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3 years ago

Lee has a mass of 40.0 kg and ice skates with a speed of 4.6 m/s.

Physics

1 answer:

faltersainse [42]3 years ago

6 0

Their common speed is 1.53 m/s.

<h3>What is speed?</h3>

Speed can be defined as the ratio of the distance to the time of a body in motion.

To calculate their final speed, we use the formula below.

Formula:

mu+m'u' = V(m+m')............. Equation 1

Where:

m = mass of the Lee
m' = mass of Mat
u = initial speed of Lee
u' = initial speed of Mat
V = Their common speed.

make V the subject of the equation

V = (mu+m'u')/(m+m')........... Equation 2

From the question,

Given:

m = 40 kg
m' = 80 kg
u = 4.6 m/s
u' = 0 m/s

Substitute these values into equation 2

V = [(40×4.6)+(80×0)]/(40+80)
V = 184/120
V = 1.53 m/s.

Hence, Their common speed is 1.53 m/s.

Learn more about speed here: brainly.com/question/3004254

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Answer:

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4 years ago

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3 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

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Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

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Now the acceleration towards the east is mathematically represented as

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substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

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substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

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The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

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