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solniwko [45]
3 years ago
15

A giant armadilo moving northward with a constant acceleration covers the distance between two points 60m apart in 6 seconds. It

s velocity as it passes the second point is 15m/s
what is the acceleration? (answer is 1.7) please explain how to solve this problem in detail.
Physics
1 answer:
Naddika [18.5K]3 years ago
8 0
Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. To determine acceleration, we need to know the initial velocity and the final velocity and the time elapsed. From the given values, we need t o calculate for the initial velocity. We use some kinematic equations. We do as follows:

 x = v0t + at^2/2
60 = v0(6) + a(6)^2/2
60 = 6v0 + 18a          (EQUATION 1)

vf = v0 + at
15 = v0 + a(6)
15 = v0 + 6a             (EQUATION 2)

Solving for v0 and a,
v0 = 5 m/s
a = 1.7 m/s^2
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If the temperature of water warming on a hot plate rises 10 degrees Celsius [°C], what is the change in temperature in units of
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Answer:

Delta_temp = 18[°F]

Explanation:

°F = 9/5*(10)

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4 0
3 years ago
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413
natima [27]

Answer:

The value of  charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force F=14.413\ N

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

r=\sqrt{x_{2}^2+y_{2}^2}

Put the value into the formula

r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}

r=0.0876\ m

We need to calculate the magnitude of the charge q₃

Using formula of net force

F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})

Put the value into the formula

14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})

(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}

\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}

q_{3}=0.0210909\times(0.0438)^2

q_{3}=40.46\times10^{-6}\ C

q_{3}=40.46\ \mu C

Hence, The value of  charge q₃ is 40.46 μC.

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