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yawa3891 [41]
1 year ago
14

A stream of doubly ionized particles (2 protons) moves at a velocity of 3.0 × 104 m/s perpendicularly to a magnetic field of 0.0

900 T. What is the magnitude of the force on the particles? (answer in scientific notation, including sign)Equations F = BqvF = BILcharge on a proton = 1.6 x 10–19C  charge on an electron = –1.6 x 10–19C
Physics
1 answer:
max2010maxim [7]1 year ago
3 0

The force on a charged particle on a magnetic field is given by:

F=Bqv\sin\theta

where B is the magnitude of the field, q is the charge of the particle, v is its velocity and θ is the angle between the field and the velocity of the particle.

In this case we have that:

• The magnitude of the field is 0.09 T.

,

• The charge of the particle is 3.2 x 10 –19C .

,

• The velocity is  3.0 × 10 4 m/s

,

• The angle between the field and the velocity is 90°, since they are perpendicular.

Plugging these we have:

\begin{gathered} F=(0.09)(3.2\times10^{-19})(3\times10^4)\sin90 \\ F=8.64\times10^{-16}\text{ N} \end{gathered}

Therefore, the force on the charge is:

F=8.64\times10^{-16}\text{N}

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grin007 [14]
KF is the covalent compound
3 0
3 years ago
Sobre un gas contenido en un cilindro provisto de un pistón se realiza un trabajo de 7000 Joules, mediante un proceso isotérmico
natita [175]

Answer:

En un proceso isotérmico, es decir, la temperatura no cambia, el trabajo puede escribirse como:

W = n*R*T*Ln(P1/P2)

Donde P1 es la presión inicial y P2 la presión final.

Donde las cantidades:

n =  número de moles

R = constante de los gases ideales

T = temperatura no cambian.

Y sabemos que la ecuación de la energía interna es:

U = C*n*R*T

Donde C es otra constante que depende del gas.

De aca, podemos concluir que ninguna de estas variables cambia en nuestro proceso, por lo que la variación de la energía interna es cero.

U2 - U1 = 0

b) Para el calor cedido o absorbido, la formula básica es:

ΔQ = C*(T2 - T1)

Donde ΔQ es el calor absorbido o cedido por el gas, C es una constante que depende del gas, T2 es la temperatura final del gas y T1 es la temperatura inicial del gas.

Como la temperatura no cambia en el proceso, entonces:

T2 = T1

ΔQ = C*(T2 - T1) = C*0 = 0

No hay calor absorbido ni cedido.

c) Podemos concluir que en un proceso isotérmico la energía interna no cambia, y no hay un intercambio de calor.

8 0
3 years ago
Which statements best describes why science changes?
MrRa [10]

Explanation:

Science changes in so many ways. In my opinion, science changes due to Climatic change.

Hope it helps.

7 0
2 years ago
Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.
Sliva [168]

To solve the problem it is necessary to take into account the concepts related to energy efficiency in the engines, the work done, the heat input in the systems, the exchange and loss of heat in the soupy the radius between the work done the lost heat ( efficiency).

By definition the efficiency of the heat engine is

\epsilon = 1- \frac{T_c}{T_h}

Where,

T_c = Temperature at the room

T_h  =Temperature of the soup

The work done is defined as,

dW = \epsilon(dQ_h)

Where Q_h represents the input heat and at the same time is defined as

dQ_h =c_v (dT_h)

Where,

c_V =Specific Heat

The change at the work would be defined then as

dW = \epsilon(dQ_h)

dW = \epsilon c_v (dT_h)

dW = (1-\frac{T_c}{T_h})c_v (dT_h)

W = \int dW = \int (1-\frac{T_c}{T_h})c_v (dT_h)

W = c_v (T_h-T_c)-c_v T_c ln(\frac{T_h}{T_c})

On the other hand we have that the heat lost by the soup is equal to

dQ_h =c_v (dT_h)

Q_h =c_v (T_h-T_c)

The ratio between both would be,

\frac{W}{Q_h} = \frac{c_v (T_h-T_c)-c_v T_c ln(\frac{T_h}{T_c})}{c_v (T_h-T_c)}

\frac{W}{Q_h} = \frac{1+ln(\frac{T_h}{T_c})}{1-\frac{T_h}{T_c}}

Replacing with our values we have,

\frac{W}{Q_h} = 1+\frac{ln(\frac{340K}{300K})}{1-\frac{340K}{300K}}

\frac{W}{Q_h} = 0.0613

Therefore the fraction of heat lost by the soup that can be turned into useable work by the engine is 0.0613.

5 0
3 years ago
A 91.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.73 r
KIM [24]

Answer:

\omega_f = 1.08 rad/s

Explanation:

As we know that there is no external torque on the system of platform, banana and the monkey

so the angular momentum of the system will remains conserved

so here we can say

L_i = L_f

I_o\omega = (I_o + m_1r_1^2 + m_2r_2^2)\omega_f

here we know that

I_o = \frac{1}{2}mR^2

I_o = \frac{1}{2}(91.1)(1.61)^2 = 118.1 kg m^2

m_1 = 9.41 kg

r_1 = \frac{4}{5}(1.61) = 1.29 m

m_2 = 21.1 kg

Now from above equation

118.1(1.73) = (118.1 + 9.41(1.29^2) + 21.1(1.61^2))\omega_f

118.1(1.73) = 188.45\omega_f

\omega_f = 1.08 rad/s

8 0
3 years ago
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