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4 years ago

13

Marcus used a toaster oven in the morning.He notices that when he plug it in and turn it on the coils inside begin to glow red w

hat transformation are taking place

1 answer:

Rasek [7]4 years ago

7 0

Answer:Conversion of electric energy to Heat energy

Explanation:Energy is a quantitative energy measured in JOULES or KILOJOULES which must be transferred to a material for a job to be done. It has also been described as the capacity to do work.

In electric toasters the ELECTRIC ENERGY FROM THE SOURCE IS TRANSFERRED INTO THE TOASTER TO BE CONVERTED TO HEAT ENERGY NEEDED TO TOAST FOODS. Other electrical appliances which converts electric energy to Heat energy includes ELECTRIC BOILERS, ELECTRIC COOKERS etc.

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Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton

solmaris [256]

Answer:

a

$\lambda = 3.68 *10^{-36} \ m$

b

$\lambda_p = 1.28*10^{-14} \ m$

Explanation:

From the question we are told that

The mass of the person is $m = 180 \ kg$

The speed of the person is $v = 1 \ m/s$

The energy of the proton is $E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J$

Generally the de Broglie wavelength is mathematically represented as

$\lambda = \frac{h}{m * v }$

Here h is the Planck constant with the value

$h = 6.62607015 * 10^{-34} J \cdot s$

So

$\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }$

=> $\lambda = 3.68 *10^{-36} \ m$

Generally the energy of the proton is mathematically represented as

$E_p = \frac{1}{2} * m_p * v^2_p$

Here $m_p$ is the mass of proton with value $m_p = 1.67 *10^{-27} \ kg$

=> $8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2$

=> $v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }$

=> $v = 3.09529 *10^{7} \ m/s$

So

$\lambda_p = \frac{h}{m_p * v_p }$

so $\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }$

=> $\lambda_p = 1.28*10^{-14} \ m$

5 0

3 years ago

Can the magnitude of a vector ever (a) be equal to one of its components, or (b) be less than one of its components? 9. Can a pa

Ber [7]

Answer:

a) the other components are zero, in the direction of one of the coordinate axes

b) the magnitude is less than the value of one of its components, it must occur when the vector is in some arbitrary direction

9) constant velocity the acceleration must necessarily be zero,

constant speed can be accelerated since it may be changing the direction of the velocity vector

Explanation:

Vectors are quantities that have modulo (scalar) direction and sense.

a) If in a vector its magnitude is equal to one d its components implies that the other components are zero, therefore the vector must be in the direction of one of the coordinate axes

b) if the magnitude is less than the value of one of its components, it must occur when the vector is in some arbitrary direction, other than the direction of the axes, that is

R² = x² + y²

where R is the magnitude of the vector e x, and are the components

9) When a particle has a constant velocity, the acceleration must necessarily be zero,

v = vo + a t

The bold letters indicate vectors If a = 0 implies that v = vo

If a particle has constant speed it can be accelerated since it may be changing the direction of the velocity vector, this type of acceleration has the name of centripetal acceleration

6 0

3 years ago

When force is perpendicular to the direction of motion, no work is done.

Vlada [557]

No work is done by THAT force.

8 0

3 years ago

John and mary are skating at an ice rink. john skates at a constant speed of 6.7 m/s, with respect to the ice surface, directly

iragen [17]

60.3Â° from due south and 5.89 m/s For this problem, first calculate a translation that will put John's destination directly on the origin and apply that translation to Mary's destination. Then the vector from the origin to Mary's new destination will be the relative vector of Mary as compared to John. So John is traveling due south at 6.7 m/s. After 1 second, he will be at coordinates (0,-6.7). The translation will be (0,6.7) Mary is traveling 28Â° West of due south. So her location after 1 second will be (-sin(28)*10.9, -cos(28)*10.9) = (-5.117240034, -9.624128762) After translating that coordinate up by 6.7, you get (-5.117240034, -2.924128762) The tangent of the angle will be 2.924128762/5.117240034 = 0.57142693 The arc tangent is atan(0.57142693) = 29.74481039Â° Subtract that value from 90 since you want the complement of the angle which is now 60.25518961Â° So Mary is traveling 60.3Â° relative to due south as seen from John's point of view. The magnitude of her relative speed is sqrt(-5.117240034^2 + -2.924128762^2) = 5.893783 m/s Rounding the results to 3 significant digits results in 60.3Â° and 5.89 m/s

8 0

3 years ago

A freight car of mass M contains a mass of sand m. At t = 0 a constant horizontal force F is applied in the direction of rolling

Veronika [31]

Answer:

Amount of linear movement

Explanation:

Our system is defined by the rate of change in mass that

leaves the car $\Delta m_ {s}$ , this happens during a time interval

$[t, t + \Delta t]$ , in addition to freight car and sand at time t.

In this way we need to define the two states:

State 1,

consider $t, m_ {c} (t) + \Delta m_ {s}$ and V.

State 2,

consider $t + \Delta t, m_ {c} (t), V + V \Delta V$

In this state is the mass of sand output, which

is composed of

$\Delta m_{s}, V + \Delta V$

In this way we define the Linear movement in x, like this:

$p_ {x} (t) = (\Delta m_ {s} + m_ {c} (t)) v$

$p_ {x} (t+\Delta t) = (\Delta m_ {s} + m_ {c} (t)) (v + \Delta v)$

$m_ {c} (t) = m_ {c, 0} - bt = m_ {c} + m_ {s} -bt$

In this way we proceed to obtain the Force

$F =\lim_{\Delta t \rightarrow 0} \frac {p_x (t + \Delta t) -p_ {x} (t)} {\Delta t}$

$F = lim_{\Delta t \rightarrow 0} m_ {c} (t) \frac {\Delta v} {\Delta t} + lim_{\Delta t \rightarrow 0} m_ {s} (t) \frac {\Delta v} {\Delta T}$

Since the mass of the second term becomes 0, the same term is eliminated, thus,

$F = m_ {c} (t) \frac {dv} {dt}$

$\int\limit ^ {v (t)} _ {v = 0} dv = \int\limit^t_0 \frac{Fdt} {m_ {c} + m_ {s} -bt}$

$V (t) = - \frac {F} {b} ln (\frac {m_c + m_s-bt} {m_c + m_s})$

3 0

3 years ago