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3 years ago

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A force of 2 kN is applied to an object to make it move 3.6 m in the direction of the force. Select the correct value of work do

ne on the object.
Answers to choose from:
1.8J
7200J
7.2J
555.6J

1 answer:

vaieri [72.5K]3 years ago

4 0

Answer:

W= F × d

W= 2kn × 3.6

W= 7.2 J

Work is measured in Joules!

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A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

3 years ago

An arrow is launched from a bow with an initial horizontal velocity of 40

Ivan

Answer:

V = (Vx^2 + Vy^2)^1/2 = (40^2 + 62^2)^1/2

V = 73.8 m/s

tan theta = Vy / Vx = 62/40 = 1.55

theta = 57.2 deg

4 0

2 years ago

An electron is moving at 2.02.0 ×× 105m/s105 m/s in the positive y direction. The magnetic force on the electron is 3.03.0 ×× 10

inn [45]

Answer:

The magnitude and direction of the magnetic field is 93.63 T in negative x direction.

Explanation:

Given;

speed of the electron in positive y direction, v = 2.0 x 10⁵ m/s

magnetic force on the electron, F in negative z direction = 3.0 x 10⁻¹² N

The magnitude of the magnetic force is given by;

F = Qv x B

B = F / Qv

$B = \frac{3*10^{-12}}{ (1.602*10^{-19})(2*10^5)}\\\\B = 93.63 \ T$

The direction of the magnetic field is is as;

Based on the direction of magnetic force (negative z direction), the charge will be directed into negative y-direction because electron is negatively charged. Thus, the direction of the magnetic field will be in the negative x-direction

$F_{(-z)}= Q_{(-y)}V* B_{(-x)}$

Therefore, the magnitude and direction of the magnetic field is 93.63 T in negative x direction.

6 0

3 years ago

A laser beam is incident at an angle of 33.0° to the vertical onto a solution of cornsyrup in water.(a) If the beam is refracted

IrinaK [193]

Answer:

1.29649

488.08706 nm

$6.14644\times 10^{14}\ Hz$

231715700.28346 m/s

Explanation:

n denotes refractive index

1 denotes air

2 denotes solution

$\lambda_0$ = 632.8 nm

From Snell's law we have the relation

$n_1sin\theta_1=n_2sin\theta_2\\\Rightarrow n_2=\dfrac{n_1sin\theta_1}{sin\theta_2}\\\Rightarrow n_2=\dfrac{1\times sin33}{sin24.84}\\\Rightarrow n_2=1.29649$

Refractive index of the solution is 1.29649

Wavelength is given by

$\lambda=\dfrac{\lambda_0}{n_2}\\\Rightarrow \lambda=\dfrac{632.8}{1.29649}\\\Rightarrow \lambda=488.08706\ nm$

The wavelength of the solution is 488.08706 nm

Frequency is given by

$f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{488.08706\times 10^{-9}}\\\Rightarrow f=6.14644\times 10^{14}\ Hz$

The frequency is $6.14644\times 10^{14}\ Hz$

$v=\dfrac{c}{n_2}\\\Rightarrow v=\dfrac{3\times 10^8}{1.29469}\\\Rightarrow v=231715700.28346\ m/s$

The speed in the solution is 231715700.28346 m/s

8 0

3 years ago

The surface of the sun has a temperature of about 5,800 Kelvin. What is the average kinetic energy of particles on the surface o

LenaWriter [7]

Answer:

273.15

Explanation:

So that's three over two times 1.38 times ten to the minus twenty-three joules per Kelvin, times 5500 degrees Celsius, the surface of the sun converted into Kelvin by adding 273.15. This works out to 1.20 times ten to the minus nineteen joules. So that's the average kinetic energy of hydrogen atoms.

5 0

3 years ago