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otez555 [7]
3 years ago
5

What two traits must a viable hypothesis have

Physics
1 answer:
I am Lyosha [343]3 years ago
8 0
<span>The hypothesis must be testable and falsifiable. The first stipulation is that the hypothesis must be able to be tested and be phrased as such. In addition, the second stipulation holds that the statement must be able to be falsified; that is, the statement can be showed to not be the case.</span>
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At a certain location, Earth has a magnetic field of 0.60 ✕ 10−4 T, pointing 75° below the horizontal in a north-south plane. A
saveliy_v [14]

Answer with Explanation:

We are given that

Magnetic field,B=0.6\times 10^{-4} T

\theta=75^{\circ}

Length of wire,l=15 m

Current,I=19 A

a.We have to find the magnitude of magnetic force and direction of magnetic force.

Magnetic force,F=IBlsin\theta

Using the formula

F=0.6\times 10^{-4}\times 15\times 19sin75

F=16.5\times 10^{-3} N

Direction=tan\theta=cot(90-75)=tan15^{\circ}

\theta=15^{\circ}

15 degree above the horizontal  in the northward direction.

5 0
3 years ago
PLEASE HELP DUE BEFORE 11:30 TODAY!!!!
user100 [1]

Answer:

D is not the a vector quantities

4 0
3 years ago
What is the average velocity of an arrow that travels 12 m [E] in 0.15 s?
sergij07 [2.7K]
Velocity is defined as Distance divided by Time.
In other words, V = D/T.

Now that we have our formula, we can solve.
Let's plug in the numbers we have.

We have 12m [East (direction not necessary when solving yet)] for our distance, and 0.15s as our time.

Divide the distance (12 /) by the time (0.15)
12 / 0.15 = 80.

Your velocity is 80 m/s [E]

I hope this helps!
6 0
3 years ago
In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this
soldier1979 [14.2K]

Answer:

1). v = - 2960526m/s

2). Toward us

3). v = - 493421m/s

4). Toward us

5). v = 1480263m/s

6).  Away from us

7). v = 3207236m/s

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm (\lambda_{0} = 121.6nm)

Redshift: \lambda_{measured} > \lambda_{0}

Blueshift: \lambda_{measured} < \lambda_{0}

Then, for this particular case it is gotten:

Star 1: \lambda_{measured} = 120.4nm

Star 2: \lambda_{measured} = 121.4nm

Star 3: \lambda_{measured} = 122.2nm

Star 4: \lambda_{measured} = 122.9nm

Star 1:

Blueshift: 120.4nm < 121.6nm

Toward us

Star 2:

Blueshift: 121.4nm < 121.6nm

Toward us

Star 3:

Redshift: 122.2nm > 121.6nm

Away from us

Star 4:

Redshift: 122.9nm > 121.6nm

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

v = c\frac{\Delta \lambda}{\lambda_{0}}  (1)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the wavelength at rest, v is the velocity of the source and c is the speed of light.

v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}}) (2)

<em>Case for star 1 \lambda_{measured} = 120.4 nm:</em>

<em></em>

v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})

v = - 2960526m/s

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 \lambda_{measured} = 121.4 nm:</em>

v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})

v = - 493421m/s

<em>Case for star 3 \lambda_{measured} = 122.2 nm:</em>

v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})

v = 1480263m/s

<em>Case for star 4 \lambda_{measured} = 122.9 nm:</em>

v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})

v = 3207236m/s

4 0
3 years ago
A ball is thrown w a speed of 30m/s at an angle of 10. When is the vertical component of the velocity equal to zero
irga5000 [103]

Now the vertical velocity of the ball thrown at an angle 10° is given as

Voy(initial vertical velocity)= 30m/s x sin 10

Voy(initial vertical velocity)= 5.2m/s

Now the ball is decelerating with an acceleration due to gravity equivalent to 9.8m/s^2.

Let Vy be the final velocity and that is equal to zero in this case.

Now

Vy= Voy- tx9.8

Where t is the time at which the vertical velocity becomes 0.

Substituting the values we get

0= 5.2-tx9.8

9.8t=5.2

t=0.53 secs



5 0
2 years ago
Read 2 more answers
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