Suppose a proton ( = 1. 67×10^−27 kg) is confined to a box of width = 1. 00×10^−14 m (a typical nuclear radius).

Find the work done in moving a particle from p to q if the magnitude and direction

exis [7]

<span>Work=32unit</span><span>...hopes thats helps</span>

3 0

3 years ago

Where is the acceleration zero?

ladessa [460]

Answer:

When acceleration is zero (that is, a = dv/dt = 0), rate of change of velocity is zero. That is, acceleration is zero when the velocity of the object is constant. so probably D

Explanation:

D

3 0

3 years ago

Claire is traveling home when her mother calls and asks her to go to her grandma's house. She is 300 meters north of her house w

m_a_m_a [10]

Answer:

you divide distance traveled by the time it takes to travel that distance. average speed. If your speed changes from 10 km/h to 6 km/h, you have a(n) ... 400 km. Suppose that the average speed your dog can run is 3 m/s. ... she drives her scooter 7 kilometres north. She stops for lunch and then drives 5 kilometres south.

i rlly dunno lma.o

Explanation:

5 0

3 years ago

A parallel-plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative pla

MissTica

Answer:

The speed of proton when it emerges through the hole in the positive plate is $2.05\times 10^5\ m/s$ .

Explanation:

Given that,

A parallel-plate capacitor is held at a potential difference of 250 V.

A A proton is fired toward a small hole in the negative plate with a speed of, $u=3\times 10^5\ m/s$

We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :

$qV=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2\\\\1.6\times10^{-19}\times250=\dfrac{1}{2}mv^2-\frac{1}{2}\cdot1.67\times10^{-27}\cdot(3\times10^{5})^{2}\\\\\dfrac{1}{2}mv^2=3.515\cdot10^{-17}\\\\v=\sqrt{\dfrac{3.515\cdot10^{-17}\cdot2}{1.67\times10^{-27}}}\\\\v=2.05\times 10^5\ m/s$

So, the speed of proton when it emerges through the hole in the positive plate is $2.05\times 10^5\ m/s$ .

5 0

3 years ago

If A > B, under what condition is |A-BI=|A|- IB|? a. Vectors A and B are in opposite directions b. Vectors A and B are in the

kupik [55]

Answer:

b) Vectors A and B are in the same direction.

Explanation:

To understand this problem we will say that vector A has a magnitude of 5 units and vector B a magnitude of 3 units. In the subtraction of vectors the initial parts of vectors always bind together. And the vector resulting from the subtraction is traced from the end of the second vector (B) to the end of the first vector (A).

The length of the resultant vector will be 5 - 3 = 2

In the attached image, we analyze case a), b), and d)

For a)

As we can see in the attached image the resultant vector has a length of 8 units.

For d)

As we can see in the attached image the resultant vector has a length of 5.83 units.

For b)

The resultant vector has a length of 2 units.

Therefore the case given in b) is true

5 0

4 years ago