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2 years ago

an athlete sprints from 150 m south of the finish line to 65 m south of the finish line in 5.0s what is his average velocity

Physics

2 answers:

MatroZZZ [7]2 years ago

6 0

Total displacement=150-65m=85m
Time=5s

$\\ \rm\longrightarrow Avg\:Velocity=\dfrac{Total\:Displacement}{Total\:Time}$

$\\ \rm\longrightarrow Avg\:Velocity=\dfrac{85}{5}$

$\\ \rm\longrightarrow Avg\:Velocity= 17m/s$

ladessa [460]2 years ago

5 0

Answer:

$displacement = 150 - 65 = 85m \: \\ time = 5sec \\ velocity = displacement/ time= \frac{85}{5} = 17 \frac{m}{s} \: south\\ thank \: you$

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If the sun in 148 million kilometers from the earth, how many minutes will it take the light from the sun to reach the earth?

Natasha2012 [34]

Answer:

about 8 mins and 20 secs

Explanation:

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3 years ago

To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a

irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

= 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

W = F x S joules

Substituting the values in the above equation

W = 1.5 x 10⁶ x 355 · sin θ

= 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J

6 0

3 years ago

a chuck wagon with an initial velocity of 4 m/s and a mass of 35 kg gets a push with 350 joules of force. what is the wagon's fi

Kitty [74]

Answer:

the final velocity of the wagon is 6 m/s.

Explanation:

Given;

initial velocity of the wagon, u = 4 m/s

mass of the wagon, m = 35 kg

energy applied to the wagon, E = 350 J

The final velocity of the wagon is calculated as;

E = ¹/₂m(v² - u²)

$m(v^2-u^2) = 2E\\\\v^2-u^2 = \frac{2E}{m} \\\\v^2 = \frac{2E}{m} + u^2\\\\v = \sqrt{\frac{2E}{m} + u^2} \\\\v = \sqrt{\frac{2(350)}{35} + (4)^2}\\\\v = 6 \ m/s$

Therefore, the final velocity of the wagon is 6 m/s.

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2 years ago

A jetliner, traveling northward, is landing with a speed of 71.9 m/s. Once the jet touches down, it has 675 m of runway in which

Leviafan [203]

Answer:

The value is $a = -3.7 \ m/s^2$

Explanation:

From the question we are told that

The landing speed is $u = 71.9 \ m/s$

The distance traveled is $d = 675 \ m$

The velocity it is reduced to is $v = 11.3 \ m/s$

Generally the average acceleration is mathematically represented as

$a = \frac{ v^2 - u^2 }{ 2 * d }$

=> $a = \frac{ 11.3^2 - 71.9^2 }{ 2 * 675 }$

=> $a = -3.7 \ m/s^2$

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3 years ago

When light of frequency equal to 2.40 × 1015 s−1 shines on the surface of silver metal, the kinetic energy of ejected electrons

sesenic [268]

Answer: 9.98 *10^-19 J

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h*f-W=Ek where h is the Planck constant and W the work function and Ek the kinetic energy. f is the frequency of light.

W=h*f-Ek=6.62*10^-34*2.4*10^15-5.9*10^-19=9.98*10^-19J

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3 years ago