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4 years ago

Find the velocity, acceleration, and speed of a particle with the given position function. r(t = t2i 6tj 4 ln t k

1 answer:

6 0

1st derivative gives velocity;
d r(t)/ dt = 2t i + 6 j + 4/t k

2nd derivative gives acceleration;
d^2 r(t)/ dt^2 = 2 i - 4/ t^2

Speed ;
Square root of (4 t^2 + 36 + 16/ t^2)

For a given time, like 2 seconds, t will be 2. And answer of speed will be scalar.

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A crate is dragged up a ramp at constant speed. The work done on the system can be accounted for by:

Alex

Answer:

The work done on the system can be accounted for by;

Both $E_g$ and $E_{int}$

Explanation:

The speed of the crate = Constant

Therefore, the acceleration of the crate = 0 m/s²

The net force applied to the crate, $F_{NET}$ = 0

Therefore, the force of with which the crate is pulled = The force resisting the upward motion of the crate

However, we have;

The force resisting the upward motion of the crate = The weight of the crate + The frictional resistance of the ramp due to the surface contact between the ramp and the crate

The work done on the system = The energy to balance the resisting force = The weight of the crate × The height the crate is raised + The heat generated as internal energy to the system

The weight of the crate × The height the crate is raised = Gravitational Potential Energy = $E_g$

The heat generated as internal energy to the system = $E_{int}$

Therefore;

The work done on the system = $E_g$ + $E_{int}$ .

6 0

3 years ago

The amount of work required for a

kirill [66]

Answer:

if I'm not wrong, 0 Joules

Explanation:

If it is not being moved at all, then no work is being done

8 0

3 years ago

G = 10 N/kg or 10 m/s2

Irina18 [472]

Answer:

a) $U_{g} = 40\,J$ , b) $\eta = 70\,\%$ , c) $v = 20\,\frac{m}{s}$

Explanation:

a) The initial potential energy is:

$U_{g} = (0.2\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (20\,m)$

$U_{g} = 40\,J$

b) The efficiency of the bounce is:

$\eta = \left(\frac{14\,m}{20\,m} \right)\times 100\,\%$

$\eta = 70\,\%$

c) The final speed of Danielle right before reaching the bottom of the hill is determined from the Principle of Energy Conservation:

$K = U_{g}$

$U_{g} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot U_{g}}{m} }$

$v = \sqrt{\frac{2\cdot (40\,J)}{0.2\,kg} }$

$v = 20\,\frac{m}{s}$

5 0

3 years ago

A 1.0 kg object is attached to a 0.50 m string. It is twirled in a horizontal circle above the ground at a speed of 5.0 m/s.

7nadin3 [17]

Given that,

Mass of the object, m = 1 kg

It moves in a circle of radius 0.5 m with a speed of 5 m/s

To find,

The direction of the acceleration.

Solution,

Whenever an object moves in a circular path, the only force that acts on its is centripetal force which is given by the formula as follows :

$F=\dfrac{mv^2}{r}$

The centripetal acceleration acts in the direction of force. It acts along the radius of the circular path. In this figure, the direction of the acceleration is shown at point d.

4 0

3 years ago

Because plate movements have raised ancient sea floors above sea level,

Elena-2011 [213]

Because plate movements have raised ancient sea floors above sea level, _________?

<span>Answer : Limestone that began as coral reefs can be found on the continents.</span>

3 0

3 years ago

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