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gizmo_the_mogwai [7]
3 years ago
5

What is the key point from the kennedy-nixon debate​

Physics
1 answer:
3241004551 [841]3 years ago
4 0

Answer:

The key point was that Kennedy challenged Nixon to a series of televised debates. It was the first televised presidential debate in American history.

In 1960, 88 % of American homes had television. About 2/3 of the electorate watched the first debate on TV. Nixon was recovering from a knee injury, he looked drained. Kennedy, meanwhile, had been resting in a hotel for an entire weekend, he looked tan and confident.

Most Americans watching the debates voted for Kennedy, most radio listeners seemed to give the edge to Nixon. hope this helps

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A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in ou
elena55 [62]

Answer:

3.2\times 10^{-7}\ m or 0.32 μm.

Explanation:

Given:

The radiations are UV radiation.

The frequency of the radiations absorbed (f) = 9.38\times 10^{14}\ Hz

The wavelength of the radiations absorbed (λ) = ?

We know that, the speed of ultraviolet radiations is same as speed of light.

So, speed of UV radiation (v) = 3\times 10^8\ m/s

Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

v=f\lambda

Now, expressing the above equation in terms of wavelength 'λ', we have:

\lambda=\frac{v}{f}

Now, plug in the given values and solve for 'λ'. This gives,

\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m

Therefore, the wavelength of the radiations absorbed by the ozone is nearly 3.2\times 10^{-7}\ m or 0.32 μm.

7 0
3 years ago
What is the answer to these questions?
Oksana_A [137]

Answer:

THES IS NOT

Explanation:

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7 0
2 years ago
In preparation for the final exam, our astronomy study group has reconvened to discuss Mercury's unique orbital properties. They
grigory [225]

Answer:

The only incorrect statement is from student B

Explanation:

The planet mercury has a period of revolution of 58.7 Earth days and a rotation period around the sun of 87 days 23 ha, approximately 88 Earth days.

Let's examine student claims using these rotation periods

Student A. The time for 4 turns around the sun is

           t = 4 88

           t = 352 / 58.7 Earth days

In this time I make as many rotations on itself each one with a time to = 58.7 Earth days

           #_rotaciones = t / to

           #_rotations = 352 / 58.7

           #_rotations = 6

therefore this statement is TRUE

student B. the planet rotates 6 times around the Sun

          t = 6 88

          t = 528 s

The number of rotations on itself is

           #_rotaciones = t / to

           #_rotations = 528 / 58.7

           #_rotations = 9

False, turn 9 times

Student C. 8 turns around the sun

           t = 8 88

           t = 704 days

the number of turns on itself is

            #_rotaciones = t / to

            #_rotations = 704 / 58.7

            #_rotations = 12

True

The only incorrect statement is from student B

6 0
2 years ago
A car goes from rest to a velocity of 108 km/h north in 10s what is the car's acceleration in m/s2
fomenos

initial velocity of the car given as

v_i = 0

final velocity is given as

v_f = 108 km/h

as we know that

1 km/h = 0.277 m/s

now we can convert final speed into m/s

v_f = 108 * 0.277 = 30 m/s

now acceleration is rate of change in velocity

a = \frac{v_f - v_i}{t}

a = \frac{30 - 0}{10}

a = 3 m/s^2

so the acceleration of the car is 3 m/s^2

7 0
3 years ago
A wave oscillates 5.0 times a second and has a speed of 4.0m/s what is the frequency of this wave
Viefleur [7K]

Answer:

The frequency of the wave is 5.0Hz

Explanation:

The frequency of a wave is the number of oscillations or revolutions made in a second.

7 0
2 years ago
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