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k0ka [10]
4 years ago
11

How can magnet pick wooden fish

Physics
2 answers:
Korolek [52]4 years ago
5 0
It would have to have a magnet on the other side
lisov135 [29]4 years ago
4 0
If you have a peice of metal on the wooden fish then you will be abel to pick up the fish with the magnet hope this helped
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What is the mass of an object that requires 100N of force in order to accelerate 10m/s2
iragen [17]

Explanation:

Hey there!

Given;

Force (f) = 100 N

Acceleration (a) = 10m/s^2.

Mass (m)=?

We have;

Force = mass * acceleration.

So,

100 = m * 10

Or, 100 = 10m

Or, m= 100/10

Therefore, mass is 10kg.

<em><u>Hope</u></em><em><u> it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

3 0
3 years ago
What did Sarah Stewart enjoy reading as a kid
Alecsey [184]

Answer:

Science fiction novels and such.

Explanation:

3 0
3 years ago
What is the "plate boundry type" of north american plate and the pacific plate near california
Fittoniya [83]
The North American one is im pretty sure divergent boundary but no so 100% on it
8 0
3 years ago
A few years ago, Serena Williams dived to hit a tennis ball right after it bounced off the ground. The ball bounced on the groun
pshichka [43]

Answer:

The initial velocity of the ball was 20 m/s

Explanation:

Please, see the figure for a description of the problem.

The initial velocity vector can be written as follows:

v0 = (v0x, v0y)

where:

v0 = initial velocity

v0x = horizontal component of the initial velocity

v0y = vertical component of the initial velocity

The position and velocity of the ball at time "t" are described by the vector "r" and "v" respectively:

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

v = (v0x, v0y + g*t)

Where:

r = position vector of the ball

x0 = initial horizontal position

t = time

y0 = initial vertical position

g = acceleration due to gravity

v = velocity vector

Considering the center of our system of reference as the point at which the ball left Serena´s racket, x0 and y0 = 0.

We know that at a time t = 1.21 s the y-component of the position vector must be 0 (see "r final" in the figure). Then:

y0 + v0y * t + 1/2 * g * t² = 0          y0= 0

v0y * 1.21 s + 1/2 * (-9.8 m/s²) * (1.21 s)² = 0

v0y = -(1/2 * (-9.8 m/s²) * (1.21 s)²) / 1.21 s

v0y = 1/2 * 9.8 m/s² * 1.21 s

v0y = 5.93 m/s

If we see in the figure the trajectory of the ball if there had been no gravity ("s"), we will notice that it is a stright line with a slope of:

Δy/Δx = (0.95m(y) + 2.63m(y)) / 11.7 m(x) = 0.31 m(y) / m(x)

This slope means that the ball will go up 0.31 m for every meter it goes right.

Then, if initially the ball goes up 5.93 m every second, it will go right

(5.93 m(y) * (1 m(x) / 0.31 m(y)) = 19.1 m(x). Then, v0x = 19.1 m/s

The vector initial velocity will be:

v0 = (19.1 m/s, 5.93 m/s)

magnitude of v0 =|v0| = \sqrt{(19.1m/s)^{2}+(5.93m/s)^{2}}= 20.0 m/s

Another way to solve this is by using the equation for velocity:

We know that when the ball passes over the net, the vertical velocity is 0. Then, we can calculate the time at which the ball passes over the net and use that time to obtain v0x from the equation for position, since we know that at that time the x-component of the position is 11.7 m.

When the ball is over the net:

v0y + g*t = 0

t = -v0y/g = -5.93 m/s/-9.8 m/s² = 0.61 s.

Notice that, since the trajectory is a parabola, knowing the final time we could easily calculate the time at which the ball passes the net by dividing that final time by 2: 1.21 s / 2 = 0.61 s

Then, using this time in the equation for position:

v0x * t = 11.7 m

v0x = 11.7 m / 0.61 s = 19.2 m/s which is aproximately the same as the obtained above.

7 0
3 years ago
Imagine that you have obtained spectra for several galaxies and have measured the observed wavelength of a hydrogen emission lin
Anna71 [15]

Answer:

Galaxy 1:

z = 0.0056

Galaxy 2:

z = 0.014

Galaxy 3:

z = 0.040

Explanation:

Spectral lines will be shifted to the blue part of the spectrum¹ if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect). The source in this particular case is represented for each of the galaxies of interest.

Hence, the redshift represents this shift of the spectral lines to red part in the spectrum of a galaxy or any object which is moving away. That is a direct confirmation of how the universe is in an expanding accelerated motion.

The redshift can be defined in analytic way by means of the 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}})

\frac{v}{c} = \frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}}  

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

Where z is the redshift.

<em>For the case of Galaxy 1: </em>

Where \lambda_{measured} = 660.0 nm and \lambda_{0} = 656.3 nm.

z = \frac{\lambda_{measured}-\lambda_{0}}{\lambda_{0}}

z = (\frac{660.0 nm - 656.3 nm}{656.3 nm})

z = 0.0056

<em>For the case of Galaxy 2: </em>

Where \lambda_{measured} = 665.8 nm and \lambda_{0} = 656.3 nm.

z = \frac{665.8 nm - 656.3 nm}{656.3 nm}

z = 0.014

<em>For the case of Galaxy 3:</em>

       

Where \lambda_{measured} = 682.7 nm and \lambda_{0} = 656.3 nm.

z = \frac{682.7 nm - 656.3 nm}{656.3 nm}

z = 0.040

Key terms:

¹Spectrum: Decomposition of light in its characteristic colors (wavelengths).

8 0
3 years ago
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