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2 years ago

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Three identical 50-kg balls are held at the corners of an equilateral triangle, 30 cm on each side. if one of the balls is relea

sed, what is the magnitude of its initial acceleration if the only forces acting on it are the gravitational forces due to the other two masses? (g = 6.67 × 10-11 n • m2/kg2)

2 answers:

Ivahew [28]2 years ago

6 0

Answer:

a= 6.42×10⁻⁸m/s²

Explanation:

Given Data

mass=m=50 kg

angle=α=30°

Gravitational Constant=G=6.67 × 10-11 Nm²/kg²)

To find

acceleration=a=?

Solution

From Gravitational Law " the force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface "

F₁ = G×m²/D²

F = 2*F₁*cos30° = 2*G*50²*cos30°/0.3² = 3.21×10⁻⁶N

a = F/m = 3.21×10⁻⁶/50

a= 6.42×10⁻⁸m/s²

natima [27]2 years ago

3 0

F1 = G*m²/D²
<span>F = 2*F1*cos30° = 2*G*50²*cos30°/0.3² = 3.21E-6 </span>

<span>a = F/m = 3.21E-6/50 = 6.42E-8

Hope this helped!
STSN</span>

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Sea breezes that occur near the shore are attributed to a difference between land and water with respect to what property?

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Answer:

a. mass density

Explanation:

<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>

When the air gets heated it becomes rarer in density and thus rises up in the atmosphere and its space is occupied by a cooler and denser air that flows to the place.

<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>

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2 years ago

Two fun-loving otters are sliding toward each other on a muddy (and hence frictionless) horizontal surface. One of them, of mass

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Answer:

(a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

Explanation:

Given that,

Mass of one otter = 8.50 kg

Speed = 6.00 m/s

Mass of other = 5.75 kg

Speed = 5.50 m/s

(a). We need to calculate the magnitude and direction of the velocity of these free-spirited otters right after they collide

Using conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v$

Put the value into the formula

$8.50\times(-6.00)+5.75\times5.50=(8.50+5.75)\times v$

$v=\dfrac{-19.375}{14.25}$

$v=-1.35\ m/s$

Negative sign shows the direction of motion of the object after collision is toward left.

(b). We need to calculate the initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}\times8.50\times(6.00)^2+\dfrac{1}{2}\times5.75\times(5.50)^2$

$K.E_{i}=239.96\ J$

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Using formula of kinetic energy

$K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Put the value into the formula

$K.E_{f}=\dfrac{1}{2}\times(8.50+5.75)\times(-1.35)^2$

$K.E_{f}=12.98\ J$

We need to calculate the mechanical energy dissipates during this play

Using formula of loss of mechanical energy

$\Delta K.E=K.E_{f}-K.E_{i}$

Put the value into the formula

$\Delta K.E=12.98-239.96$

$\Delta K.E=-226.98\ J$

Negative sign shows the loss of mechanical energy

Hence, (a). The magnitude and direction of the velocity of the otters after collision is 1.35 m/s toward left.

(b). The mechanical energy dissipates during this play is 226.98 J.

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Anni [7]

Answer:

$\mathbf{1.51\times10^{-15}N}$

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1 year ago