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gregori [183]
3 years ago
6

The Zero Gravity Research Facility at the NASA Glenn Research Center includes a

Physics
1 answer:
denis23 [38]3 years ago
4 0

Answer: (a) t = 5.44 sec

(b) vf = 53.31 m/s

(c) s = 5.0m

Explanation: from the question, given data

the Height of the tower, h = 145m

from question

(a)

the initial velocity, v₁ = 0 m/s

s = v₁t + 1/2 gt²

-145 m = 0(t) + 1/2 (-9.8t²)

  t² = 145/4.9

   t² = 29.59

    t = 5.44 sec

(b)

the speed of the sphere at the bottom of the tower is

vf² = vi² +2as

vf² = 0 + 2(-9.8 × -145)

vf² = 2842

vf = 53.31 m/s

(c)

when caught, the sphere experiences a deceleration of;

   a = -29.0g

the time it would take to decelerate becomes;

vf = vi + at

0 = (53.31) + (-29 ×9.8)t

where t = 53.31 / 284.2

t = 0.1876 sec

∴ the distance travelled during the deceleration becomes;

vf² = vi² + 2as

s = (vf² - vi²) / 2a

s = (0 - 53.31²) / 2×-29×9.8

s = -2841.9561 / -568.4

s = 4.99 ≈ 5.0m

i hope this helps, cheers

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When a wave enters a new medium from an angle, both the speed and the ________ change
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Answer:

B: Amplitude

Explanation:

When a wave travels from one medium to the other from an angle, the things that change are amplitude, wavelength, intensity and velocity.

The frequency doesn't change because the frequency depends upon the source of the wave and not the medium by which the wave is propagated.

8 0
2 years ago
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A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro
svetoff [14.1K]

Answer:

The equation of motion is x(t)=-\frac{1}{3} cos4\sqrt{6t}

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is 32ft/s^2

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet x=\frac{4}{12} =\frac{1}{3}feet

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

W=kx ⇒ k=\frac{W}{x}

The spring constant , k=\frac{24}{1/3}

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    m\frac{d^2x}{dt} +kx=0

    \frac{3}{4} \frac{d^2x}{dt} +72x=0

  \frac{d^2x}{dt} +96x=0

Auxiliary equation is, m^2+96=0

                                 m=\sqrt{-96}

                               =\frac{+}{} i4\sqrt{6}

Thus , the solution is x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}

                                 x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2  4\sqrt{6} cos4\sqrt{6t}

The mass is released from the rest x'(0) = 0

                    =-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2 4\sqrt{6} cos4\sqrt{6(0)} =0

                                                    c_2 4\sqrt{6} =0

                                     c_2=0

Therefore , x(t)=c_1 cos 4\sqrt{6t}

Since , the mass is released from the rest from 4 inches

                    x(0)= -4 inches

c_1 cos 4\sqrt{6(0)} =-\frac{4}{12} feet

   c_1=-\frac{1}{3} feet

Therefore , the equation of motion is  -\frac{1}{3} cos4\sqrt{6t}

7 0
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Which is a correct step in a scientific experiment involving acids and bases?
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A 1.5m wire carries a 8A current when a potential difference of 67v is applied what is the resistance of the wire
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Data is inappropriate

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                              R  = ρl/A            

            where R= resistance ; Ω

                       l = length of wire ; m

                       A = area of wire ; m²

                       ρ = resistivity ; Ω-m

But in general ohms law is

                    V = I R

                    R = V/I ;

but here we also calculate "R" from length of wire in which the current is flowing.

I hope it is helpful to you.

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