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ss7ja [257]
3 years ago
13

19. A 0.0340 kg bullet traveling at 120 m/s embeds itself in a 1.24 kg wooden block which

Physics
1 answer:
Blababa [14]3 years ago
3 0

Answer: x ≈ 36.3 cm

Explanation:

Conservation of momentum during the collision

0.0340(120) + 1.24(0) = (0.0340 + 1.24) v

v = 3.2025 m/s

The kinetic energy of the block/bullet mass will convert to spring potential

½kx² = ½mv²

x = √(mv²/k)

x = √(1.274(3.2025²) / 99.0)

x = 0.363293... ≈ 36.3 cm

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How many neutrons are in an atom of carbon with a mass number of 13
BabaBlast [244]

C is the answer to your question.

4 0
3 years ago
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0
3 years ago
What tension would you need to make a middle c (261.6 hz) fundamental mode on a 1 m string (for example, on a harp)? the linear
Allisa [31]
The frequency of middle C on a string is
f = 261.6 Hz.

The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
   = 0.002 kg/m

The length of the string is L = 1 m.

Let T =  the tension in the string (N).
The velocity of the standing wave is
v= \sqrt{ \frac{T}{\rho} }

In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
\sqrt{ \frac{T}{\rho} } =f \lambda = fL \\\\ \frac{T}{\rho} = (fL)^{2} \\\\ T = \rho (fL)^{2}

From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
   = 136.87 N

Answer: 136.9 N (nearest tenth)

4 0
3 years ago
At a distance of 8 m, the sound intensity of one speaker is 66 dB. If we were to place 3 speakers in a circle of radius 8 m, wha
ludmilkaskok [199]

Answer:

dβ = 70. 77 dβ

Explanation:

The intensity of sound in decibels is

         dβ = 10 log I/I₀

let's look for the intensity of this signal

         I / I₀ = 10 dβ/10

         I / I₀ = 3.981 10⁶

the threshold intensity of sound for humans is I₀ = 1 10⁻¹² W / m²

         I = 3.981 10 ⁶ 1 10⁻¹²

         I = 3,981 10⁻⁶ W / m²

It is indicated that 3 cornets are placed in the circle, for which total intensity is

        I_total - 3 I

        I_total = 3  3,981 10⁻⁶

        I_total = 11,943 10⁻⁶ W / m²

let's reduce to decibels

      dβ = 10 log (11,943 10⁻⁶/1 10⁻¹²)

      dβ = 10  7.077

      dβ = 70. 77 dβ

3 0
3 years ago
The fully loaded bus accelerates uniformly from rest to a speed of 14 m / s. The time taken to reach a speed of 14 m / s is 20 s
tino4ka555 [31]

Answer:

0.7m/s^2

Explanation:

acceleration=(final-initial velocity)/time

x=(14-0)/20

x=14/20

x=0.7

3 0
3 years ago
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