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SVETLANKA909090 [29]
3 years ago
5

Vesna Vulovic survived the longest fall on record without a parachute when her plane exploded and she fell 5 miles, 733 yards. W

hat is this distance in meters?
Physics
1 answer:
Valentin [98]3 years ago
8 0

Answer:

8717 meters.

Explanation:

We need to know the conversion factors. We know that:

1 mile = 1609.34 meters

1 yard = 0.9144 meters

This means that:

\frac{1609.34 meters}{1 mile}=1

\frac{0.9144 meters}{1 yard}=1

It is convenient to leave the units we want at the end in the numerator so the ones in the denominator cancel out with the ones we want to remove, as will be seen in the next step.

We will convert first the miles, then the yards, and add them up.

5miles=5miles\frac{1609.34 meters}{1 mile}=8046.7meters

733yards=733yards\frac{0.9144 meters}{1 yard}=670.2552meters

So total distance is the sum of these,  8717 meters.

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A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

7 0
3 years ago
Calculate the speed of a gamma ray with a frequency of 3.0 x 10^19 Hz and a wavelength of 1.0 x 10^-11 m.
Ulleksa [173]

Answer:

Speed of gamma rays = 3 x 10⁸ m/s

Explanation:

Given:

Frequency of gamma ray = 3 x 10¹⁹ Hz

Wavelength of gamma rays = 1 x 10⁻¹¹ meter

Find:

Speed of gamma rays

Computation:

Velocity = Frequency x wavelength

Speed of gamma rays = Frequency of gamma ray x Wavelength of gamma rays

Speed of gamma rays = [3 x 10¹⁹][1 x 10⁻¹¹]

Speed of gamma rays = 3 x [10¹⁹⁻¹¹]

Speed of gamma rays = 3 x [10⁸]

Speed of gamma rays = 3 x 10⁸ m/s

6 0
3 years ago
Sphere A has a charge of +2 x 10 to the power of -6 coulomb and is brought into contact with a similar sphere, B which has a cha
Setler [38]
When sphere A and B are brought in contact and separated, charge on each sphere becomes [2x10^-6 + (-4x10^-6)]/ 2 = -1x10^-6 C.

That is, charge is equally separated and is the average of charges on both spheres. The reason behind equal charge on both spheres after separation is, when they are kept in contact, their potential difference becomes same.
7 0
3 years ago
You want to slide a 0.39 kg book across a table. If the coefficient of kinetic friction is .21, what force is required to move t
uranmaximum [27]
Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together.  This is done because the friction force is going to have to be compensated for.  We will need that much more force than we otherwise would to achieve the desired acceleration:

F_{NoFric}=ma=0.39kg \times0.18 \frac{m}{s^2}  =0.0702N

The friction force will be given by the normal force times the coefficient of friction.  Here the normal force is just its weight, mg

F_{Fric}=0.39kg \times 9.8 \frac{m}{s^2} \times 0.21=0.803N

Now the total force required is:

0.0702N+0.803N=0.873N

5 0
3 years ago
On the Moon, the acceleration due to the effect of gravity is only about 1/6 of that on Earth. An astronaut whose weight on Eart
Ulleksa [173]
A = 94.22 Newtons

b = 58.16 kg

Gravity on the moon is 1.62 m/s^2
8 0
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