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tekilochka [14]

2 years ago

12

HELP ME OUT PLEASE!!!!!

2 answers:

masya89 [10]2 years ago

7 0

Answer:

B and D is the correct answer.

Explanation:

Nataliya [291]2 years ago

3 0

Answer:

B and D is the answer

Explanation:

I took the assignment <3

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Part 1: A rope has one end tied to a vertical support. You hold the other end so that the rope is horizontal. If you move the en

Anit [1.1K]

1) c. 2 m/s

Explanation:

The relationship between frequency, wavelength and speed of a wave is

$v=\lambda f$

where

v is the speed

$\lambda$ is the wavelength

f is the frequency

For the wave in this problem,

f = 4 Hz

$\lambda=0.5 m$

So, the speed is

$v=(0.5 m)(4 Hz)=2 m/s$

2) a. 2.8 m/s

The speed of the wave on a string is given

$v=\sqrt{\frac{T}{\mu}}$

where

T is the tension in the string

$\mu$ is the linear mass density

In this problem, we have:

$T=2 \cdot 4 N=8 N$ (final tension in the rope, which is twice the initial tension)

$\mu = 1 kg/m$ --> mass density of the rope

Substituting into the formula, we find

$v=\sqrt{\frac{8 N}{1 kg/m}}=2.8 m/s$

5 0

3 years ago

Which energy transformation occurs in an

Mnenie [13.5K]

A motor takes in electricity from a cell or power source and then uses it to move components.

Therefore the answer is (1) Electrical to Mechanical

5 0

3 years ago

Read 2 more answers

A 10 g bullet is fired into, and embeds itself in, a 2 kg block attached to a spring with a force constant of 19.6 n/m and whose

dmitriy555 [2]

Answer:

$x=0.478\ m$ is the compression in the spring

Explanation:

Given:

mass of the bullet, $m=10\ g=0.01\ kg$

mass of block, $M=2\ kg$

stiffness constant of the spring, $k=19.6\ N.m^{-1}$

initial velocity of the spring just before it hits the block, $u=300\ m.s^{-1}$

<u>Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:</u>

$m.u=(M+m).v$

$0.01\times 300=(2+0.01)\times v$

$v=1.4925\ m.s^{-1}$

Now this kinetic energy of the combined mass gets converted into potential energy of the spring.

$\rm Kinetic\ energy=Spring\ potential\ energy$

$\frac{1}{2} (M+m).v^2=\frac{1}{2} k.x^2$

$\frac{1}{2} \times (2+0.01)\times 1.4925^2=\frac{1}{2} \times 19.6\times x^2$

$x=0.478\ m$ is the compression in the spring

4 0

3 years ago

Read 2 more answers

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago

Convert the orbital period 27.32 days into units of seconds

maks197457 [2]

Answer:

$2360448seconds$

Explanation:

Dimensional analysis is attached.

3 0

2 years ago