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worty [1.4K]
3 years ago
14

For a moving object,When. The force acting on the object varies directly with the object's acceleration. When a force of 64 N ac

ts on a certain object, the acceleration of the object is 8 m/s2. If the acceleration of the object becomes 9 m/s2, what is the force?
Mathematics
1 answer:
Tems11 [23]3 years ago
5 0

Answer:

72 N

Step-by-step explanation:

From the question.

Force is directly proportional to acceleration, with the mass of the object constant.

F α a

F = ka

F₁/a₁ = F₂/a₂....................... Equation 1

Where F₁ = Initial force, a₁ = initial acceleration. F₂ = final force, a₂ = Final acceleration.

make F₂ the subject of the equation

F₂ = (F₁/a₁)a₂................ Equation 2

Given; F₁ = 64 N, a₁ = 8 m/s², a₂ = 9 m/s²

Substitute these values into equation 2

F₂ = (64/8)(9)

F₂ = 72 N

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In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track
Aloiza [94]

Answer:

Approximately 0.31\; \rm m, assuming that g = 9.81\; \rm N \cdot kg^{-1}.

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned} .

Weight of the slide:

\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}.

Normal force between the sled and the slope:

\begin{aligned}F_{\rm N} &= W\cdot  \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}.

Calculate the kinetic friction between the sled and the slope:

\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}.

Assume that the sled and its passenger has reached a height of h meters relative to the base of the slope.

Gain in gravitational potential energy:

\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}.

Distance travelled along the slope:

\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}.

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}.

In other words, the sled and its passenger would have lost (approximately) ((137 + 68.1)\, h)\; {\rm J} of energy when it is at a height of h\; {\rm m}.

The initial amount of energy that the sled and its passenger possessed was \text{KE} = 63\; {\rm J}. All that much energy would have been converted when the sled is at its maximum height. Therefore, when h\; {\rm m} is the maximum height of the sled, the following equation would hold.

((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}.

Solve for h:

(137 + 68.1)\, h = 63.

\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}.

Therefore, the maximum height that this sled would reach would be approximately 0.31\; \rm m.

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Yesterday morning, the temperature was -11. The temperature increased 7 by noon and then decreased 4 by nightfall. What was the
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Answer:

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Step-by-step explanation:

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Given the formula A = 5h (B + b); solve for B.<br> 2
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Answer:

B = A/5h - b; You could use

B = (A - 5hb)/5h  This just puts everything over a common denominator.

Step-by-step explanation:

A = 5h (B + b)          Divide both sides by 5h

A/5h = B + b             Subtract b from both sides.

A/5h - b = B

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