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SpyIntel [72]
3 years ago
15

Suppose that Lisa walks her dog around the block for a little exercise. The block is 1 mile around. If she walks around the bloc

k once, what is her displacement? What is her distance?
Physics
1 answer:
natita [175]3 years ago
7 0
Displacement only measure how far between the starting and ending point. In this case, Lisa walks around the block as a circle so the starting point is the same as the ending point. Thus, displacement is 0mile.
On the other hand, distance measures exactly how far she walks. In this case, the distance is 1 mile, same as the perimeter of the block.
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"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two firs
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Answer:

A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if

the dispersion is great

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What is the mass of an object that is hanging 12.6 m above the surface of the earth and has a
satela [25.4K]

Answer:

22.05 Kg

Explanation:

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8750 J of heat are applied to a piece of aluminium causing a 56C increase in its
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3 years ago
Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87
marshall27 [118]

Answer:

a)  \lambda=0.935\ \textup{m}

b) f=36.19\approx 36\ \textup{Hz}

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

L= n\frac{\lambda}{2}

where,

\lambda = wavelength

on substituting the values, we get

1.87= 4\times \frac{\lambda}{2}

or

\lambda=0.935\ \textup{m}

b) Speed of the wave (v) in the string is given as:

v =f\lambda

also,

v=\sqrt\frac{T}{(\frac{m}{L})}

equating both the formula for 'v' we get,

f\lambda=\sqrt\frac{T}{(\frac{m}{L})}

on substituting the values, we get

f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}

or

f=\frac{33.84}{0.935}

or

f=36.19\approx 36\ \textup{Hz}

5 0
2 years ago
A child of mass m is standing at the edge of a carousel. Both the carousel and the child are initially stationary. The carousel
suter [353]

Answer:

the angular velocity of the carousel after the child has started running =

\frac{2F}{mR} \delta t

Explanation:

Given that

the mass of the child = m

The radius of the disc = R

moment of inertia I = \frac{1}{2} mR^2

change in time = \delta \ t

By using the torque around the inertia ; we have:

T = I×∝

where

R×F = I × ∝

R×F = \frac{1}{2} mR^2∝

F = \frac{1}{2} mR∝

∝ = \frac{2F}{mR}           ( expression for angular  angular acceleration)

The first equation of motion of rotating wheel can be expressed as :

\omega = \omega_0  + \alpha  \delta t

where ;

∝ = \frac{2F}{mR}    

Then;

\omega = 0+ \frac{2F}{mR} \delta t

\omega =  \frac{2F}{mR} \delta t

 

∴ the angular velocity of the carousel after the child has started running =

\frac{2F}{mR} \delta t

7 0
3 years ago
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