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3 years ago

A skier starting from rest skis straight down a slope 50 meters long in 5.0 seconds. What is the magnitude of the acceleration.

1 answer:

6 0

His average speed all the way is

   (50 meters) / (5 sec) = 10 m/s .

But, if the acceleration is uniform all the way down, then

Average speed = (1/2) (start speed + end speed)

Start speed = 0.
So
   Average speed = (1/2) (0 + end speed)

   = (1/2) of end speed .

   10 m/s = (1/2) of end speed

   End speed (at the bottom) = 20 m/s .

Magnitude of acceleration = (change in speed) / (time for the change)

      = (20 m/s) / (5 sec)

   =    4 m/s² .

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Find the first three harmonics of a string of linear mass density 2. 00 g/m and length 0. 600 m when the tension in it is 50. 0

gavmur [86]

The first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

<h3>Velocity of the wave</h3>

The velocity of the wave is calculated as follows;

v = √T/μ

where;

T is tension
μ is mass per unit length = 2 g/m = 0.002 kg/m

v = √(50/0.002)

v = 158.1 m/s

<h3>First harmonic or fundamental frequency of the wave</h3>

f₀ = v/λ

where;

λ is the wavelength = 2L

f₀ = v/2L

f₀ = 158.1/(2 x 0.6)

f₀ = 131.8 Hz

<h3>Second harmonic of the wave</h3>

f₁ = 2f₀

f₁ = 2(131.8 Hz)

f₁ = 263.6 Hz

<h3>Third harmonic of the wave</h3>

f₂ = 3f₀

f₂ = 3(131.8 Hz)

f₂ = 395.4 Hz

Thus, the first three harmonics of the string are 131.8 Hz, 263.6 Hz and 395.4 Hz.

Learn more about harmonics here: brainly.com/question/4290297

#SPJ1

6 0

1 year ago

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

marusya05 [52]

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.

3 0

2 years ago

In a certain process, the energy change of the system is 250 \rm kJ. The process involves 480 \rm kJ of work done by the system.

alisha [4.7K]

Answer:

$\Delta Q=-230kJ$

Explanation:

Using the first law of thermodynamics:

$\Delta U=\Delta Q-W$

Where $\Delta U$ is the change in the internal energy of the system, in this case $\Delta U=250kJ$ , $\Delta Q$ is the heat tranferred, and $W$ is the work, $W=-480kJ$ with a negative sign since the work is done by the system.

From the previous equation we solve for heat, because it is the unknown variable in this problem

$\Delta Q=\Delta U +W$

And replacing the known values:

$\Delta Q=250kJ +(-480kJ)$

$\Delta Q=250kJ -480kJ$

$\Delta Q=-230kJ$

The negative sign shows us that the heat is tranferred from the system into the surroundings.

3 0

3 years ago

A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m

erastovalidia [21]

Answer:

a) W = 46.8 J and b) v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

W = ΔK = $k_{f}$ -K₀

a) work is the scalar product of force by distance

W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

W = F d cos θ

W = 39.0 1.20 cos 0

W = 46.8 J

b) zero initial kinetic language because the package is stopped

W - $W_{fr}$ = $k_{f}$ -K₀

W - fr d= ½ m v² - 0

W - μ N d = ½ m v

on the horizontal surface using Newton's second law

N-W = 0

N = W = mg

W - μ mg d = ½ m v

v² = (W -μ mg d) 2/m

v = √(W -μ mg d) 2/m

v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3 ]

v = √(31.63 2/4.3)

v = 3.84 m/s

8 0

3 years ago

En el MCU, la aceleración es: a) constante b) Varía en módulo, direccion, y sentido c) constante solo en el módulo d) constante

JulijaS [17]

If you put it into English in the comments I would be more then happy to help you! Thank you!

7 0

3 years ago