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3 years ago

Before a rifle is fired, the linear momentum of the bullet-rifle system is zero.

Physics

1 answer:

marusya05 [52]3 years ago

3 0

Hi there!

II. Linear momentum of the system is zero.

This is an example of a RECOIL collision. With the Law of Conservation of Momentum, momentum remains constant before and after the collision.

Thus, the total momentum would also be equivalent to zero after the collision.

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I have a problem in this questions?

photoshop1234 [79]

Answer:

8.46E+1

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = 39 C

Charge 2 (q₂) = –53 C

Force (F) of attraction = 26×10⁸ N

Electrical constant K) = 9×10⁹ Nm²/C²

Distance apart (r) =?

The distance between the two charges can be obtained as follow:

F = Kq₁q₂ / r²

26×10⁸ = 9×10⁹ × 39 × 53 / r²

26×10⁸ = 1.8603×10¹³ / r²

Cross multiply

26×10⁸ × r² = 1.8603×10¹³

Divide both side by 26×10⁸

r² = 1.8603×10¹³ / 26×10⁸

r² = 7155

Take the square root of both side

r = √7155

r = 84.6 m

r = 8.46E+1 m

7 0

2 years ago

What is force ? <br>question for spammers hehe..

love history [14]

Answer:

Force: Strength or energy as an attribute of physical action or movement.

6 0

3 years ago

Read 2 more answers

Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t

storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

$T=2\pi\sqrt{\frac{I}{K}}$

$I=mR^{2}$

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

$\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}$

As dimensions of new balance wheel are one-third of their original values

$R_{new}=\frac{R}{3}$

$\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega$

5 0

3 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.37 times a second. A tack is stuck in the tire a

zhannawk [14.2K]

Answer:

The tangential speed of the tack is 8.19 m/s.

Explanation:

The wheel rotates 3.37 times a second that means wheel complete 3.37 revolutions in a second. Therefore, the angular speed ω of the wheel is given as follows:

$\omega =3.37rev/s \times(\frac{2\pi rad}{1s} )\\\\=21.174rad/s$

Use the relation of angular speed with tangential speed to find the tangential speed of the tack.

The tangential speed v of the tack is given by following expression

v = ω r

Here, r is the distance to the tack from axis of rotation.

Substitute 21.174 rad/s for ω, and 0.387 m for r in the above equation to solve for v.

v = 21.174 × 0.387

v = 8.19m/s

Thus, The tangential speed of the tack is 8.19 m/s.

4 0

3 years ago

Read 2 more answers

A clarinet sounds as a closed pipe. if a clarinet sounds a note with a pitch of 375 hz, what are the frequencies of the lowest t

mote1985 [20]

So mathematical harmonics are based around a divergent set of fractions. Sigma(1/n)
with the 1st harmonic being... well 1, or 1 full wavelength.The second harmonic is exactly 1/2 the wavelength of the 1st with the third being 1/3 the wavelength. As Wavelengths go down, frequencies go up in a perfect ratio.

Second Harmonic has double the Frequency of the 1st or base note. Third Harmonic is triple and so on.

So the Harmonic set of 375 is.
1. 375
2. 375×2=750
3. 375×3= 1125
.
.
.
etc (: I hope this helps.

8 0

3 years ago