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NeTakaya
3 years ago
7

what is the momentum of a child and a wagon if the total mass of the child and wagon is 22kg and the velocity is 1.5m/s?

Physics
1 answer:
antoniya [11.8K]3 years ago
7 0

Answer:

33 kg m/s

Explanation:

The momentum of an object is given by:

p=mv

where

m is the mass of the object

v is the velocity of the object

In this problem, the total mass of the child and the wagon is m =22 kg, while the velocity is v = 1.5 m/s, therefore the momentum is

p=mv=(22 kg)(1.5 m/s)=33 kg m/s

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Answer:The object characteristics

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Light of wavelength 550 nm comes into a thin slit and produces a diffraction pattern on a board 8.0 m away. The first minimum da
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Answer:

Width of the slit will be equal to 1.47 mm

Explanation:

We have given wavelength of the light \lambda =550nm=550\times 10^{-9}m

Distance D = 8 m

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So x=3\times 10^{-3}m

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For the first order wavelength is equal to \lambda =\frac{x}{D}\times a, here a width of slit

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3 years ago
what is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 3 m/sec
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Data:
m (mass) = 1 Kg
s (speed) = 3 m/s
Kinetic energy = ? (Joule)


Formula (Kinetic energy)
E_{k} =  \frac{m*s^2}{2}

Solving:
E_{k} = \frac{m*s^2}{2}
E_{k} =  \frac{1*3^2}{2}
E_{k} =  \frac{1*9}{2}
E_{k} =  \frac{9}{2}
\boxed{\boxed{E_{k} = 4.5\:J}}\end{array}}\qquad\quad\checkmark
3 0
3 years ago
Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
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Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

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d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

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k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

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x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

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F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

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