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sergij07 [2.7K]
2 years ago
6

Red, Yellow, Green, and Purple are in a physical fitness race, pulling different amounts of weights. Considering the equations f

or Newton’s second law (using force, mass, acceleration), who would MOST LIKELY win if each competitor pulls with a force of 570 N? Explain why.
• Red has 8 weights with a total mass of 200 kg.
• Yellow has 6 weights with a total mass of 100 kg.
• Green has 7 weights with a total mass of 250 kg.
• Purple has 10 weights with a total mass of 300 kg.
Physics
1 answer:
solong [7]2 years ago
4 0

Answer:

a

Explanation:

plz make me brainliest

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What holds the moon in place, orbiting around the Earth
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The Earth's gravity keeps the Moon orbiting us. It keeps changing the direction of the Moon's velocity. This means gravity makes the Moon accelerate all the time, even though its speed remains constant.

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Match the correct sentence together.
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I’m pretty sure you times them so 1 with A, 2 with e, 3 with C, and 4 with B
6 0
1 year ago
Read 2 more answers
A bucket of water of mass 10 kg is pulled at constant velocity up to a platform 30 meters above the ground. This takes 10 minute
Rudik [331]

Answer:

W = 2352 J

Explanation:

Given that:

  • mass of the bucket, M = 10 kg
  • velocity of pulling the bucket, v = 3m.min^{-1}
  • height of the platform, h = 30 m
  • time taken, t = 10 min
  • rate of loss of water-mass, m = 0.4 kg.min^{-1}

Here, according to the given situation the bucket moves at the rate,

v=3 m.min^{-1}

The mass varies with the time as,

M=(10-0.4t) kg

Consider the time interval between t and t + ∆t. During this time the bucket moves a distance

∆x =  3∆t meters

So, during this interval change in work done,

∆W = m.g∆x

<u>For work calculation:</u>

W=\int_{0}^{10} [(10-0.4t).g\times 3] dt

W= 3\times 9.8\times [10t-\frac{0.4t^{2}}{2}]^{10}_{0}

W= 2352 J

5 0
3 years ago
A thin beam of light of wavelength 50 μm (in the infrared portion of the em spectrum) goes through
Brut [27]

Solution :

Given :

Wavelength of the thin beam of light, λ = 50 μm

Distance of the screen from the slit, D = 3.00 m

Width of the fringe, Δy = ±8.24 mm

Therefore, width of the slit is given by :

$d=\frac{n \lambda D}{\Delta y}$

  $=\frac{2 \times 50 \times 10^{-9} \times 3}{2 \times 8.24 \times 10^{-3}}$

  = 0.000018203 m

  = 0.0182 mm

  = 0.018 mm

The intensity of light is given by :

$I=I_0\left(\frac{\sin \beta /2}{\beta/ 2}\right)^2$   , where $\beta=\frac{2 \pi D \sin \theta}{\lambda}$

$I=I_0\left(\frac{\sin \frac{\pi d \sin \theta}{\lambda}}{\frac{\pi d \sin \theta}{\lambda}}\right)^2$

$I=I_0\left(\frac{\sin \frac{\pi d y}{D\lambda}}{\frac{\pi d y}{D\lambda}}\right)^2$

Now, $\frac{dy}{D \lambda} = \frac{8.24 \times 10^{-3}\times 0.018 \times 10^{-3}}{4 \times 50\times 10^{-9}\times 4}$

               = 0.1854

               ≈ 0.18

$I=I_0\left(\frac{\sin 0.56}{0.56}\right)^2$

  $=I_0 \times 0.81$

  = 2  x0.81

  $= 1.62 \ W/m^2$

 

6 0
3 years ago
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