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alex41 [277]
3 years ago
6

One of the contests at the school carnival is to throw a spear at an underwater target lying flat on the bottom of a pool. The w

ater is 1.20 m deep. You're standing on a small stool that places your eyes 3.10 m above the bottom of the pool. As you look at the target, your gaze is 30∘below horizontal. At what angle below horizontal should you throw the spear in order to hit the target?Part AYour raised arm brings the spear point to the level of your eyes as you throw it, and over this short distance you can assume that the spear travels in a straight line rather than a parabolic trajectory.
Physics
1 answer:
sdas [7]3 years ago
4 0

Answer:

35.67°

Explanation:

Given that:

angle of glance(i) = 30°

Depth of water d = 1.20 m

The height of the observer above the water is h  = 3.10 m - 1.20 m

= 1.90 m

Refractive Index of water (n) = 1.33

Using Snell's Law at the water air interface;

n₁ × sin (90- i) = n₂ × sin (90 - r)

1 × cos (i) = 1.33 cos (r)

r = cos ⁻¹ (cos 30/1.33)

r = 49.4

D = h/tan (i)

D = 1.9 / tan (30)

D = 3.291 m

D' = d/tan (r)

D' = 1.2 m/ tan (49.4)

D' = 1.0285 m

∴ the angle at which the spear is to be thrown is :

x = tan⁻¹ [(h+d)/(D+D')]

x = tan⁻¹  [(1.9+ 1.2)/(3.291+1.0285)]

x = tan⁻¹  [3.1/4.3195]

x = 35.67°

∴ At an angle of  35.67° below horizontal  is required for  you to  throw the spear in order to hit the target.

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You toss a conductive open ring of diameter d = 1.75 cm up in the air. The ring is flipping around a horizontal axis at a rate o
Mama L [17]

Answer:

The maximum emf induced in the ring

= (2.882 × 10⁻⁷) V

Explanation:

According to the law of electromagnetic induction, the emf induced in the ring is given by

E = N BA w sin wt

The maximum emf induced is

E = N BA w

B = 30.5 μT = (30.5 × 10⁻⁶) T

A = (πD²/4)

D = 1.75 cm = 0.0175 m

A = (π×0.0175²/4) = 0.000240625 m²

Nw = 2π × 6.25 = 39.29 rad/s

E = 30.5 × 10⁻⁶ × 0.000240625 × 39.29

E = (2.882 × 10⁻⁷) V

Hope this Helps!!!

8 0
3 years ago
An inductor in an LC circuit has a maximum current of 2.4 A and a maximum energy of 56 mJ.
Harrizon [31]

Answer:

The energy stored in the capacitor, when the current in the inductor is 1.2 A, is 41.6 mJ.

Explanation:

In a LC oscillating circuit, the energy is stored in the electric field (between the plates of the capacitor) and in the magnetic field (surrounding the wires of the inductor).

At any time, the sum of both energies can be expressed as follows:

E = 1/2 Q² / C   +  1/2 L I²

In this type of circuit, energy oscillates, which means that it is exchanging between both fields all time.

When the capacitor is completely discharged, all the energy is stored in the magnetic field, and at that time, the current is maximum.

The total energy, when I is maximum, can be written as follows:

E = 1/2 L I² (1)

In our case, when I= 2.4A, E= 56 mJ.

So, we can find out the value of L, which will allow us to know the value of the magnetic energy at any time, having the value of the instantaneous current.

Solving for L in (1):

L = 2 *.56 mJ / (2.4)² A² = 20 mH

The next step is getting the value of the energy stored in the inductor, when I = 1.2 A, as follows:

Em = 1/2 *20 mH.* (1.2)² A² = 14.4 mJ

As the total energy must be always the same, i.e., 56 mJ, the energy stored in the capacitor, assuming no losses, must be the difference between the total energy and the one stored in the magnetic field:

Ec = 56 mJ - 14.4 mJ = 41.6 mJ

3 0
3 years ago
The distance recorded for riding a motorcycle on its rear wheel without stopping is more than 320 km! Suppose the rider in this
antiseptic1488 [7]

Answer:

<h3>14.97m/s</h3>

Explanation:

Given

Initial velocity of the car u = 8m/s

Distance travelled by the rider S = 40m

Acceleration a = 2m/s²

Required

rider's velocity after the acceleration v

Using the equation of motion

v² = u²+2as

v² = 8²+2(2)(40)

v² = 64+160

v² = 224

v = √224

v = 14.97m/s

Hence the rider's velocity after the acceleration is 14.97m/s

5 0
3 years ago
A car in an amusement park ride rolls without friction around a track (Fig. P7.42). The hB car starts from rest at point A at a
MrRissso [65]

Answer:

h>\dfrac{5}{2}R

Explanation:

Given that

Height = h

Radius = R

From energy conservation

KE_A+U_A=KE_B+U_B

At point B

The minimum speed to complete the   the circle

V_B=\sqrt{gR}\ m/s

So the kinetic energy at point B

KE_B=\dfrac{1}{2}mV^2

KE_B=\dfrac{1}{2}mgR

KE_A+U_A=KE_B+U_B

0+mgh=\dfrac{1}{2}mgR+2mgR

Without falling off at the top (point B)

0+mgh>\dfrac{1}{2}mgR+2mgR

mg(h-2R)>\dfrac{1}{2}mgR

g(h-2R)>\dfrac{1}{2}gR

h>\dfrac{5}{2}R

6 0
3 years ago
1. A large turbine has an initial angular momentum of 6700 kgm^2/s. A storm is rolling in and the wind picks up. 8 seconds later
LekaFEV [45]

Answer:

262.5 Nm

Explanation:

Torque is the rate of change of angular momentum.

Hence, we have

\tau = \dfrac{\Delta L}{t}

Δ<em>L</em> is the change in angular momentum.

Using values in the question,

\tau = \dfrac{8800-6700 \text{ kg m}^2\text{/s}}{8\text{ s}}  = 262.5 \text{ Nm}

4 0
3 years ago
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