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3 years ago

What is the mass of a cow moving at a velocity of 0.5 m/s with 25J of kinetic energy? Show work

Physics

2 answers:

Karo-lina-s [1.5K]3 years ago

5 0

Answer:

200 kg

Explanation:

Free_Kalibri [48]3 years ago

3 0

Answer:

m= 50

Explanation:

m=f/a this is the formula for mass

m= 25/0.5 you substitute the numbers with their variables they you divide them to get the mass

m= 50

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As viewed from our location, the stars of the Big Dipper canbe connected with imaginary lines to form the shape of a pot with ac

jasenka [17]

Answer: California

Explanation:

The Big Dipper has eight stars in it. Seven are visible at a glance, while the eighth is a visible double star that is just detectable with the naked eye in an area with clear "seeing" and with good vision.

To first observe an easily recognizable change in the shape formed by these stars (to an observer without a telescope), someone can look high in the sky toward the northeast at a certain time of the year and you can't miss the Big Dipper at around 8 p.m from mid-northern latitudes which is California

5 0

4 years ago

Coherent light that contains two wavelengths, 660 nm (red) and 470 nm (blue), passes through two narrow slits that are separated

11111nata11111 [884]

Answer:

2.64 mm

Explanation:

We are given that

Distance between two slits=0.310= $0.31\times 10^{-3} m$

1 mm= $10^{-3} m$

Distance between slit and screen=4.3 m

Wavelength of red light = $\lambda_1=660 nm=660\times 10^{-9} m$

1 nm= $10^{-9}m$

Wavelength of blue light= $\lambda_2=470 nm=470\times 10^{-9} m$

We have to find the distance on the screen between the first order bright fringe for each wavelength.

We know that

The distance between the first order bright fringes on the screen is given by

$\Delta y=\frac{Rm}{d}\Delat \lambda$

Where

R=Distance between screen and slits

m=Order of fringe=1

d=distance between two slits

$\Delta \lambda=$ Difference in wavelength of two light source

Substitute the values then we get

Distance between the first order bright fringes on the screen for two sources = $\frac{4.3\times 1}{0.31\times 10^{-3}}(660-470)\times 10^{-9}=2.6 \times 10^{-3} m=2.64 mm$