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4 years ago

If the sun was in space why is space dark?

1 answer:

8 0

At night, when that part of Earth is facing away from the Sun, space looks blackbecause there is no nearby bright source of light, like the Sun, to be scattered. If you were on the Moon, which has no atmosphere, the sky would be black both night and day.

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According to the Law of Reflection, the angle of incidence the angle of reflection. O A. is greater than B. is less than C. equa

Amanda [17]

Answer:

C. Equals

Explanation:

Law of reflection Equals the angle of incidence

8 0

3 years ago

Orchestra instruments are commonly tuned to match an A-note played by the principal oboe. The Baltimore Symphony Orchestra tunes

Nina [5.8K]

Answer:

Δλ = 3*10⁻³ m.

Explanation:

At any wave, there exists a fixed relationship between the speed of the wave, the wavelength, and the frequency, as follows:

$v = \lambda* f (1)$

where v is the speed, λ is the wavelength and f is the frequency.

Rearranging terms, we can get λ from the other two parameters, as follows:

$\lambda = \frac{v}{f} (2)$

Since v is constant for sound at 343 m/s, we can find the different wavelengths at different frequencies, as follows:

$\lambda_{1} =\frac{v}{f_{1}} = \frac{343m/s}{440(1/s)} = 0.779 m (3)$

$\lambda_{2} =\frac{v}{f_{2}} = \frac{343m/s}{442(1/s)} = 0.776 m (4)$

The difference between both wavelengths, is just the difference between (3) and (4):

$\Delta \lambda = \lambda_{1} - \lambda_{2} = 0.779 m - 0.776m = 3e-3 m (5)$

⇒ Δλ = 3*10⁻³ m.

6 0

3 years ago

find the mass of a ball on a roof 30 meters high, if the ball's gravitational potential energy is 58.8 joules

Sidana [21]

We are given the gravitational potential energy and the height of the ball and is asked in the problem to determine the mass of the ball. the formula to be followed is PE = mgh where g is the gravitational acceleration equal to 9.81 m/s^2. substituting, 58.8 J = m*9.8 m/s^2 * 30 m; m = 0.2 kg.

3 0

3 years ago

Read 2 more answers

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

Mariana [72]

a) $F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

$R-Fsin \theta - mg = 0$ (1)

where

R is the normal reaction of the floor (upward)

$F sin \theta$ is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

$F cos \theta - \mu_k R = 0$ (2)

where

$F cos \theta$ is the horizontal component of the force F (forward)

$\mu_k R$ is the force of friction (backward)

From (1) we get

$R=Fsin \theta +mg$

And substituting into (2)

$F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

b) $\mu_s=cot(\theta)$

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

$R-Fsin \theta - mg = 0$ (1)

$F cos \theta - \mu_s R = 0$ (2)

In this second case, we want to find the critical value of $\mu_s$ such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, $F cos \theta$ .

Therefore, using the same procedure as before,

$R=Fsin \theta +mg$

$F cos \theta - \mu_s (Fsin \theta +mg) = 0$

And solving for $\mu_s$ ,

$F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}$

Now we analyze the expression that we found. We notice that if the force applied F is very large, $F sin \theta >> mg$ , therefore we can rewrite the expression as

$\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)$

So, this is the critical value of the coefficient of static friction.

8 0

3 years ago

A tetrameric protein dissociates into dimers when the detergent sodium dodecyl sulfate (SDS) is added to a solution of the prote

KatRina [158]

Answer:

At the dimer-dimer interface there might be acting non-covalent forces (van der waals, Hidrogene bridges, hydrophobic forces)

At the monomer-monomer interface there might be covalent forces acting (disulfide bridges).

Explanation:

On the SDS-PAGE application works by disrupting non-covalent bonds in the proteins, and so denaturing them. Therefore, the disulfide bridges won´t be disrupted, so the monomers will remain bounded.

5 0

4 years ago