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masya89 [10]
3 years ago
13

The force of friction between an object and the surface upon which it is sliding is 14N and the coefficient of friction between

them is 0.72. What is the weight of the object? ​
Physics
1 answer:
Nat2105 [25]3 years ago
5 0

Answer:

14 x 0.27 = 3.78 is your answer

Explanation:

the question is asking for the weight of the object so you multiply and get 3.79

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viktelen [127]

Given:

Momentum of the dog (p) = 120.5 kg m/s

Speed of the dog (v) = 5 m/s

To Find:

Mass of the dog (m)

Concept/Theory:

\underline{\underline{ \bf{\Large{Momentum}}}}

  • It is defined as the quantity of motion contained in a body.
  • It is measured as the product of mass of the body and it's speed.
  • It is represented by p.
  • It's SI unit is kg m/s
  • Mathematical Representation/Equation of Momentum: \boxed{ \bf{p = mv}}

Answer:

By using equation of momentum, we get:

\rm \longrightarrow m =  \dfrac{p}{v}  \\  \\  \rm \longrightarrow m =  \dfrac{120.5}{5}  \\  \\  \rm \longrightarrow m = 24.1 \: kg

\therefore Mass of the dog (m) = 24.1 kg

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2 years ago
How much work is required for a 74 kg sprinter to accelerate from rest to 2.2 m/s?
My name is Ann [436]

Answer:

179.1 kg

Explanation:

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2 years ago
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Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha
krek1111 [17]

Answer:

F=18.58\times 10^{-11}\ N

\theta=30.276^{\circ}

Explanation:

Given:

mass of first particle, m_1=6.7\ kg

mass of second particle, m_2=5.1\ kg

mass of third particle, m_3=3.7\ kg

coordinate position of first particle in meters, (x_1,y_1)\equiv(4.2,0)

coordinate position of second particle in meters, (x_2,y_2)\equiv(0,2.8)

coordinate position of third particle in meters, (x_3,y_3)\equiv(0,0)

<u>Now, gravitational force on particle 3 due to particle 1:</u>

F_{31}=G\frac{m_1.m_3}{r_{31}^2}

F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}

F_{31}=9.37\times 10^{-11}\ N

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

F_{32}=G\frac{m_2.m_3}{r_{21}^2}

F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}

F_{32}=16.05\times 10^{-11}\ N

towards positive X axis.

<u>Now the net force</u>

F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}

F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}

F=18.58\times 10^{-11}\ N

<em>For angle in counterclockwise direction from the +x-axis</em>

tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}

\theta=30.276^{\circ}

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Answer: It's C

Explanation:

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coldgirl [10]

Explanation:

Suppose you want to shine a flashlight beam down a long, straight hallway. Just point the beam straight down the hallway -- light travels in straight lines, so it is no problem. What if the hallway has a bend in it? You could place a mirror at the bend to reflect the light beam around the corner. What if the hallway is very winding with multiple bends? You might line the walls with mirrors and angle the beam so that it bounces from side-to-side all along the hallway. This is exactly what happens in an optical fiber.

The light in a fiber-optic cable travels through the core (hallway) by constantly bouncing from the cladding (mirror-lined walls), a principle called total internal reflection. Because the cladding does not absorb any light from the core, the light wave can travel great distances.

However, some of the light signal degrades within the fiber, mostly due to impurities in the glass. The extent that the signal degrades depends on the purity of the glass and the wavelength of the transmitted light (for example, 850 nm = 60 to 75 percent/km; 1,300 nm = 50 to 60 percent/km; 1,550 nm is greater than 50 percent/km). Some premium optical fibers show much less signal degradation -- less than 10 percent/km at 1,550 nm.

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