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1. A partikle initially beated at the origin has an acceleration of a=3.00jm/s^2 and an instial velocity of v_(i)=5001m/s. Find (a) The vector position and velocity at any time and (b) The coordinates and speed of the particle at t=2.00s 2. A forice overrightarrow (F)=6i+2jN acts on a particle that undergoes a displacement Delta overrightarrow (r)=(31+D)m Find (a) the work deno by the force on the particle and (b) the angle between overrightarrow (F) and Delta overrightarrow (r) 3. A 500-kg object placed on a frictionless, horvontal table is connected to a strung that passes over a pulley and then is fastened to a hanging 9.00-kg object, as in Figure below a. Draw free-body diagrams of both objects. b. Find the acceleration of the two objects and c The tension in the string. 4 An older model car accelerates from rest to speed 2v in S seconds. A newer, more powerful car accelerates from rest to 3times the speed of old car within 5 seconds. What is the ratio of the power of the newer car to that of the older car?

Pergunta

1. A partikle initially beated at the origin has an acceleration of
a=3.00jm/s^2 and an instial
velocity of v_(i)=5001m/s. Find
(a) The vector position and velocity at any time and
(b) The coordinates and speed of the particle at t=2.00s
2. A forice overrightarrow (F)=6i+2jN acts on a particle that undergoes a displacement
Delta overrightarrow (r)=(31+D)m
Find (a) the work deno by the force on the particle and (b) the angle between
overrightarrow (F) and Delta overrightarrow (r)
3. A 500-kg object placed on a frictionless, horvontal table is connected to a strung that passes
over a pulley and then is fastened to a hanging 9.00-kg object, as in Figure below
a. Draw free-body diagrams of both objects.
b. Find the acceleration of the two objects and
c The tension in the string.
4 An older model car accelerates from rest to speed 2v in S seconds. A newer, more powerful
car accelerates from rest to 3times the speed of old car within 5 seconds. What is the ratio of
the power of the newer car to that of the older car?

1. A partikle initially beated at the origin has an acceleration of a=3.00jm/s^2 and an instial velocity of v_(i)=5001m/s. Find (a) The vector position and velocity at any time and (b) The coordinates and speed of the particle at t=2.00s 2. A forice overrightarrow (F)=6i+2jN acts on a particle that undergoes a displacement Delta overrightarrow (r)=(31+D)m Find (a) the work deno by the force on the particle and (b) the angle between overrightarrow (F) and Delta overrightarrow (r) 3. A 500-kg object placed on a frictionless, horvontal table is connected to a strung that passes over a pulley and then is fastened to a hanging 9.00-kg object, as in Figure below a. Draw free-body diagrams of both objects. b. Find the acceleration of the two objects and c The tension in the string. 4 An older model car accelerates from rest to speed 2v in S seconds. A newer, more powerful car accelerates from rest to 3times the speed of old car within 5 seconds. What is the ratio of the power of the newer car to that of the older car?

Solução

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RenanProfissional · Tutor por 6 anos

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1. (a) The vector position \(\vec{r}(t)\) and velocity \(\vec{v}(t)\) of the particle at any time can be found using the equations of motion. Given the initial velocity \(\vec{v}_i = 5.00 \, \hat{i} \, \text{m/s}\) and acceleration \(\vec{a} = 3.00 \, \hat{j} \, \text{m/s}^2\), we have:<br /><br />- Velocity: \(\vec{v}(t) = \vec{v}_i + \vec{a}t = 5.00 \, \hat{i} + 3.00t \, \hat{j} \, \text{m/s}\).<br /><br />- Position: \(\vec{r}(t) = \vec{r}_i + \vec{v}_i t + \frac{1}{2} \vec{a} t^2 = (5.00t) \, \hat{i} + \left(\frac{1}{2} \times 3.00 \times t^2\right) \, \hat{j} = 5.00t \, \hat{i} + 1.50t^2 \, \hat{j} \, \text{m}\).<br /><br />(b) At \(t = 2.00 \, \text{s}\):<br /><br />- Velocity: \(\vec{v}(2.00) = 5.00 \, \hat{i} + 3.00 \times 2.00 \, \hat{j} = 5.00 \, \hat{i} + 6.00 \, \hat{j} \, \text{m/s}\).<br /><br />- Position: \(\vec{r}(2.00) = 5.00 \times 2.00 \, \hat{i} + 1.50 \times (2.00)^2 \, \hat{j} = 10.0 \, \hat{i} + 6.00 \, \hat{j} \, \text{m}\).<br /><br />- Speed: \(|\vec{v}(2.00)| = \sqrt{(5.00)^2 + (6.00)^2} = \sqrt{25 + 36} = \sqrt{61} \approx 7.81 \, \text{m/s}\).<br /><br />2. (a) The work done by the force is given by the dot product of the force and displacement vectors: \(W = \overrightarrow{F} \cdot \Delta \overrightarrow{r}\).<br /><br />Given \(\overrightarrow{F} = 6\hat{i} + 2\hat{j} \, \text{N}\) and \(\Delta \overrightarrow{r} = (3\hat{i} + D\hat{j}) \, \text{m}\), the work done is:<br /><br />\[ W = (6\hat{i} + 2\hat{j}) \cdot (3\hat{i} + D\hat{j}) = 6 \times 3 + 2 \times D = 18 + 2D \, \text{J}. \]<br /><br />(b) The angle \(\theta\) between \(\overrightarrow{F}\) and \(\Delta \overrightarrow{r}\) can be found using the dot product formula:<br /><br />\[ \cos \theta = \frac{\overrightarrow{F} \cdot \Delta \overrightarrow{r}}{|\overrightarrow{F}| |\Delta \overrightarrow{r}|}. \]<br /><br />Calculate magnitudes:<br /><br />\[ |\overrightarrow{F}| = \sqrt{6^2 + 2^2} = \sqrt{40}, \]<br />\[ |\Delta \overrightarrow{r}| = \sqrt{3^2 + D^2} = \sqrt{9 + D^2}. \]<br /><br />Then,<br /><br />\[ \cos \theta = \frac{18 + 2D}{\sqrt{40} \sqrt{9 + D^2}}. \]<br /><br />3. (a) Free-body diagrams:<br /><br />- For the 500-kg object on the table: Tension \(T\) acts horizontally.<br />- For the 9.00-kg hanging object: Weight \(mg\) acts downward, and tension \(T\) acts upward.<br /><br />(b) To find the acceleration \(a\), use Newton's second law for both objects:<br /><br />For the 500-kg object: \(T = 500a\).<br /><br />For the 9.00-kg object: \(mg - T = 9.00a\).<br /><br />Solving these equations simultaneously:<br /><br />\[ 9.00g - 9.00a = 500a, \]<br />\[ 9.00 \times 9.81 = 509a, \]<br />\[ a = \frac{88.29}{509} \approx 0.173 \, \text{m/s}^2. \]<br /><br />(c) The tension in the string \(T\) is:<br /><br />\[ T = 500a = 500 \times 0.173 = 86.5 \, \text{N}. \]<br /><br />4. The power ratio of the newer car to the older car can be determined using the definition of power as the rate of doing work or the rate of change of kinetic energy.<br /><br />For the older car, the final speed is \(2v\) in \(S\) seconds. The kinetic energy change is proportional to \((2v)^2\).<br /><br />For the newer car, the final speed is \(3 \times 2v = 6v\) in 5 seconds. The kinetic energy change is proportional to \((6v)^2\).<br /><br />The power ratio is:<br /><br />\[ \text{Power ratio} = \frac{\frac{1}{2} m (6v)^2 / 5}{\frac{1}{2} m (2v)^2 / S} = \frac{36v^2 / 5}{4v^2 / S} = \frac{36S}{20} = \frac{9S}{5}. \]
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