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CaHeK987 [17]
3 years ago
15

A proton and an electron traveling with the same velocity enter a uniform electric field. compared to the acceleration of the pr

oton, the acceleration of the electron is
Physics
1 answer:
kenny6666 [7]3 years ago
6 0
The acceleration of the electron is larger than the acceleration of the proton.

The reason for this is that the mass of the electron is smaller (about 1000 times smaller) than the mass of the proton. The two particles have same charge (e), so they experience the same force under the same electric field E:
F=eE 
However, according to Newton's second law, the force is the product between the mass particle, m, and its acceleration, a:
F=ma 
which can be rewritten as
a= \frac{F}{m}
we said that the force exerted on the two particles, F, is the same, while the mass of the electron is smaller: therefore, from the last formula we see that the acceleration of the electron will be larger than that of the proton.

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A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does
astra-53 [7]

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus =  14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

               = 24.25 × 10⁶  N/m²

Volume = \dfrac{4}{3} \pi r^3

         \dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}

Bulk modulus is equal to

B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }

now

B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }

B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }

(\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

7 0
3 years ago
An athlete stretches a spring an extra 40.0 cm beyond its initial length. how much energy has he transferred to the spring, if t
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The energy transferred to the spring is given by:
U= \frac{1}{2}kx^2
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x is the elongation of the spring with respect its initial length

Let's convert the data into the SI units:
k=52.9 N/cm = 5290 N/m
x=40.0 cm=0.4 m

so now we can use these data inside the equation ,to find the energy transferred to the spring:
U= \frac{1}{2}kx^2= \frac{1}{2}(5290 N/m)(0.4m)^2=423.2 J
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Answer:

I attached a PFD with the answer to your question and roughly 50 others involving acceleration and how to calculate it. This PDF gives you the answers to all the question whilst showing you an in-depth explanation on how they got the answer. Hopefully that helps

Explanation:

May I have brainliest please? :)

Download pdf
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