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Vlad [161]
3 years ago
5

1. Which object is farthest from the origin at t=2sec

Physics
1 answer:
Stolb23 [73]3 years ago
6 0

Answer:

that one i know only pe not that sorry again

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Lindsay is planning a flight from St. Catharines to Hamilton, which lies due west of St. Catharines. Her aircraft flies at a spe
olga nikolaevna [1]

Lindsay has to fly this plane towards this direction [W 12.5° S] to get to Hamilton.

From this question, the plane is still up in the air.

We have wind blowing in [W 60° N ]

To solve the problem we have to make use of the sine rule

\frac{SinA}{a}=\frac{SinB}{b} =\frac{SinC}{c}

We put the values in the equation, we have:

50/Sinθ = 200/sin60°

The next step is to cross multiply

50 x sin60° = 200Sinθ

50 x 0.8660 = 200sinθ

We make Sin θ the subject

Sine θ = 43.30/200

sine θ = 0.2165

we find the value of θ

θ = sine⁻¹(0.2165)

θ = 12.50

So Lindsay has to fly this plane towards this direction

[W 12.5° S]

Here is a similar question brainly.com/question/13338067?referrer=searchResults

7 0
3 years ago
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I WILL MARK BRAINLIEST
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Sound waves require a medium. For example - think about space. there is no air or liquid because it is a vacuum, which is why sounds don’t travel through space.
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A force of 660 n stretches a certain spring a distance of 0.300 m. what is the potential energy of the spring when a 70.0 kg mas
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A force of 660 n stretches a certain spring a distance of 0.300 m. what is the potential energy of the spring when a 70.0 kg mass hangs vertically from it?
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I need help, ASAP i’m failing and i have no clue what’s going on in my AP physics class at all.
garri49 [273]
What’s the question or problem ?
6 0
3 years ago
An airplane is moving at 350 km/hr. If a bomb is
bogdanovich [222]

Answers:

a) -171.402 m/s  

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final height

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb's initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity </h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
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