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Ksju [112]
3 years ago
10

A penguin slides at a constant velocity of 3.57 m/s down an icy incline. The incline slopes above the horizontal at an angle of

9.85°. At the bottom of the incline, the penguin slides onto a horizontal patch of ice. The coefficient of kinetic friction between the penguin and the ice is the same for the incline as for the horizontal patch. How much time is required for the penguin to slide to a halt after entering the horizontal patch of ice?
Physics
1 answer:
igomit [66]3 years ago
8 0

Answer:t=0.81 s

Explanation:

Given

Penguin slides down with constant velocity of 3.57 m/s

as the Penguin Slides with constant velocity therefore F_{net} is zero on Penguin

F_{net}=mg\sin \theta -f_r

f_r=friction Force

f_r=\mu mg

\mu =coefficient of Kinetic friction

mg\sin \theta =\mu mg

\tan \theta =\mu

\mu =\tan 9.85=0.45

after reaching on floor final velocity of penguin will be zero after time t

thus

v=u+at

here a=\mu g

a=0.45\times 9.8=4.41  (deceleration)

0=3.57-4.41\times t

t=\frac{3.57}{4.41}=0.809

t=0.81 s

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A mass hanging from a spring undergoes vertical simple harmonic motion.1) Where in the motion is the velocity equal to zero?At t
tamaranim1 [39]

Answer:

At the highest point and at the lowest point the velocity of the mass hung on a spring = 0

Explanation:

Simple Harmonic Motion ( S.H.M) : Simple harmonic motion can be defined as a type of motion were a body vibrates or moves to and fro along a straight line under the influence of a force, so that the acceleration of the body towards a fixed point (equilibrium position) is proportional to its distance or displacement from that point. Examples of  bodies undergoing simple harmonic motion are

<em>⇒ The motion of a mass hung on a spring.</em>

<em>⇒ The motion of a simple pendulum</em>

<em>⇒ The motion of a loaded test - tube in a liquid.</em>

Motion of a mass hung on a spring:When a mass is hung to one end spring and other end is firmly clamped to a rigid support.(i)When the mass is in motion, (ii)it pulled down to its lowest point, passes through it equilibrium position (iii) goes to its highest point.

<em>(1) At the lowest and the highest point during the motion of a mass hung a spring, the velocity = 0</em>

<em>(2) At the equilibrium point or unstretched position the velocity is maximum</em>

<em></em>

<em></em>

4 0
3 years ago
Suppose a plot of 1/λ vs f has a slope of 0.728, what is the wave velocity? Give only the numerical answer with 4 decimal place
Charra [1.4K]

Answer:

The wave velocity is 1.3736

Explanation:

A plot of inverse of wavelength (λ) versus frequency (f), can be analyzed as follows;

The vertical axis will be 1/λ and the horizontal axis will be f

slope = (Δ1/λ)/(Δf)

        = 1/(fλ)

∴ 1/(fλ) = 0.728

From wave equation, V = fλ

where;

V is the wave velocity

f is the frequency of the wave

λ is the wavelength

1/(fλ) = 0.728

1/(V) = 0.728

V = (0.728)⁻¹

V = 1.3736

Therefore, the wave velocity is 1.3736

3 0
3 years ago
Read 2 more answers
two skateboarders of mass 50 kg and 60 kg push each other with force 70N.what is the acceleration of each skaters
Free_Kalibri [48]

Answer:

0 m/s²

Explanation:

Since each skater pushes the other with a force of 70 N, according to Newton's third law, there is an equal reaction and thus the other pushes back with a force of 70 N in the other direction, so we have forces of +70N and -70 N respectively. So, the net force on each skateboarder is F = + 70 N + (-70 N) = + 70 N - 70 N = 0 N.

Since force, F = ma where a = acceleration and m = mass,

a = F/m.

So, since for each skater, F = 0N,

a = 0 N/m

= 0 m/s²

So, the acceleration of each skater is 0 m/s²

5 0
3 years ago
Two rams run toward each other. One ram has a mass of 49 kg and runs west
olga nikolaevna [1]

Answer: (d)

Explanation:

Given

Mass of the first ram m_1=49\ kg

The velocity of this ram is v_1=-7\ m/s

Mass of the second ram m_2=52\ kg

The velocity of this ram v_2=9\ m/s

They combined after the collision

Conserving the momentum

\Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow 49\times (-7)+52\times (9)=(52+49)v\\\Rightarrow v=\dfrac{125}{101}\ m/s \quad[\text{east}]

Momentum after the collision will be

\Rightarrow 101\times \dfrac{125}{101}=125\ kg-m/s\ \text{East}

Therefore, option (d) is correct

4 0
2 years ago
How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenl
spin [16.1K]

Answer:

L=0.654 m

Explanation:

<u>Concepts and Principles  </u>

1- The speed of sound in air is expressed as a function of the temperature of air as follows:  

 v=(331 m/s)√(1+T_C/273°C)                        (1)

where 331 m/s is the speed of sound in air at temperature 0°C and Tc is the temperature of air in Celsius.  

<u>Standing Wave Patterns in Pipes:  </u>

A pipe open at both ends can have standing wave patterns with resonant frequencies:  

f=v/λ=nv/2L                     n=1,2,3.........

where v is the speed of sound in air.  

<u>Given Data </u>

f_1 (fundamental frequency of the flute) = 262 Hz

T (temperature of the air) = 20°C  

The flute is open at both ends.  

<u>Required Data </u>

We are asked to determine the length of the tube.  

<u>Solution</u><u>  </u>

The speed of sound in air at temperature T = 20°C is found from Equation (1):

 v=(331 m/s)√(1+T_C/273°C)  

 =342.91 m/s

The fundamental frequency of the flute is found by substituting n = 1 into Equation (2):  

f=v/2L

Solve for L:  

L=v/2f_1

L=0.654 m

7 0
3 years ago
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