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OLEGan [10]
3 years ago
13

g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before

interacting. From this information, find the time interval required for the strong interaction to occur.
Physics
1 answer:
AnnyKZ [126]3 years ago
5 0

Answer:

Time, t=1.07\times 10^{-23}\ s

Explanation:

Given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2\times 10^{-15}\ m before interacting.

Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,

t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s

So, the time interval is 1.07\times 10^{-23}\ s

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A man is standing on the edge of a 20.0 m high cliff. He throws a rock horizontally with an initial velocity of 10.0 m/s.
kherson [118]

Answer:

<em>a. The rock takes 2.02 seconds to hit the ground</em>

<em>b. The rock lands at 20,2 m from the base of the cliff</em>

Explanation:

Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.

The time taken by the object to hit the ground is calculated by:

\displaystyle t=\sqrt{\frac{2h}{g}}

The range is defined as the maximum horizontal distance traveled by the object and it can be calculated as follows:

\displaystyle d=v.t

The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.

a,

The time taken by the rock to reach the ground is:

\displaystyle t=\sqrt{\frac{2*20}{9.8}}

\displaystyle t=\sqrt{4.0816}

t = 2.02 s

The rock takes 2.02 seconds to hit the ground

b.

The range is calculated now:

\displaystyle d=10\cdot 2.02

d = 20.2 m

The rock lands at 20,2 m from the base of the cliff

5 0
3 years ago
Which list of properties of alternation current is the most likely reason it was chosen over direct current to provide electrici
Paraphin [41]

Answer:

- Direct current is a current in which electrons flow in one direction only

- Alternating current is a current in which the direction of the electron flow reverses periodically - so, half a cycle forward, half a cycle backward

There are several advantages of using alternating currents for the transmission of electricity across a country, over large distances. The main advantages are:

- The voltage of alternating currents can be easily increased/decreased by using transformers. For instance, a transformer is used at the beginning of the transmission line to increase the voltage (electricity is transmitted at high voltage in order to reduce dissipated power), and then another transformer is used before the electricity enters the houses, in order to decrease the voltage. Transformers only work with alternating currents.

- It is easy to interrupt the flow of an alternating current, because its value naturally becomes zero every half a cycle, so this is useful in case the current must be interrupted.

8 0
3 years ago
If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle
ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0
2 years ago
Which type of wave does the illustration depict?
Naddika [18.5K]
A transverse wave. A wave is a disturbance that transmits energy from one place to another by the particles of the medium.
4 0
2 years ago
Read 2 more answers
A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 43 kg barbell straight up fro
Tasya [4]

Answer:

A. 231.77 J

B. 5330.71 J

C. 46 donuts

Explanation:

A. To lift the barbell once, she will have to extend it the full length of her arm. The work done will then be:

W = F * d

Where the force is the weight of the barbell.

F = m * g

Hence, the work done in lifting the barbell is:

W = m * g * d

W = 43 * 9.8 * 0.55

W = 231.77 J

B. If she does 23 repetitions, the total energy she expend will be equal to the Potential energy when the barbell is lifted multiplied by 23:

E = 23 * m * g * d

E = 23 * 231.77

E = 5330.71 J

C. 1 Joule = 4.184 calories

5330.71 Joules = 5330.71 * 4.184 = 22303.69

If 1 donut contains 490 calories, the number of donuts she will need will be:

N = 22303.69/490 = 45.5 donuts or 46 donuts

5 0
3 years ago
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