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meriva
3 years ago
6

a parking lot is going to be 60m wide and 240m long which dimensions could be used please please help

Physics
2 answers:
Wewaii [24]3 years ago
4 0

Answer:

A = 20x80

Explanation: DID IT ON APEX

Bezzdna [24]3 years ago
3 0

Answer: The area of the parking lot is 14,400 meters squared.

Explanation:

We have the dimensions of the parking lot.

60m by 240m

The units used here are meters.

Now, if we want to know the area of the parking lot is equal to the product between the length and the width:

A = 60m*240m = 14,400 m^2

The area of the parking lot is 14,400 meters squared.

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A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and
Leya [2.2K]

Answer:

A) 0.9844 s

B) x2 = 0.4587 m

C) v = 6.657 m/s

Explanation:

We are given;

Height of take off point above pool; x1 = 1.8 m

Initial take off velocity; u = 3 m/s

Final velocity at highest point before free fall; v = 0 m/s

B) To find the highest point above the board her feet reaches means the distance from take off to the top of the motion just before free fall.

Thus, we will be using equation of motion and we have;

v² = u² + 2gs

Now, let s = x2 which will be the distance between take off and the top before free fall.

So;

v² = u² + 2g(x2)

Now,since the motion is against gravity, g will be negative.

Thus;

v² = u² + 2(-9.81)(x2)

Plugging in the relevant values to give;

0² = 3² - (19.62x2)

19.62(x2) = 9

x2 = 9/19.62

x2 = 0.4587 m

A) We want to find how long her feet is in air.. It means we want to find out the time to get to a distance of x1 and also the time to achieve the distance (x1 + x2) on free-fall.

Thus, using equation of motion;

v = u + gt

Again, g = -9.81

Thus;

0 = 3 - 9.81t1

9.81t1 = 3

t1 = 3/9.81

t1 = 0.3058 s

Now, for the time taken to achieve the distance (x1 + x2) on free-fall, we will use the formula;

s = ut + ½gt²

Where s = (x1 + x2) = 1.8 + 0.4587 = 2.2587 m

And now, u = 0 m/s because the start of the free fall is from maximum height with velocity of 0 m/s. Again, g = - 9.81 m/s²

Thus;

2.2587 = 0 - ½(-9.81)(t2)²

2.2587 = 4.905(t2)²

(t2)² = 2.2587/4.905

(t2)² = 0.4605

t2 = √0.4605

t2 = 0.6786 s

Thus, total time of feet in air = t1 + t2 = 0.3058 + 0.6786 = 0.9844 s

C) Velocity when feet hit the water would be given by;

v = u + gt

Where u = 0 m/s and t = t2 = 0.6786

Since it's in direction of gravity, g = 9.81 m/s

v = 0 + (0.6786 × 9.81)

v = 6.657 m/s

4 0
3 years ago
When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin
mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}

where n=1,2,3 or ...n

L=Length of string

T=Tension

\mu =Mass per unit length

When string is clamped at mid-point

Effecting length becomes L'=0.5 L

Thus new Frequency becomes

f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz      

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How does the digestive system help the muscular system?
Sveta_85 [38]

Answer:

I would say the answer is A... but I'm not so sure ....

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2 years ago
If a net force is greater than 0 Newtons, what is the force?
goldenfox [79]

Answer:disequilibrium

Explanation:

When the net force is not zero it is called disequilibrium

5 0
3 years ago
Debbie places two shopping carts in a cart Corral. she pushes the first cart, which then pushes a second cart. what force is bei
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When Debbie pushes the first cart she is using an applied force. An applied force is created when someone or something pushes another thing using, of course, an applied force. Now, when the second cart is being pushed by the first cart, this is also an applied force. You can tell because the first cart is being pushed using forced and this causes the second cart to be pushed using some of the force that is being transmitted to the first cart.


Debbie exerts applied force on the first cart. The first cart exert applied force on the second cart.



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