A plane, diving with constant speed at an angle of 53.0° with the vertical, releases a projectile at an altitude of730 m. The pr
ojectile hits the ground 5.00 s after release. (a) What is the speed of the plane? (b) How far doesthe projectile travel horizontally during its flight? What are the (c) horizontal and (d) vertical components of itsvelocity just before striking the ground?
1 answer:
Answer:
A) V=198.75m/s
B) 793.7mts
C) Vx=158.72m/s
D) Vy=-168.71m/s
Explanation:
In order to answer the first question we need to calculate the initial speed of the projectile because it's the same speed of the plane.
The velocity has to be negative because is going down so, Voy=-119.61m/s
now that we have the Y component of velocity we can get the plane velocity by:
the angle with the horizontal is ∝=-90+53=-37
Now that we have the Initial velocity we can calculate the horizontal displacement:
Because the horizontal velocity remains the same, it is given by:
In order the calculate the Y componet we will use the next formula:
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Answer:
10.2 Watt
Explanation:
= number of turns in flat coil = 160
= area = 0.20 m²
B₀= initial magnetic field = 0.40 T
= final magnetic field = - 0.40 T
Change in magnetic field is given as
ΔB = B - B₀ = - 0.40 - 0.40 = - 0.80 T
= time taken for the magnetic field to change = 2.0 s
Induced emf is given as
= 12.8 volts
= Resistance of the coil = 16 Ω
Power is given as
= 10.2 Watt
1) In the first case, the correct answer is
<span>A.Wavelengths measured would match the actual wavelengths emitted.
In fact, the stars are not moving relative to Earth, so there is no shift in the measured wavelength.
2) In this second case, the correct answer is
</span><span>A.Wavelengths measured would be shorter than the actual wavelengths emitted.
</span>in fact, since the stars in this case are moving towards the Earth, then apparent frequency of their emitted light will be larger than the actual frequency, because of the Doppler effect, according to the formula:
where f0 is the actual frequency, f' the apparent frequency, c the speed of light and vs the velocity of the source (the stars) relative to the obsever (Earth). Vs is negative when the source is moving towards the observer, so the apparent frequency f' is larger than the actual frequency f0. But the wavelength is inversely proportional to the frequency, so the apparent wavelength will be shorter than the actual wavelength.
<span>A.Wavelengths measured would match the actual wavelengths emitted.
In fact, the stars are not moving relative to Earth, so there is no shift in the measured wavelength.
2) In this second case, the correct answer is
</span><span>A.Wavelengths measured would be shorter than the actual wavelengths emitted.
</span>in fact, since the stars in this case are moving towards the Earth, then apparent frequency of their emitted light will be larger than the actual frequency, because of the Doppler effect, according to the formula:
where f0 is the actual frequency, f' the apparent frequency, c the speed of light and vs the velocity of the source (the stars) relative to the obsever (Earth). Vs is negative when the source is moving towards the observer, so the apparent frequency f' is larger than the actual frequency f0. But the wavelength is inversely proportional to the frequency, so the apparent wavelength will be shorter than the actual wavelength.