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2 years ago

If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Physics

1 answer:

vovikov84 [41]2 years ago

7 0

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is $u = 7.73 \ m/s$

Explanation:

From the question we are told that

The height at which he let go of the brief case is h = 130 m

The time taken before the the brief case hits the water is t = 6 s

Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as

$s = h+ u t + 0.5 gt^2$

Here s is the distance covered by the bag at sea level which is zero

$0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $u = \frac{-130 + (0.5 * 9.8 * 6^2) }{6}$

=> $u = 7.73 \ m/s$

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Stels [109]

The hoop is attached.

Consider that the friction force is given by:
F = μ·N
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We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·g·sinθ

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Now, solve for θ:
θ = tan⁻¹(2·μ)
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Therefore, the maximum angle you could ride down without worrying about skidding is 61°.

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2 years ago

A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the

likoan [24]

Answer:

The separation between the first two minima on either side is 0.63 degrees.

Explanation:

A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:

$a\sin \theta_n=n\lambda$

with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:

for the first minimum

$a\sin \theta_1=(1)\lambda$

solving for θ1:

$\theta_1=\arcsin (\frac{\lambda}{a})=\arcsin (\frac{550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_1=0.63 degrees$

for the second minimum:

$a\sin \theta_2=(2)\lambda$

$\theta_2=\arcsin (\frac{2\lambda}{a})=\arcsin (\frac{2*550\times10^{-9}}{0.05\times10^{-3}})$

$\theta_2=1.26 degrees$

So, the angular separation between them is the rest:

$\Delta \theta =1.26-0.63$

$\Delta \theta=0.63$

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2 years ago

The 20-lb cabinet is subjected to the force F = (3 + 2t) lb, where t is in seconds. If the cabinet is initially moving down the

Andre45 [30]

Answer:

t₁ = 0.95 s

Explanation:

In this chaos we must use the definition of Newton's second law

F = m a = m dv / dt

dv = F dt / m

Let's replace and integrate, let's take the upward direction of the plane as positive, the force is positive

dv = ∫ (3 + 2t) dt / m

v = (3 t + 2 t²/ 2) /m

Let's evaluate between the lower limit t = 0 v = -6 ft / s (going down) to the upper limit t = t and v = 0

0 - (-6) = (3 (t- 0) + (t² -0)) / m

t² + 3t -6m = 0

Let's look for the mass

W = mg

m = W / g

m = 20/32

m = 0.625 slug

Let's solve the second degree equation

t² + 3t -3.75 = 0

t = (-3 ± √ (32 + 4 1 3.75)) / 2

t = (-3 ± 4,899) / 2

t₁ = 0.95 s

t₂ = -3.95 s

We take the positive time

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2 years ago

To practice Problem-Solving Strategy 25.1 Power and Energy in Circuits. A device for heating a cup of water in a car connects to

jeyben [28]

Answer:

The time to boil the water is 877 s

Explanation:

The first thing we must do is calculate the external resistance (R) of the circuit, from the description we notice that it is a series circuit, by which the resistors are added

V = i (r + R)

We replace we calculate

r + R = V / i

R = v / i - r

R = 10/12 -0.04

R = 0.793 Ω

We calculate the power supplied

P = V i = I² R

P = 12² 0.793

P = 114 W

This is the power dissipated in the external resistance

We use the relationship, that power is work per unit of time and that work is the variation of energy

P = E / t

t = E / P

t = 100 10³/114

t = 877 s

The time to boil the water is 877 s

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2 years ago

A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. T

den301095 [7]

Answer:

0.266 m

Explanation:

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P= mv where p is momentum, m is mass and v is the speed of an object. In this case

$m_pv_p=v_c(m_p+m_b)$ where sunscripts p and b represent putty and block respectively, c is common velocity.

Substituting the given values then

3*8=v(15+3)

V=24/18=1.33 m/s

The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy

$0.5(m_p+m_b)v_c^{2}=0.5kx^{2}$ where k is spring constant and x is the compression of spring. Substituting the given values then

$(3+15)*1.33^{2}=450*x^{2}\\x\approx 0.266 m$

7 0

2 years ago