All categories

3 years ago

A proton and an electron are placed in an electric field. Which undergoes the greater acceleration?

Physics

1 answer:

iren2701 [21]3 years ago

6 0

Newton's 2nd law:

Fnet = ma

Fnet is the net force acting on an object, m is the object's mass, and a is the acceleration.

The electric force on a charged object is given by

Fe = Eq

Fe is the electric force, E is the electric field at the point where the object is, and q is the object's charge.

We can assume, if the only force acting on the proton and electron is the electric force due to the electric field, that for both particles, Fnet = Fe

Fe = Eq

Eq = ma

a = Eq/m

We will also assume that the electric field acting on the proton and electron are the same. The proton and electron also have the same magnitude of charge (1.6×10⁻¹⁹C). What makes the difference in their acceleration is their masses. A quick Google search will provide the following values:

mass of proton = 1.67×10⁻²⁷kg

mass of electron = 9.11×10⁻³¹kg

The acceleration of an object is inversely proportional to its mass, so the electron will experience a greater acceleration than the proton.

You might be interested in

When finding net force, why must you know the directions of the forces acting on an object?

Mariulka [41]

<h3>One reason we should know what the directions of the forces acting on an object is so we can know if we have to add or subtract. same = add together, opposite = subtract from each other. Also if we don't pay attention to the direction the Net force will be wont be accurate. There will be factors that will upset the calculation.So we must know the direction of the two forces because we have to know if we are adding or subtracting and if the answer is accurate. </h3>

<em>I hope this helps!.</em>

8 0

3 years ago

An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the (1) maximum air temperature and (2)

weqwewe [10]

Answer:

(1) The maximum air temperature is 1383.002 K

(2) The rate of heat addition is 215.5 kW

Explanation:

T₁ = 17 + 273.15 = 290.15

$\frac{T_2}{T_1} =r_v^{k - 1} =18^{0.4} =3.17767$

T₂ = 290.15 × 3.17767 = 922.00139

$\frac{T_3}{T_2} =\frac{v_3}{v_2} = r_c = 1.5$

Therefore,

T₃ = T₂×1.5 = 922.00139 × 1.5 = 1383.002 K

The maximum air temperature = T₃ = 1383.002 K

(2)

$\frac{v_4}{v_3} =\frac{v_4}{v_2} \times \frac{v_2}{v_3} = \frac{v_1}{v_2} \times \frac{v_2}{v_3} = 18 \times \frac{1}{1.5} = 12$

$\frac{T_3}{T_4} =(\frac{v_4}{v_3} )^{k-1} = 12^{0.4} = 2.702$

Therefore;

$T_4 = \frac{1383.002}{2.702} =511.859 \ k$

$Q_1 = c_p(T_3-T_2)$

Q₁ = 1.005(1383.002 - 922.00139) = 463.306 kJ/jg

Heat rejected per kilogram is given by the following relation;

$c_v(T_4-T_1)$ = 0.718×(511.859 - 290.15) = 159.187 kJ/kg

The efficiency is given by the following relation;

$\eta = 1-\frac{\beta ^{k}-1}{\left (\beta -1 \right )r_{v}^{k-1}}$

Where:

β = Cut off ratio

Plugging in the values, we get;

$\eta = 1-\frac{1.5 ^{1.4}-1}{\left (1.5 -1 \right )18^{1.4-1}}= 0.5191$

Therefore;

$\eta = \frac{\sum Q}{Q_1}$

$\therefore 0.5191 = \frac{150}{Q_1}$

Heat supplied = $\frac{150}{0.5191} = 288.978 \ hp$

Therefore, heat supplied = 215491.064 W

Heat supplied ≈ 215.5 kW

The rate of heat addition = 215.5 kW.

7 0

3 years ago

True or False : Waves lose most of their energy when traveling across the ocean.

klemol [59]

Answer:

true

Explanation:

that's the answear

3 0

2 years ago

Read 2 more answers

Two vehicles, a car and a truck, leave an intersection at the same time. the car heads east at an average speed of 50 miles pe

olya-2409 [2.1K]

The car heads east at an average speed of 50 miles per hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per hour from the intersection point towards South.

The distance of car from the intersection point after t hours is $50t$ .