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raketka [301]
2 years ago
10

--->Two aircraft P and Q are flying at the same speed. 300 m/s, The direction along which P is flying is at right angles to t

he
direction along which Q is flying. Find the magnitude of the
velocity of the aircraft P relative to aircraft Q.

pls guys I need answer so answer to me if you know the answer pls pls pls​
Physics
1 answer:
REY [17]2 years ago
5 0

Answer:

The magnitude of the velocity of the aircraft P relative to aircraft Q is zero

Explanation:

The velocity of the two aircraft, P & Q, v = 300 m/s

The angle of the direction between them, Ф = 90°

The magnitude of the velocity of aircraft P relative to aircraft Q is given by the formula

                                  <em> V = v cos Ф </em>

Substituting the values in the above equation

                                   v = 300 x cos 90°

                                      = 300 x 0

                                      = 0

Since the aircraft are at right angles, the velocity of one aircraft relative to the other is zero.

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A balloon of mass M is floating motionless in the air. A person of mass less than M is on a rope ladder hanging from the balloon
Alex73 [517]

Answer:

Down with a speed less than v

Explanation:

Let the person's mass be represented by m

The mass of the balloon = M

Total momentum is conserved thus F = external force = 0

when the person starts to climb the ladder, external force is still equal to zero.

We know that change is p = F * change in t

in this equation, F is the external force = 0

Hence change in p = 0

This means that total momentum is conserved thus F = external force = 0

(owing to the fact that exterior forces on the system are balanced)

Since all external force was zero before they began to climb the ladder, as they climb all external forces will still = 0

Thus, As the person starts moving

mv + MV = 0

or mv = MV

V = mv/M

Since M > m, m/M will definitely be a number that is less than 1,

Hence, V = (a number less than 1)v.

This means that V < v, or the balloon moves down a speed V which is less than v.

4 0
2 years ago
An athlete prepares to throw a 2.0-kilogram discus. His arm is 0.75 meters long. He spins around several times with the discus a
wlad13 [49]
Centripetal Force (Fcp) = ?

His arm length = Radius (R) = 0.75 m

Discus velocity = Linear Velocity (V) = 5 m/s

Discus mass (m) = 2 kg

Centripetal Acceleration (Acp) = V^2/R or W^2 x R
In this case i will use the V^2/R formula, because it uses the discus velocity (V).

fcp = m \times acp \\ fcp = m \times {v}^{2} \div r \\ fcp = 2 \times {5}^{2} \div 0.75 \\ fcp = 2 \times 25 \div 0.75 \\ fcp = 50 \div 0.75
fcp = 66.666... = 66 \: newtons

Answer: Last option, 66 N.
7 0
1 year ago
Read 2 more answers
You place a light bulb 8 cm in front of a concave mirror. You then move a sheet of paper back and forth in front of the mirror.
Alika [10]

sorry - late reply...just stumbled across tis...hope u can still use it :)


By the mirror equation: 1/di + 1/do = 1/f

<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>


and do = distance to object = +8cm


Substitute and solve for f, the focal length

<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
</span>
5 0
2 years ago
A child on a playground swing is swinging back and forth (one complete oscillation) once every four seconds, as seen by her fath
Sati [7]

Answer:

A

Explanation:

Solution:-

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- Ignoring Doppler Effect.

- So if the relative speeds of two objects in motion i.e ( swing and spaceship) are positive then the time frame of reference for both object relative to other other decreases. So in other words if spaceship approaches the swing i.e relative velocity is positive then the time period of oscillation observed would be less than actual i.e less than 4 seconds.

- Similarly, if spaceship moves away from the swing i.e relative velocity is negative then the time period of oscillation observed would be more than actual i.e more than 4 seconds.

4 0
1 year ago
Some hydrogen gas is enclosed within a chamber being held at 200∘c with a volume of 0.0250 m3. the chamber is fitted with a mova
Mrac [35]

Answer: The final volume V₂ of the container is  0.039 m³.

Explanation:

Since the temperature is constant, the gas would expand isothermally.

For isothermal expansion,

P₁V₁=P₂V₂

Where, P₁ and P₂ are the initial and final pressure and V₁ and V₂ are initial and final volume.

It is given that:

V₁ = 0.0250 m³

P₁ = 1.5 × 10⁶ Pa

P₂ = 0.950 × 10⁶ Pa

V₂ = ?

⇒ 1.5 × 10⁶ Pa × 0.0250 m³ = 0.950 × 10⁶ Pa × V₂

⇒V₂ = 0.039 m³

Hence, the final volume V₂ of the container is  0.039 m³.

4 0
1 year ago
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