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kati45 [8]
3 years ago
9

sledge (including load) weighs 5000 N. It is pulled on level snow by a dog team exerting ahorizontal force on it. The coefficient

of kinetic friction between sledge and snow is 0.05. Howmuch work is done by the dog team pulling the sledge 1000 m at constant speed?
Physics
1 answer:
lara31 [8.8K]3 years ago
6 0

Answer:

2.5 x 10^{5} J

Explanation:

weight = 5,000 N

coefficient of friction = 0.05

distance = 1000 m

how much work is done by the dogs pulling the sledge

work done = force x coefficient of friction x distance

work done = 5000 x 0.05 x 1000 = 2.5 x 10^{5} J

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Answer:

Speed = 0.296m/2

Period = 0.203 s

Explanation:

If by 'long' you mean the wavelength of the waves, then the wavelength \lambda=0.06m.

The frequency f of the waves is 14.8 waves every 3 seconds or

f=14.8/3 =4.33Hz.

Now the relationship between wavelength \lambda, frequency f and speed v of the waves is:

v=\lambda f

We put in the values \lambda=0.06m and f=4.933Hz and get:

\boxed{v=0.06*4.922=0.296m/s}

Now the period T is just the inverse of the frequency, or

T=\frac{1}{f}

\boxed{T=\frac{1}{4.933}=0.203\:seconds }

4 0
3 years ago
Approximately what is the smallest detail observable with a microscope that uses ultraviolet light of frequency 1. 72 x 1015 hz?
postnew [5]

The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light  , is calculated as follows: 3×10^{8} / 1.72×10^{15} or approximately 1.74×10^{-7}m.

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            λ = wavelength of light

Given data = f = 1.72×10^{15}Hz

Therefore, λ = 3×10^{8} / 1.72×10^{15}

                  λ = 1.74×10^{-7}m

The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light  , is calculated as follows: 3×10^{8} / 1.72×10^{15} or approximately 1.74×10^{-7}m.

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2 years ago
what is the force that opposes the movement of two surfaces that are in contact and are sliding over each other? and it is not f
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Answer:

Friction

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7 0
2 years ago
Read 2 more answers
A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is droppe
pentagon [3]

Answer:

(A) l = 0.39 m      

(B)  l =0.38 m  

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

         potential energy of the object = mg(1.05 + l)  

         potential energy of the spring = 0.5 x k x l^{2}  (k= spring constant)

 therefore we now have

              mg(1.05 + l)  = 0.5 x k x l^{2}

              1.6 x 9.8 x (1.05 + l)  = 0.5 x 300 x l^{2}

               15.68 (1.05 + l) = 150 x l^{2}

                   16.5 + 15.68l = 150l^{2}

l = 0.39 m        

(B)   with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}        

     1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l)  = 0.5 x 300 x l^{2}

         (16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}        

          16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

            15.71 + 14.93l = 150^{2}

            l =0.38 m  

(C)   where g = 1.63 m/s^{2} and neglecting air resistance

      the equation mg(1.05 + l)  = 0.5 x k x l^{2} now becomes

        1.6 x 1.63 x (1.05 + l)  = 0.5 x 300 x l^{2}

        2.608 (1.05 +l) = 0.5 x 300 x l^{2}

        2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

6 0
3 years ago
Allison wants to determine the density of a bouncing ball. which metric measurements must she use?
lubasha [3.4K]
Density depends on mass and volume so option D is correct answer. Hope this helps!
3 0
3 years ago
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