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3 years ago

Which two spheres are interacting when there is a rainbow ? Please help make you brainiest.

1 answer:

7 0

Answer: Honestly miss im not sure but i am pretty sure it is The HYDROSPHERE AND THE GEOSPEHER. I figured it out. Pls give me brainlest!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Explanation:

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a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleratio

Harman [31]

Answer:

The angular acceleration is $\alpha = 3.235 \ rad/s ^2$

Explanation:

From the question we are told that

The moment of inertia is $I = 0.034\ kg \cdot m^2$

The net torque is $\tau = 0.11\ N \cdot m$

Generally the net torque is mathematically represented as

$\tau = I * \alpha$

Where $\alpha$ is the angular acceleration so

$\alpha = \frac{\tau }{I}$

substituting values

$\alpha = \frac{0.1 1}{ 0.034}$

$\alpha = 3.235 \ rad/s ^2$

6 0

3 years ago

The seeds were sown (change the voice)

MaRussiya [10]

Answer:

He sowed the seeds

Explanation:

While the question should have given the person who did the sowing for example the seeds were sown by him/ her/ the farmer/ or any name. Therefore, voice is given in passive and to change passive voice to active voice then the sentence will read as follows assuming that the seeds were sown by him

He sowed the seeds

7 0

2 years ago

What is the sound intensity level in a car when the sound intensity is 0.525 μW/m2 ? Use I0 = 1.00×10−12 W/m2 for the reference

never [62]

Answer:

The sound intensity level in the car is 57.2 dB.

Explanation:

Sound intensity level in decibels, β = 10 log (I/I₀); where I = 0.525 × 10⁻⁶ W/m², I₀ = 1.0 × 10⁻¹² W/m²

β (dB) = 10 log ((0.525 × 10⁻⁶)/(1.0 × 10⁻¹²)) = 10 × 5.72 = 57.2 dB

Hope this Helps!!!

8 0

3 years ago

The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 459 nm. What is t

spin [16.1K]

Answer:

2.7067 eV

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

$\lambda_0$ = Threshold wavelength = 459 nm

Work function is given by

$W_0=\frac{hc}{\lambda_0}\\\Rightarrow W_0=\frac{6.626\times 10^{-34}\times 3\times 10^8}{459\times 10^{-9}}\\\Rightarrow W_0=4.33072\times 10^{-19}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$4.33072\times 10^{-19}\ J=4.33072\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}\ eV=2.7067\ eV$

The work function W0 of this metal is 2.7067 eV

4 0

3 years ago

An airplane with an airspeed of 120 km/h has a heading of 30 degree east of North in a wind that is blowing toward the east at 6

lesya692 [45]

Answer:

Explanation:

Velocity of plane relative to ground V_pg = ?

Given the velocity in vector form ,

velocity of plane relative to air V_pw = 120 cos30 i + 120sin30j

V_wg = 60 i

V_pg = V_pw +V_wg

= 120 cos30 i + 120sin30j + 60i

= 164 i + 60 j

magnitude

=251 km / h

8 0

3 years ago