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2 years ago

4. An applied force of 6.2 N acts on a 2.1-kg object, pushing it horizontally across a surface

1 answer:

5 0

The frictional force on the body is 0.25 N the net force on the body is 5.95 N while the acceleration produced is 2.83 m/s^2

<h3>What is the net force?</h3>

The net force is used to describe the resultant force that acts on the object.

The frictional force that acts on the object s obtained from;

Ff = 0.15 * 2.1-kg * 9.8 m/s^2

Ff = 0.25 N

The net force that acts on the body = F - Ff where F is the applied the force

= 6.2 N - 0.25 N

Net force = 5.95 N

Now;

Net force = ma

a= Net force /m

a = 5.95 N/2.1-kg

a = 2.83 m/s^2

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A sock with a mass of 0.03 kg is stuck to the inside of a clothes dryer spins

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Answer:

15.71 m/s

Explanation:

We are given;

Time; t = 0.2 s

Radius; r = 0.5 m

The circumference will give us the distance covered.

Formula for circumference is 2πr

Thus; Distance = 2πr = 2 × π × 0.5 = π

Linear speed = distance/time = π/0.2 = 15.71 m/s

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4 years ago

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The phosphorus cycle is the biogeochemical cycle that describes the movement of phosphorus through the lithosphere,

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3 years ago

Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .

aliya0001 [1]

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s

Let $S_{0}$ be the speed of the river's current given as 1.00 m/s.

Note that this speed is the magnitude of the velocity which is a vector quantity.

The direction of the swimmer is upstream.

Hence the resultant velocity is given as, $S_{R}$ = S — S 0 $S_{0}$

$S_{R}$ = 1.25 — 1

$S_{R}$ = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

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3 years ago

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

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4 years ago