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vekshin1
3 years ago
15

Bullets spin when shot from a rifle or handgun. What causes this spinning?

Physics
1 answer:
tamaranim1 [39]3 years ago
7 0

Answer:

The spark from the primer ignites the gunpowder. Gas converted from the burning powder rapidly expands in the cartridge. The expanding gas forces the bullet out of the cartridge and down the barrel with great speed. The rifling in the barrel causes the bullet to spin as it travels out of the barrel.

Explanation:

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The potential difference across a variable resistor is 11V and the current flowing through it is 0.4A.
klemol [59]
The resistance is 27.5 ohms
5 0
4 years ago
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Your friend says that inertia is a force that keeps things in their place, either at rest or in motion. Do you and your discussi
Galina-37 [17]

Answer:

yes i agree

Explanation:

because law of inertia state that object remain at rest or in motion unless external force apply on it

8 0
3 years ago
If an atomic nucleus were the size of a dime how far away might one of its electrons be?
mihalych1998 [28]
The radius of a nucleus of hydrogen is approximately r_{n1}=1\cdot 10^{-15}m, while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: r_{e1}=5.3 \cdot 10^{-11}m

The radius of a dime is approximately r_{n2} = 9\cdot 10^{-3}m: if we assume that the radius of the nucleus is exactly this value, then we  can find how far is the electron by using the proportion
r_{n1}:r_{e1}=r_{n2}:r_{e2}
from which we find
r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.
6 0
3 years ago
2. At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a n
AleksandrR [38]

Answer:

No

Explanation:

Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):

A = 12 - 12t

B = 9t

Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t

d = \sqrt{A^2 + B^2} = \sqrt{(12 - 12t)^2 + (9t)^2}

For the ships to sight each other, distance must be 5 or smaller

d \leq 5

\sqrt{(12 - 12t)^2 + (9t)^2} \leq 5

(12 - 12t)^2 + (9t)^2 \leq 25

144t^2 - 288t + 144 + 81t^2 - 25 \leq 0

225t^2 - 288t + 119 \leq 0

(15t)^2 - (2*15*9.6)t + 9.6^2 + 26.84 \leq 0

(15t^2 - 9.6)^2 + 26.84 \leq 0

Since (15t^2 - 9.6)^2 \geq 0 then

(15t^2 - 9.6)^2 + 26.84 > 0

So our equation has no solution, the answer is no, the 2 ships never sight each other.

8 0
4 years ago
Work-Energy Theorem: A 4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects en
allsm [11]

The same braking force does work on these objects to slow them down. The work done is equal to their change in kinetic energy:

FΔx = 0.5mv²

F = force, Δx = distance traveled, m = mass, v = speed

Isolate Δx:

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Calculate Δx for each object.

Object 1: m = 4.0kg, v = 2.0m/s

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Object 2: m = 1.0kg, v = 4.0m/s

Δx = 0.5(1.0)(4.0)²/F = 8/F

The two objects travel the same distance before stopping.

4 0
3 years ago
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