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Volgvan
3 years ago
8

What are the colors that have lower energy than the color blue

Physics
1 answer:
lubasha [3.4K]3 years ago
3 0
The colour that have lower energy than blue colour is colour red.
There is also a reason behind it.
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Which of the following is a<br> fossil fuel?<br> A. Coal<br> B. Wind energy<br> C. Biomass energy
almond37 [142]

A. Coal

is the correct answer

5 0
3 years ago
Read 2 more answers
A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif
Serggg [28]

Answer:

Height of the rocket be one minute after liftoff is 40.1382 km.

Explanation:

v(t)=-gt-v_e\times \ln \frac{m-rt}{m}

v = velocity of rocket at time t

g = Acceleration due to gravity =9.8 m/s^2

v_e = Constant velocity relative to the rocket = 2,900m/s.

m = Initial mass of the rocket at liftoff = 29000 kg

r = Rate at which fuel is consumed = 170 kg/s

Velocity of the rocket after 1 minute of the liftoff =v

t = 1 minute = 60 seconds'

Substituting all the given values in in the given equation:

v(60)=-9.8 m/s^2\times 60 s-2,900m/s\times \ln (\frac{29,000 kg-170 kg/s\times 60 s}{2,9000 kg})

v(60) = 668.97 m/s

Height of the rocket = h

Velocity=\frac{Displacement}{time}

668.97 m/s=\frac{h}{60 s}

h=668.97 m/s\times 60 s=40,138.2 m = 40.1382 km

Height of the rocket be one minute after liftoff is 40.1382 km.

4 0
3 years ago
Is this snow or water
fredd [130]
I think its snow. Yup.
4 0
3 years ago
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When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

  = 4.59\times 10^{-19} \ J

or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

  = 2.87 \ eV

(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

5 0
3 years ago
A single circular loop of wire of radius 0.75 m carries a constant current of 3.0 A. The loop may be rotated about an axis that
NISA [10]

Answer:

B = 0.8 T

Explanation:

It is given that,

Radius of circular loop, r = 0.75 m

Current in the loop, I = 3 A

The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.

When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.

We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :

\tau=NIAB\sin\theta

B is magnetic field

B=\dfrac{\tau}{NIA\sin\theta}\\\\B=\dfrac{1.8}{1\times \pi \times (0.75)^2\times 3\times \sin(25)}\\\\B=0.8\ T

So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.

6 0
3 years ago
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