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V125BC [204]
3 years ago
8

An air craft heads north at 320 km/hr relative to the wind. the wind velocity is 80km/hr from the north. find the relative veloc

ity of the air craft ralative to the ground
​
Physics
1 answer:
Gnoma [55]3 years ago
7 0

Answer:

Relative to the ground, the velocity of the aircraft is 240 km/hr

Explanation:

Relative velocity is different from normal velocity;

When 2 objects are moving in opposite directions towards each other, they will appear to be faster than they actually are;

This is known as the relative velocity;

The information tells us we have the aircraft moving 320 km/hr northwards relative to the wind;

The wind is in the opposite direction at 80 km/hr;

R = relative velocity of the aircraft

v = actual velocity of the aircraft

w = velocity of the wind

R = v + w

Note: if the wind was moving in the same direction, the formula would be R = v - w

320 = v + 80

v = 320 - 80

v = 240

The velocity relative to the ground is simply the actual velocity as the ground doesn't move;

So, relative to the ground, the velocity of the aircraft is simply 240 km/hr

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Explanation:

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8 0
3 years ago
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A skydiver jumps from an airplane that is moving at 50 meters per second at a height of 1,000 meters what describes the skydiver
kkurt [141]

Answer:

11060M  Joules, where M is the mass of the diver in kg

Explanation:

Mass of the skydiver missing, we're assuming it's M.

It's total energy is the sum of the contribution of his kinetic energy (K)- since he's moving at 50 m/s, and it's potential energy (U), since he's subject to earth gravity.

Energy is the sum of the two, so E = K+U= \frac 12 M v^2 + Mgh = M (\frac 12 \cdot 50^2 + 9.81\cdot 1000) = M ( 1250 + 9810) = 11060\cdot M

7 0
2 years ago
How do u do 5? Show me plz
nikitadnepr [17]
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6 0
3 years ago
17. A volleyball weighs about 300 grams.
atroni [7]

Answer:

PE = 44.1 J

Explanation:

Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)

We have:

  • 1 kilograms = 1000 grams.

We convert it using a rule of 3, replacing, simplifying units and solving:

  • \boxed{\bold{x=\frac{gr*1\ kg}{1000\ gr}=\frac{300\ gr*1\ kg}{1000\ gr}=\frac{300\ kg}{1000}=\boxed{\bold{0.3\ kg}}}}

==================================================================

Earth's gravity is known to be 9.8 m/s², so we have:

Data:

  • m = 0.3 kg
  • g = 9.8 m/s²
  • h = 15 m
  • PE = ?

Use formula of potencial energy:

  • \boxed{\bold{PE=m*g*h}}

Replace and solve:

  • \boxed{\bold{PE=0.3\ kg*9.8\frac{m}{s^{2}}*15\ m}}
  • \boxed{\boxed{\bold{PE=44.1\ J}}}

Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.

The potential energy of the volleyball is <u>44.1 Joules.</u>

Greetings.

8 0
3 years ago
A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm
andrew11 [14]

Answer:

2.8 cm

Explanation:

y_1 = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm

Fringe width is also given by

\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}

For second order

\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1

Distance between two second order minima is given by

y_2=2\beta_2

\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm

The distance between the two second order minima is 2.8 cm

8 0
3 years ago
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