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padilas [110]
3 years ago
9

A charged particle is observed traveling in a circular path in a uniform magnetic field. If the particle had been traveling four

times as fast, the radius of the circular path would be:
Physics
1 answer:
8090 [49]3 years ago
8 0

Answer:

The new radius of the trajectory of the particle is four times the previous radius

Explanation:

In order to know what is the radius of the trajectory of the charged particle, if its speed is four times as fast, you take into account the following formula, which describes the radius of a charged particle in a magnetic field:

r=\frac{mv}{qB}          (1)

If the speed of the particle is for time as fast, that is, v' = 4v, you obtain, in the equation (1):

r'=\frac{mv'}{qB}=\frac{m(4v)}{qB}=4\frac{mv}{qB}=4r

The new radius of the trajectory of the particle is four times the previous radius

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.2, A car starting from rest has an acceleration of
riadik2000 [5.3K]

Answer: 2.5 m/s and 6.25 m

Explanation:

u = 0

a = 0.5 m/s²

t = 5 s

v = u + at

=  0 + 0.5 × 5

= <u>2.5 m/s</u>

s = ut + 1/2 at²

= 1/2 × 2.5 × 5

=<u> 6.25 m</u>

7 0
4 years ago
Math the following to definitions
Leona [35]
Hi,
The answer should be:
Vector 5
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Mass 2
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Hope this helps!
6 0
3 years ago
The same net force is applied to two bodies. The first body acquires half of the acceleration of the second body. What is the re
Yanka [14]

The mass of the first body is double the mass of the second body.

This all revolves around Newtons 2nd law of motion that states F=m*a.

If we understand that body one has the same force applied but half of the acceleration than body two, we can easily understand that it has to do something with the objects mass. This is because mass and acceleration in newtons 2nd law are inversely proportional. This means if and objects force stays the same, but it’s mass doubles, then the acceleration must be half its original value. And if the acceleration doubles and the force stays constant than the mass must have halved from its original value. It sounds quite complicated but it’s very simple. Hopefully this helps!!!

4 0
3 years ago
Una atracción de feria consiste en lanzar un trineo de 2 kg por una rampa ascendente que forma un ángulo de 30º con la horizonta
Arada [10]

Answer:

La velocidad con la que se debe lanzar el trineo es aproximadamente 9.96 m/s

Explanation:

La masa del trineo, m = 2 kg

El ángulo de inclinación del trineo, θ = 30 °

El coeficiente de fricción, μ = 0,15

La altura a la que debe ascender el trineo, h = 4 m

La reacción normal del trineo en la rampa, N = m · g · cos (θ)

La fuerza de fricción F_f = N × μ

Dónde;

g = Th aceleración debida a la gravedad ≈ 9.81 m/s²

∴ F_f  = 0.15 × 2 kg × 9.81 m/s² × cos (30°) ≈ 2.55 N

La longitud de la rampa que se mueve el trineo, l = h/(sin(θ))

∴ l = 4/(sin(30°)) = 8

La longitud de la rampa que se mueve el trineo, l = 8 m

El trabajo realizado sobre la fricción, W_f = F_f × l

W_f = 2.55 × 8 ≈ 20.4

El trabajo realizado sobre la fricción, W_f ≈ 20.4 Julios

El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. = m·g·h

∴ P.E. = 2 kg × 9.81 m/s² × 4 m ≈ 78.48 Joules

El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. ≈ 78.48 Joules

El trabajo total requerido para levantar el trineo 4 metros por la rampa, W_t = W_f + P.E.

W_t = 20.4 J + 78.84 J ≈ 99.24 J

Energía cinética, K.E. = 1/2 × m × v²

La energía cinética necesaria para mover el trineo por la rampa, K.E. se da de la siguiente manera;

K.E. = 1/2 × 2 kg × v² ≈ 99.24 J

∴ v² ≈ 99.24 J/(1/2 × 2 kg)

La velocidad con la que se debe lanzar el trineo para que ascienda 4 metros por la rampa, v ≈ 9.96 m/s

4 0
3 years ago
75000×(151÷100)².WHAT IS ITS VALUE ???​
Lady_Fox [76]

Answer:

171007  1/2

Explanation:

it's a mixed number

3 0
3 years ago
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