Scientific question must be

Fiora starts riding her bike at 20 mi/h. After a while, she slows down to 14 mi/h, and maintains that speed for the rest of the

goblinko [34]

Answer:

3.5 hours

Explanation:

Speed = distance/time

Let the distance that Fiora biked at 20 mi/h through be x miles and the time it took her to bike through that distance be t hours at 20 mi/h

Then, the rest of the distance that she biked at 14 mi/h is (112 - x) miles

And the time she spent biking at 14 mi/h the rest of the distance = (6.5 - t) hours

Her first biking speed = 20 mph = 20 miles/hour

Speed = distance/time

20 = x/t

x = 20 t (eqn 1)

Her second biking speed = 14 mph = 14 miles/hour

14 = (112 - x)/(6.5 - t)

112 - x = 14 (6.5 - t)

112 - x = 91 - 14t (eqn 2)

Substitute for x in (eqn 2)

112 - 20t = 91 - 14t

20t - 14t = 112 - 91

6t = 21

t = 3.5 hours

x = 20t = 20 × 3.5 = 70 miles.

(112 - x) = 112 - 70 = 42 miles

(6.5 - t) = 6.5 - 3.5 = 3 hours

Meaning that she travelled at 20 mi/h for 3.5 hours.

4 0

3 years ago

A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the

liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0

3 years ago

When a flat slab of transparent material is placed under water, the critical angle for light traveling from the slab into water

Novosadov [1.4K]

Answer:

(a) 40.6 degree

Explanation:

When refraction takes place from slab to water, the critical angle is 60 degree.

Use Snell's law

refractive index of water with respect to slab

$\mu _{w}^{s}=\frac{Sin60}{Sin90}$

$\frac{\mu _{w}}{\mu _{s}}=0.866$

$\frac{1.33}{\mu _{s}}=0.866$

μs = 1.536

Now for slab air interface, the critical angle is C.

$\mu _{a}^{s}=\frac{SinC}{Sin90}$

$\frac{\mu _{a}}{\mu _{s}}=\frac{SinC}{Sin90}$

1 / 1.536 = Sin C

C = 40.6 degree

3 0

3 years ago

two people, each with a mass of 70 kg, are wearing inline skates and are holding opposite ends of a 15m rope. One person pulls f

Zina [86]

Answer:

7.75 s

Explanation:

Newton's second law:

∑F = ma

35 N = (70 kg) a

a = 0.5 m/s²

Given v₀ = 0 m/s and Δx = 15 m:

Δx = v₀ t + ½ at²

(15 m) = (0 m/s) t + ½ (0.5 m/s²) t²

t = 7.75 s

5 0

3 years ago

Four velcro-lined air-hockey disks collide with each other in a perfectly

Reil [10]

Answer:

The magnitude of the final velocity is approximately 0.526 m/s in approximately the direction of 8.746° East of South

Explanation:

The given collision parameters are;

The kind of collision experienced by the four velcro-lined air-hockey disk = Inelastic collision

The mass of the first disk, m₁ = 50.0 g

The velocity of the first disk, v₁ = 0.80 m/s West = -0.8·i

The mass of the second disk, m₂ = 60.0 g

The velocity of the second disk, v₂ = 2.50 m/s North = 2.5·j

The mass of the third disk, m₃ = 100.0 g

The velocity of the third disk, v₃ = 0.20 m/s East = 0.20·i

The mass of the fourth disk, m₄ = 40.0 g

The velocity of the fourth disk, v₄ = 0.50 m/s South = -0.50·j

Therefore, the total initial momentum of the four velcro-lined air-hockey disk, $\Sigma P_{initial}$ is given as follows;

$\Sigma P_{initial}$ = m₁·v₁ + m₂·v₂ + m₃·v₃ + m₄·v₄ = 50.0×(-0.80·i) + 60.0×(2.50·j) + 100 × (0.20·i) + 40.0 × (-0.50·j)

∴ $\Sigma P_{initial}$ = -40·i + 150·j + 20·i - 20·j = -20·i + 130·j

∴ $\Sigma P_{initial}$ = -20·i + 130·j

By the law of conservation of linear momentum, we have;

$\Sigma P_{initial} = \Sigma P _{final}$ = -20·i + 130·j

Therefore, given that the collision is perfectly inelastic, the disks move as one after the collision and the four masses are added to form one mass, "m", m = m₁ + m₂ + m₃ + m₄ = 50.0 + 60.0 + 100.0 + 40.0 = 250.0

∴ m = 250.0 g

Let, "v" represent the final velocity of the four disks moving as one after the collision

We have;

$\Sigma P _{final}$ = m × v = 250.0 × v = -20·i + 130·j

∴ v = -20·i/250 + 130·j/250 = -0.08·i + 0.52·j

The final velocity of the four disks after collision, v = -0.08·i + 0.52·j

The magnitude of the final velocity, $\left | v \right |$ = √((-0.08)² + (0.52)²) ≈ 0.526

$\left | v \right |$ ≈ 0.526 m/s

The direction of the final velocity, θ = arctan(0.52/(-0.08)) ≈ -81.254°

The direction of the final velocity, θ ≈ -81.254° which is 8.746° East of South

4 0

3 years ago