Ask question

Ask question

All categories

Juliette [100K]

3 years ago

10

Item 19 Question 1 A Ferris wheel has a radius of 75 feet. You board a car at the bottom of the Ferris wheel, which is 10 feet a

bove the ground, and rotate 255º counterclockwise before the ride temporarily stops. How high above the ground are you when the ride stops? Round your answer to the nearest foot.

1 answer:

sergey [27]3 years ago

4 0

Answer:

$y = 104.4 ft$

Explanation:

As we know that we board in the car of ferris wheel at the bottom position

So we will have

final height of the car at angular displacement given as

$y = y_o + R + R sin(270 - 255)$

$y = y_o + R + R sin 15$

here we know that

$y_o = 10 ft$

$R = 75 ft$

so we have

$y = 10 + 75 + 75 sin15$

$y = 104.4 ft$

You might be interested in

Does A= 6.4 B=12 C=12.2 Form a RIGHT TRIANLGE?

Lina20 [59]

To answer this question, you must remember the equation:

a²+b²= c²

(6.4)² + (12)²= (12.2)²

<span>40.96 + 144 = 184.96
</span> (12.2)² = <span>148.84
</span>
184.96 ≠ 148.84

This cannot be a triangle

hope this helps

3 0

3 years ago

A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w

Scilla [17]

Answer:

$v_f = 20 m/s$

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2$

now for pure rolling condition we will have

$v = R\omega$

also we have

$I = mR^2$

now we will have

$KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}$

$KE = mv^2$

now by work energy theorem we can say

$W = KE_f - KE_i$

$842 J = mv_f^2 - mv_i^2$

$842 = 3(v_f^2) - 3\times 11^2$

now solve for final speed

$v_f = 20 m/s$

3 0

3 years ago

The electric field strength E₀ is measured at a perpendicular distance R from an infinitely large, thin sheet that contains a un

Troyanec [42]

Answer:

Explanation:

E=(σ/ε0)

As noted by Dirac the field is the same no matter how far you are from the sheet. When your charge covers a conducting plane, as in your case, the field is, D/eo ,(D is charge density). Because the field inside the conductor (no matter how thin) is zero. The only time the field is, D/2eo, is when you have just a sheet of charge, by itself, not on a conducting plane."

5 0

3 years ago

A cue ball, moving with 9.0 N·s of momentum strikes the nine-ball at rest. The nine-ball moves off with 2.0 N·s in the original

lisov135 [29]

Answer:

P = 7.28 N.s

Explanation:

given,

initial momentum of cue ball in x- direction,P₁ = 9 N.s

momentum of nine ball in x- direction, P₂ = 2 N.s

momentum in perpendicular direction i.e. y - direction,P'₂ = 2 N.s

momentum of the cue after collision = ?

using conservation of momentum

in x- direction

P₁ + p = x + P₂

p is the initial momentum of the nine balls which is equal to zero.

9 + 0 = x + 2

x = 7 N.s

momentum in x-direction.

equating along y-direction

P'₁ + p = y + P'₂

0 + 0 = y + 2

y = -2 N.s

the momentum of the cue ball after collision is equal to resultant of the momentum .

$P = \sqrt{x^2+y^2}$

$P = \sqrt{7^2+(-2)^2}$

P = 7.28 N.s

the momentum of the cue ball after collision is equal to P = 7.28 N.s

7 0

3 years ago

Two charges that are separated by one meter exert 1-n forces on each other. if the magnitude of each charge is doubled, the forc

Goshia [24]

The electrostatic force between the two charges is
$F=k_E \frac{q_1 q_2}{r^2}$
where q1 and q2 are the magnitudes of the two charges, and r the distance between them.

We can see from the formula that F is proportional to the product between the two charges:
$F \sim q_1 q_2$
so, if the magnitude of each charge is doubled, the new force will get a factor 4:
$F' \sim (2 q_1 )(2 q_2 )=4 q_1 q_2 =4 F$
So, the new force will be 4 times the original force:
$F' = 4 \cdot 1N= 4N$

5 0

3 years ago