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Soloha48 [4]
3 years ago
6

If a net force of 100 N is applied to an object with a mass of 10 kg what will be the acceleration experienced by the mass? Show

work.
Physics
1 answer:
dusya [7]3 years ago
3 0

Explanation:

∑F = ma

100 N = (10 kg) a

a = 10 m/s²

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A rubber bullet of mass m=0.025\,\mathrm{kg}m=0.025kg traveling at velocity v_0 = 50\,\mathrm{m/s}v 0 ​ =50m/s hits an iron bloc
Pavel [41]

Answer: 0.001 m

Explanation:

In order to get the máximum height reached by the iron block, we can use the energy conservation principle, as all the kinetic energy impressed upon the iron blck by the rubber bullet during the collision, becomes gravitational potential energy, as follows:

½ m v2 = m. g. h (1)

We don´t know the value of v, but if we look to the collision, and we asume no external forces act during it, the total momentum must be conserved.

The initial momentum, as the block is at rest, is just the one due to the rubber bullet:

P1 = mb . vb = 0.025 kg. 50 m/s = 1.25 kg. m/s

The final momemtum, is just the sum of the one due to the bullet (after being bounced back) and the one for the iron block:

P2 = mb . vfb  + mib . vib= 0.025 Kg. (-35 m/s) + 15 Kg.vib

As we have already said, P1 = P2, so we can write the following equation:

0.025 Kg. (-35 m/s) + 15 Kg. vib = 1.25 Kg. m/s.

Solving for vib, we have:

vib = 0.14 m/s

Now, we can replace this value in the equation (1) above:

½ . 15 Kg. (0.14)2  (m/s)2 = 15 Kg. 9.8 m/s2. H

Solving for H, we have:

H = 0.001 m

3 0
3 years ago
The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig
Mariulka [41]

Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

  • the upper and the lower channels are at the same pressure (the atmospheric pressure).
  • the velocity of water in the upper channel is zero (v₁ = 0 m/s).
  • y₁ = 2.00 m  (height of the upper channel)
  • y₂ = 0.00 m  (height of the lower channel)
  • g = 9.81 m/s²
  • ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A  ⇒   A = Q/v ⇒   A = Q/v₂

⇒   A = (350 m³/s)/(6.264 m/s)

⇒   A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4   ⇒   D = 2*√(A/π)

⇒   D = 2*√(55.873 m²/π)

⇒   D = 8.434 m ≈ 8.45 m

Now, the length of the pipe can be obtained as follows

L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

6 0
2 years ago
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What r three positive results of a variation within a population that occur due to natural selection
Phoenix [80]

Answer:

Traits, evolution, adaptive

4 0
3 years ago
What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne
irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance h= 1790 km = 1.790\times10^{6}\ m

Magnetic field B=4\times10^{-8}\ T

Mass of proton m=1.673\times10^{-21}\ Kg

Radius of earth R =6.38\times10^{6}\ m

Radius of orbit r=R+h

r=6.38\times10^{6}+1.790\times10^{6}

r=8170000\ m

We need to calculate the speed

Using formula of magnetic field

Bvq=\dfrac{mv^2}{r}

v=\dfrac{Bqr}{m}

Put the value into the formula

v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}

v=31.25\ m/s

Hence, The velocity is 31.25 m/s and direction is toward west.

6 0
3 years ago
What is the massof the largest ruby?
alexandr402 [8]
I think the answer is 2283g
4 0
3 years ago
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