Aiko jumped rope for 20 min and stopped at 8:05 when did she start

A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d

Alenkasestr [34]

Answer:

μ = Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

Distance 1:

- Ff*d₁ = Ef - Ei

⇒ - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ

⇒ - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ (I)

Distance 2:

- Ff*d₂ = Ef - Ei

⇒ - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒ μ*g*d₂ = 0.5*v² (II)

If we apply (I) + (II)

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ (III)

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁ ⇒ vf² = 2*g*Sin θ*d₁ = v²

then (from the equation III) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒ μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒ μ = Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK ⇒ - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁ ⇒ d₂ = Sin θ*d₁ / μ

3 0

3 years ago

One card is selected at random from a deck of cards. Determine the probability that the card selected is a 10. The probability t

Daniel [21]

Answer:

1:13

Step-by-step explanation:

A standard deck has 52 cards not including the joker. each suit has one 10. there are total four "10" in a deck of playing cards. 4/52 = 1/13 or 1:13

6 0

3 years ago

Read 2 more answers

Someone please help?

Digiron [165]

The answer to your question is,

$3.84, you take $69.20 and divide it by 18 and there is your answer!

-Mabel <3

4 0

3 years ago

Two skiers are 30 kilometers apart and head towards each other. They meet in 2 hours. If the second skier is 5 kilometers per ho

DENIUS [597]

Answer:

Speed of first skier = 5 km/h

Speed of second skier = 10 km/h

Step-by-step explanation:

Given:

Distance between skiers = 30 km

Time after which they meet = 2 hours

Second skier is 5 km/h faster than the first skier.

To find speed of each skier.

Solution:

Let the speed of first skier be in km/h = $x$

Distance covered in km 2 hours will be = $Speed\times time=x\times 2=2x\ km$

Speed of second skier in km/h can be given as = $x+5$

Distance traveled by second skier after 2 hours will be = $Speed\times time=(x+5)\times 2=(2x+10)\ km$ [Using distribution]

Since, the skiers were 30 km apart initially, so the total distance covered by both of them when they meet after 2 hours will be = 30 km

Thus, we have:

$2x+2x+10=30$

Solving for $x$

$4x+10=30$

Subtracting both sides by 10.

$4x+10-10=30-10$

$4x=20$

Dividing both sides by 4.

$\frac{4x}{4}=\frac{20}{4}$

$x=5$

Speed of first skier = 5 km/h

Speed of second skier = $5+5$ = 10 km/h

8 0

3 years ago

PLEASE HELP!

leva [86]

Answer:

B. 56°

Step-by-step explanation:

We are given that m∠R is 66° and m∠T is 122°.

We can apply the supplementary rule since ∠S and ∠T are a linear pair. So, we can use ∠T to find ∠S through 180° - 122° = 58°.

Now, we can use ∠R and ∠S to find ∠Q.

66° + 58° = 124°

180° - 124° = 56°

8 0

3 years ago