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Mandarinka [93]
2 years ago
5

13. An aircraft heads North at 320 km/h rel:

Physics
1 answer:
AURORKA [14]2 years ago
7 0

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

v' = v + v_a

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

v_a is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

v_a = -80 km/h (since the air is moving from North)

Therefore, we find

v'=+320 + (-80) = +240 km/h (north)

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

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The diagram below depicts all the forces acting on an object. Use both vector resolution and vector
maksim [4K]

Answer:

A

Explanation:

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3 0
2 years ago
The muzzle velocity of a 50.0g shell leaving a 3.00 kg rifle is 400m/s what is the recoil velocity of the rifle
serg [7]

Here if we consider bullet and gun as a system then we can say that momentum of the system will remain conserved

so we will have

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now we know that

m_1 = 50 g = 0.050 kg

m_2 = 3 kg

v_{1i} = v_{2i} = 0

v_{1f} = 400 m/s

now we will have

0.050(0) + 3(0) = 0.050(400) + 3(v)

20 + 3v = 0

v = - \frac{20}{3} = - 6.67 m/s

so gun will recoil with speed 6.67 m/s

6 0
2 years ago
Liam throws a water balloon horizontally at 8.2 m/s out of a window 18 m from the ground.
Alecsey [184]

Time taken by the water balloon to reach the bottom will be given as

h = \frac{1}{2} gt^2

here we know that

h = 18 m

g = 9.8 m/s^2

now by the above formula

18 = \frac{1}{2}*9.8* t^2

18 = 4.9 t^2

t = 1.92 s

now in the same time interval we can say the distance moved by it will be

d = v_x * t

d = 8.2 * 1.92 = 15.7 m

so it will fall at a distance 15.7 m from its initial position

5 0
2 years ago
A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along
babymother [125]

Answer:

Explanation:

height of pole = 15 ft

height of man = 6 ft

Let the length of shadow is y .

According to the diagram

Let at any time the distance of man is x.

The two triangles are similar

\frac{y-x}{y}=\frac{6}{15}

15 y - 15 x = 6 y

9 y = 15 x

y=\frac{5}{3}x

Differentiate with respect to time.

\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}

As given, dx/dt = 4 ft/s

\frac{dy}{dt}=\frac{5}{3}\times 4

\frac{dy}{dt}=\frac{20}{3} ft/s

6 0
2 years ago
Which of the following can be reduced to a single number in standard form?
raketka [301]

Complete question is;

Which of the following can be reduced to a single number in standard form?

A) 3√3 + 5√8

B) 2√5 + 5√45

C) √7 + √9

D) 4√2 + 3√6

Answer:

Only option B) 2√5 + 5√45 can be reduced to a single number

Explanation:

A) For 3√3 + 5√8;

Let's simplify it to get;

3√3 + 5√(4 × 2)

From this, we get;

3√3 + (5 × 2)√2 = 3√3 + 10√2

This is 2 numbers and not a single number. Thus it can't be reduced to a single number in standard form.

B) 2√5 + 5√45

Simplifying to get;

2√5 + 5√(9 × 5)

This gives;

2√5 + (5 × 3)√5 = 2√5 + 15√5

Adding the surds gives;

17√5.

This is a single number and thus can be reduced to a single number

C) For √7 + √9

Simplifying, to get;

√7 + 3.

This is 2 numbers and not a single number. Thus it can't be reduced to a single number in standard form.

D) 4√2 + 3√6

Thus can't be simplified further because both numbers inside the square root don't have factors that are perfect squares.

Thus, it remains 2 numbers and not a single number and can't be reduced to a single number in standard form.

6 0
2 years ago
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