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Paha777 [63]
4 years ago
14

During his recent skydiving adventure, Luke Skyfaller has reached a terminal speed of 9.8 m/s as he approach the ground with his

parachute. During an attempt to snap one last photo with his camera Luke fumbled it from a height of 52.8 m above the ground determine the speed with which the camera hits the ground.
Physics
1 answer:
Serggg [28]4 years ago
6 0

Answer:

The correct solution is "33.62 m/s".

Explanation:

The given values are:

Speed

u = 9.8 m/s

Height

s = 52.8 m

As we know,

⇒ v^2 = u^2 + 2as

On putting the estimated values, we get

⇒      =(9.8)^2+2\times 9.8\times 52.8

⇒      =96.04+1034.88    

⇒      =\sqrt{1130.92}

⇒      =33.62 \ m/s

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Sound waves cannot carry energy through. A water B air C a mirror D a vacuum
Ber [7]
I looked up the question and got D- a vacuum
3 0
3 years ago
A solenoid that is 66.1 cm long has a cross-sectional area of 13.8 cm2. There are 1410 turns of wire carrying a current of 8.01
Ymorist [56]

Answer:

a) Energy density of the magnetic field, u = 183.46 J/m³

b) Total energy, E = 0.167 J

Explanation:

a) Number of turns in the solenoid, N = 1410

Area, A = 13.8 cm² = 0.00138 m²

Current, I = 8.01 A

Length of the solenoid, l = 66.1 cm = 0.661 m

Energy density, u is given by the formula u = \frac{B^2}{2 \mu_{0} }

Where B is the magnetic field

The magnetic field in a solenoid is given by the formula, B = \frac{N \mu_{0} I }{l}

B = \frac{1410 * 8.01* \mu_{0}  }{0.661}

B = 17086.38 \mu_0 T

u = \frac{(17086.38 \mu_0)^2}{2 \mu_{0} }\\u = \frac{291944527.29 \mu_0^2}{2 \mu_{0} }\\u = \frac{291944527.29 \mu_0}{2  }\\u = \frac{291944527.29 * 4\pi * 10^{-7} }{2  }\\u = 183.46 J/m^3

b) The total energy = Energy density * Volume

E = u V

Volume = Area * Length

V = Al = 0.00138 * 0.661

V = 0.00091218 m³

E = 183.46 * 0.00091218

E = 0.167 J

4 0
3 years ago
A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot
FromTheMoon [43]

Answer:

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

Explanation:

To solve the problem/ we first write the differential equation governing the motion. So,

m\frac{d^{2} x}{dt^{2} } = -kx \\ m\frac{d^{2} x}{dt^{2} } + kx = 0\\\frac{d^{2} x}{dt^{2} } + \frac{k}{m} x = 0

with m = 1 slug and k = 9 lb/ft, the equation becomes

\frac{d^{2} x}{dt^{2} } + \frac{9}{1} x = 0\\\frac{d^{2} x}{dt^{2} } + 9 x = 0

The characteristic equation is

D² + 9 = 0

D = ±√-9 = ±3i

The general solution of the above equation is thus

x(t) = c₁cos3t + c₂sin3t

Now, our initial conditions are

x(0) = -1 ft and x'(0) = -√3 ft/s

differentiating x(t), we have

x'(t) = -3c₁sin3t + 3c₂cos3t

So,

x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)

x(0) = c₁cos(0) + c₂sin(0)

x(0) = c₁ × (1) + c₂ × 0

x(0) = c₁ + 0

x(0) = c₁ = -1

Also,

x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)

x'(0) = -3c₁sin(0) + 3c₂cos(0)

x'(0) = -3c₁ × 0 + 3c₂ × 1

x'(0) = 0 + 3c₂

x'(0) = 3c₂ = -√3

c₂ = -√3/3

So,

x(t) = -cos3t - (√3/3)sin3t

Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)

where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.

Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3

x(t) = (2√3/3)sin(3t + 4π/3)

So, the velocity  v(t) = x'(t) = (2√3)cos(3t + 4π/3)

We now find the times when v(t) = 3 ft/s

So (2√3)cos(3t + 4π/3) = 3

cos(3t + 4π/3) = 3/2√3

cos(3t + 4π/3) = √3/2

(3t + 4π/3) = cos⁻¹(√3/2)

3t + 4π/3 = ±π/6 + 2kπ    where k is an integer

3t  = ±π/6 + 2kπ - 4π/3

t  = ±π/18 + 2kπ/3 - 4π/9

t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9

t = π/18 - 4π/9 + 2kπ/3  or -π/18 - 4π/9 + 2kπ/3

t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3

Since t is not less than 0, the values of k ≤ 0 are not included

So when k = 1,

t = 5π/18 and π/6. So,

t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number

6 0
3 years ago
What force needs to touch something to affect it please help xx
Montano1993 [528]
On <span>Contact Forces: </span>
<span>Frictional Force </span>
<span>Tension Force </span>
<span>Normal Force </span>
<span>Air Resistance Force </span>
<span>Applied Force </span>
<span>Spring Force </span>

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<span>Gravitational Force </span>
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7 0
3 years ago
Suppose that the metal cylinder in the last problem has a mass of 0.10 kg and the coefficient of static friction between the sur
andrew11 [14]

Answer:

The speed maximum speed is 0.49ms^{-1}

Explanation:

The centrifugal force always acts on the cylinder and move away the rotating platform from the  rotational axis. so the centripetal force provide by the frictional force:

Therefore

\frac{mv^{2} }{r} =u_{s} mg        

coefficient of static friction: u_{s} =0.12

mass of the cylinder: m=0.10kg\

distance of the cylinder from the turntable: r=0.20m

\frac{mv^{2} }{r} =u_{s} mg

cross multiply to find v

v^{2} =u_{s} rg\\v=\sqrt{u_{s}rg }  \\v=\sqrt{0.12*0.20*9.80}\\ v=0.49m/s

4 0
3 years ago
Read 2 more answers
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