What do you call the type of rock from which a soil forms? A. lichen B. clay C. sand D. parent material

The magnitude of the voltage induced in a conductor moving through a stationary magnetic field depends on the _______ and the __

Virty [35]

The correct answer is: Option (D) length, speed

Explanation:

According to Faraday's Law of Induction:

ξ = Blv

Where,

ξ = Emf Induced

B = Magnetic Induction

l = Length of the conductor

v = Speed of the conductor.

As you can see that ξ (Emf/voltage induction) is directly proportional to the length and the speed of the conductor. Therefore, the correct answer will be Option (D) Length, Speed

3 0

3 years ago

Read 2 more answers

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

3 0

2 years ago

A ray of red light in air is incident at an angle of 30. on a

Vlad [161]

Answer:

20 degrees.

Explanation:

From Snell’s law of refraction:

sinθ1•n1 = sinθ2•n2

where θ1 is the incidence angle, θ2 is the refraction angle, n1 is the refraction index of light in medium1, and n2 is the refraction index for virgin olive oil. The incidence angle of the red light is θ1 = 30 degrees.

The red light is in air as medium1, so n1 (air) = 1.00029

So, to find θ2, the refracted angle:

sinθ1•1.00029 = sinθ2•1.464

sin(30)•1.00029 / 1.464 = sinθ2

0.5•1.00029 / 1.464 = sinθ2

sinθ2 = 0.3416291

θ2 = arcsin(0.3416291)

θ2 = 19.976 degrees

To the nearest degree,

θ2 = 20 degrees.

8 0

2 years ago

A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10r, where a and fare in m/s^2

xz_007 [3.2K]

Answer:

Mistake in question

The correct question

A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10t , where a and t are in m/s² and seconds, respectively. After reaching a speed of 125 m/s, a parachute is deployed to help slow down the dragster. Knowing that this deceleration is defined by the relationship a = - 0.02v², where v is the velocity in m/s, determine (a) the total time from the beginning of the race until the car slows back down to 10 m/s, (b) the total distance the car travels during this time.

Explanation:

Given the function

a = 50 —10t

The car started from rest u = 0

And it accelerates to a speed of 125m/s

Then, let find the time in this stage

Acceleration can be modeled by

a = dv/dt

Then, dv/dt = 50—10t

Using variable separation to solve the differentiation equation

dv = (50—10t)dt

Integrating both sides

∫ dv = ∫ (50—10t)dt

Note, v ranges from 0 to 125seconds, so we want to know the time when it accelerate to 125m/s. So t ranges from 0 to t'

∫ dv = ∫ (50—10t)dt

v = 50t —10t²/2. Equation 1

[v] 0<v<125 = 50t —10t²/2 0<t<t'

125—0 = 50t — 5t² 0<t<t'

125 = 50t' — 5t'²

Divide through by 5

25 = 10t' — t'²

t'² —10t' + 25 = 0

Solving the quadratic equation

t'² —5t' —5t' + 25 = 0

t'(t' —5) —5(t' + 5) = 0

(t' —5)(t' —5) = 0

Then, (t' —5) = 0 twice

Then, t' = 5 seconds twice

So, the car spent 5 seconds to get to 125m/s.

The second stage when the parachute was deployed

We want to the time parachute reduce the speed from 125m/s to 10m/s,

So the range of the velocity is 125m/s to 10m/s. And time ranges from 0 to t''

The function of deceleration is give as

a = - 0.02v²

We know that, a = dv/dt

Then, dv/dt = - 0.02v²

Using variable separation

(1/0.02v²) dv = - dt

(50/v²) dv = - dt

50v^-2 dv = - dt

Integrate Both sides

∫ 50v^-2 dv = -∫dt

(50v^-2+1) / (-2+1)= -t

50v^-1 / -1 = -t

- 50v^-1 = -t

- 50/v = - t

Divide both sides by -1

50/v = t. Equation 2

Then, v ranges from 125 to 10 and t ranges from 0 to t''

[ 50/10 - 50/125 ] = t''

5 - 0.4 = t''

t'' = 4.6 seconds

Then, the time taken to decelerate from 125s to 10s is 4.6 seconds.

So the total time is

t = t' + t''

t = 5 + 4.6

t = 9.6 seconds

b. Total distanctraveleded.

First case again,

We want to find the distance travelled from t=0 to t = 5seconds

a = 50—10t

We already got v, check equation 1

v = 50t —10t²/2 + C

v = 50t — 5t² + C

We add a constant because it is not a definite integral

Now, at t= 0 v=0

So, 0 = 0 - 0 + C

Then, C=0

So, v = 50t — 5t²

Also, we know that v=dx/dt

Therefore, dx/dt = 50t — 5t²

Using variable separation

dx = (50t —5t²)dt

Integrate both sides.

∫dx = ∫(50t —5t²)dt

x = 50t²/2 — 5 t³/3 from t=0 to t=5

x' = [25t² — 5t³/3 ]. 0<t<5

x' = 25×5² — 5×5³/3 —0

x' = 625 — 208.333

x' = 416.667m

Stage 2

The distance moved from

t=0 to t =4.6seconds

a = -0.002v²

We already derived v(t) from the function above, check equation 2

50/v = t + C.

When, t = 0 v = 125

50/125 = 0 + C

0.4 = C

Then, the function becomes

50/v = t + 0.4

50v^-1 = t + 0.4

Now, v= dx/dt

50(dx/dt)^-1 = t +0.4

50dt/dx = t + 0.4

Using variable separation

50/(t+0.4) dt = dx

Integrate both sides

∫50/(t+0.4) dt = ∫ dx

50 In(t+0.4) = x

t ranges from 0 to 4.6seconds

50In(4.6+0.4)—50In(4.6-0.4) = x''

x'' = 50In(5) —50In(4.2)

x'' = 8.72m

Then, total distance is

x = x' + x''

x = 416.67+8.72

x = 425.39m

The total distance travelled in both cases is 425.39m

5 0

2 years ago

Read 2 more answers

Which of these is not an example of

prohojiy [21]

Answer:

A) The process that give us day and night.

Explanation:

From the answer choices provided the one that is not an example of this is the process that give us day and night. This is because, the day/night cycle occurs as the Earth spins on it's axis. This cause one side of the Earth to be facing the Sun, which is the side that is currently experiencing day time while the other side is experiencing night time. As the Earth spins the cycles repeat. This is not an example of revolving.

5 0

3 years ago