

alex41 [277]

Answer:

New moon

Explanation:

A solar eclipse can only take place at the phase of new moon, when the moon passes directly between the sun and Earth and its shadows fall upon Earth's surface.

4 0

4 years ago

Read 2 more answers

An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference

kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$ (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

$K=\frac{1}{2}mv^2$ (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm

6 0

4 years ago

1. The large wheels on the train have a mass of 0.5 kg, while the small wheels have a mass of 0.15 kg. The train body, and child

blsea [12.9K]

Answer: 60.975 Joules

KE = 1/2 mv^2

=> 1/2 * 40.65 * 3

=> 60.975 J

5 0

2 years ago