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inessss [21]
2 years ago
14

A plane initially traveling at 200 m/s due west experiences a 10 m/s head wind coming from the opposite direction. A). What will

be the relative velocity of the plane from the frame of reference of a bystander in the ground looking at the plane when it experiences the head wind current? B) Will the plane be late, on time or early after the wind current?
Physics
1 answer:
Hitman42 [59]2 years ago
8 0
Ok so it would be late and the relative velocity would be 190 m/s because 200 m/s - 10 m/s is 190 m/s. Hope this helps.
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Two satellites revolve around the Earth. Satellite A has mass m and has an orbit of radius r. Satellite B has mass 6m and an orb
melomori [17]

Answer:

aaaaa

Explanation:

M = Mass of the Earth

m = Mass of satellite

r = Radius of satellite

G = Gravitational constant

F=G\frac{Mm}{r^2}

F=G\frac{M6m}{r_b^2}

G\frac{Mm}{r^2}=G\frac{M6m}{r_b^2}\\\Rightarrow \frac{1}{r^2}=\frac{6}{r_b^2}\\\Rightarrow \frac{r_b^2}{r^2}=6\\\Rightarrow \frac{r_b}{r}=\sqrt{6}\\\Rightarrow r_b=2.44948r

r_b=2.44948r

8 0
2 years ago
As a blacksmith heats a piece of iron, the iron glows red, then yellow, then white. The iron provides a demonstration of which p
Nitella [24]
The answer is D. Blackbody radiation. The piece of iron glows red because its temperature is around 1000 K, then yellow because its temperature is around 2800 K, and then white because its temperature is around 5500K. This shows that the spectrum of the radiation is determined by absolute temperature, as when the temperature of a blackbody radiator increases, the peak of the radiation curve moves to shorter wavelengths. 
6 0
2 years ago
11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0
Greeley [361]

Answer:

Explanation:

position of centre of mass of door from surface of water

= 10 + 1.1 / 2

= 10.55 m

Pressure on centre of mass

atmospheric pressure + pressure due to water column

10 ⁵ + hdg

= 10⁵ + 10.55 x 1000 x 9.8

= 2.0339 x 10⁵ Pa

the net force acting on the door (normal to its surface)

= pressure at the centre x area of the door

= .9 x 1.1 x 2.0339 x 10⁵

= 2.01356 x 10⁵ N

pressure centre will be at 10.55 m below the surface.

When the car is filled with air or  it is filled with water , in both the cases pressure centre will lie at the centre of the car .

7 0
2 years ago
Corey, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 850 N for the first 3.0 s after the eleva
lyudmila [28]

Answer:

v₂ = 2.568 m/s

Explanation:

given,

mass of Corey = 95 Kg

reading of sale for first 3 s when elevator start to move = 850 N

scale reading for the next 3.0 s = 930 N

Gravitation force acting =

  F = m g

  F = 95 x 9.8

  F = 931 N

using newtons second law, due to movement of elevator

    F_{net} = m a

 W - N = m a₁

931- 850 = 95 x a₁

    a₁ = 0.852 m/s²

now,

velocity calculation

v₁ = a₁t

v₁ = 0.852 x 3 = 2.557 m/s

now, For second case

931 - 930 = 95 x a₂

a₂ = 0.011 m/s²

now, velocity after 4 s

v₂ = v₁ + a₂ t

v₂ = 2.557+ 0.011 x (4 - 3)

(4-3) because velocity after 3 second is calculate we need to calculate velocity after 4 s from beginning.

v₂ = 2.557 + 0.011

v₂ = 2.568 m/s

velocity of the elevator is equal to v₂ = 2.568 m/s

7 0
2 years ago
The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

5 0
2 years ago
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