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emmainna [20.7K]

3 years ago

9

Based electromagnetic spectrum? on its surface temperature of 6,000 K, most photons that leave the Sun's surface lie in which re

gion of the________
A) microwave
B) infrared
C) visible
D) ultraviolet
E) X-ray

1 answer:

masha68 [24]3 years ago

7 0

Answer:

option C

Explanation:

given,

Temperature of the surface = 6000 K

using formula to calculate $\lambda_{max}$

$\lambda_{max} = \dfrac{0.0029\ K.m}{T}$

$\lambda_{max}$ is the maximum wavelength emission in meter.

T is the temperature in kelvin.

$\lambda_{max} = \dfrac{0.0029}{6000}$

$\lambda_{max} = \dfrac{0.0029}{6000}$

$\lambda_{max} = 4.8333 \times 10^{-7}$

$\lambda_{max} = 4833\ A^0$

the wavelength is in Visible range

The Correct answer is option C

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A charge is moving in a magnetic field that points to the left. What direction can the charge move and experience no magnetic fo

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The answer is left and right.

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Why did the “yielders” conform in Asch’s experiment?

svet-max [94.6K]

Answer:

Asch's experiment showed that about 75% of people were "yielders" who conformed and 25% were "independent" who didn't conform. Asch concludes that people ignored reality and gave an incorrect answer in order to follow the rest of the group.

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3 years ago

A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff

strojnjashka [21]

(a) $6.43\cdot 10^5 J$

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

$E=K+U$

The initial kinetic energy is:

$K=\frac{1}{2}mv^2$

where m = 58.0 kg is the mass of the projectile and $v=140 m/s$ is the initial speed. Substituting,

$K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J$

The initial potential energy is given by

$U=mgh$

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

$U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J$

So, the initial mechanical energy is

$E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J$

(b) $-1.67 \cdot 10^5 J$

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

$K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J$

while the potential energy is

$U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J$

So, the mechanical energy is

$E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J$

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

$W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J$

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

$W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J$

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

$E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J$

And this is only kinetic energy:

$E=K=\frac{1}{2}mv^2$

So, we can solve to find the final speed:

$v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s$

4 0

3 years ago

Which is true regarding diffusion and osmosis? Question 3 options:

Scilla [17]

Answer:

Option (A)

Explanation:

Diffusion is defined as the transfer of particles such as ions or molecules from a higher concentration region to a lower concentration region.

Osmosis is a special type of diffusion in which the solvents such as water are allowed to move from a region of higher to a lower concentration through a layer of a semi-permeable membrane. This is an energy-independent process and occurs along the concentration gradient.

Thus, the correct answer is option (A) that is true for both osmosis and diffusion.

4 0

3 years ago

A car tire rotates with an average angular speed of 29 rad/s. In what time interval will the tire rotate 3.5 times?

meriva

Answer:

"0.758315 sec" is the appropriate choice.

Explanation:

The given values are:

Angular speed,

= 29 rad/s

or,

= $\frac{29}{2 \pi} \ rps$

Tire rotates,

= 3.5 times

Now,

The time interval will be:

⇒ $\Delta t=\frac{3.5}{\frac{29}{2 \pi} }$

⇒ $=\frac{2 \pi\times 3.5}{29}$

⇒ $=\frac{21.99}{29}$

⇒ $=0.758315 \ sec$

8 0

3 years ago