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lana66690 [7]
3 years ago
13

A skydiver jumps from an airplane traveling 120. km/hr at a height of 1000. m. If his parachute must open 500. m above the groun

d, how long does he have for the chute to open?
Physics
1 answer:
riadik2000 [5.3K]3 years ago
6 0

Answer:

The parachute to open in 15 sec.

Explanation:

Given that,

Velocity = 120 km/hr

Height = 1000 m

If his parachute must open 500. m above the ground.

We need to calculate the time for parachute to open

Using formula of time

v=\dfrac{d}{t}

t=\dfrac{d}{v}

Where, d = distance

t = time

v = velocity

Put the value into the formula

t=\dfrac{500}{120\times\dfrac{5}{18}}

t=15\ sec

Hence, The parachute will be open in 15 sec.

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telo118 [61]

Answer:

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3 0
3 years ago
What might happen if a person’s ear canal was blocked?
mylen [45]

Answer:

D. Sounds would be harder to hear.

Explanation:

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7 0
3 years ago
Do the math: How many seconds would it take an echo sounder’s ping to make the trip from a ship to the Challenger Deep (10,994 m
Lera25 [3.4K]

Answer:

<h2>14.66secs</h2>

Explanation:

Given the formula for calculating the depth in metres expressed as

depth in meters = ½ (1500 m/sec × Echo travel time in seconds)

Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.

10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds

10,994 = 750 * Echo travel time in seconds

Dividing both sides by 750;

Echo travel time in seconds = 10,994 /750

Echo travel time in seconds ≈ 14.66secs (to two decimal places)

Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back

6 0
3 years ago
The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and
Contact [7]

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

d=\sqrt{(x-x_o)^2+(y-y_o)^2}   (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m

hence, the displacement of the airplane is 3.45*10^4 m

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3 years ago
I need help to figure out how to solve this problem and solve it!!!
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