Describe and contrast inversion and eversion?

A busy chipmunk runs back and forth along a straight line of acorns that has been set out between its burrow and a nearby tree.

dmitriy555 [2]

<u>Answer:</u>

Chipmunk's average acceleration during the 2.07 s time interval = 0.1256 $m/s^2$

<u>Explanation:</u>

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

In this case we need to find acceleration value, we have initial velocity = 1.47 m/s, final velocity = 1.73 m/s and time taken = 2.07 seconds.

Substituting

1.73 = 1.47 + a * 2.07

a = 0.1256 $m/s^2$

Chipmunk's average acceleration during the 2.07 s time interval = 0.1256 $m/s^2$

7 0

4 years ago

The angular velocity of a 755-g wheel 15.0 cm in diameter is given by the equation ω(t) = (2.00 rad/s2)t + (1.00 rad/s4)t 3. (a)

Lunna [17]

Answer:

a

The number of radians turned by the wheel in 2s is $\theta= 8\ radians$

b

The angular acceleration is $\alpha =14 rad/s^2$

Explanation:

The angular velocity is given as

$w(t) = (2.00 \ rda/s^2)t + (1.00 rad /s^4)t^3$

Now generally the integral of angular velocity gives angular displacement

So integrating the equation of angular velocity through the limit 0 to 2 will gives us the angular displacement for 2 sec

This is mathematically evaluated as

$\theta(t ) = \int\limits^2_0 {2t + t^3} \, dt$

$= [\frac{2t^2}{2} + \frac{t^4}{4}] \left\{ 2} \atop {0}} \right.$

$= [\frac{2(2^2)}{2} + \frac{2^4}{4}] - 0$

$= 4 +4$

$\theta= 8\ radians$

Now generally the derivative of angular velocity gives angular acceleration

So the value of the derivative of angular velocity equation at t= 2 gives us the angular acceleration

This is mathematically evaluated as

$\frac{dw}{dt} = \alpha (t) = 2 + 3t^2$

so at t=2

$\alpha (2) = 2 +3(2)^2$

$\alpha =14 rad/s^2$

7 0

3 years ago

Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

sergij07 [2.7K]

Answer:

A) 50.0 g Al

Explanation:

We can calculate the temperature change of each substance by using the equation:

$\Delta T=\frac{Q}{mC_s}$

where

Q = 200.0 J is the heat provided to the substance

m is the mass of the substance

$C_s$ is the specific heat of the substance

Let's apply the formula for each substance:

A) m = 50.0 g, Cs = 0.903 J/g°C

$\Delta T=\frac{200}{(50)(0.903)}=4.4^{\circ}C$

B) m = 50.0 g, Cs = 0.385 J/g°C

$\Delta T=\frac{200}{(50)(0.385)}=10.4^{\circ}C$

C) m = 25.0 g, Cs = 0.79 J/g°C

$\Delta T=\frac{200}{(25)(0.79)}=10.1^{\circ}C$

D) m = 25.0 g, Cs = 0.128 J/g°C

$\Delta T=\frac{200}{(25)(0.128)}=62.5^{\circ}C$

E) m = 25.0 g, Cs = 0.235 J/g°C

$\Delta T=\frac{200}{(25)(0.235)}=34.0^{\circ}C$

As we can see, substance A) (Aluminium) is the one that undergoes the smallest temperature change.

7 0

3 years ago

A 69 g particle is moving to the left at 25 m/s . How much net work must be done on the particle to cause it to move to the righ

LekaFEV [45]

Answer:11.59 J

Explanation:

Given

mass of Particle $m=69 gm$

Initially Particle moves towards left $v_1=25 m/s$

Final velocity of Particle is towards Right $v_2=31 m/s$

According to Work Energy theorem

Work done by all the Forces=change in Kinetic Energy

Work done by Force $=\frac{mv_2^2}{2}-\frac{mv_1^2}{2}$

$W=\frac{69\times 10^{-3}}{2}\left [ 31^-25^2\right ]$

$W=\frac{69\times 10^{-3}}{2}\left [ 961-625\right ]$

$W=11.59 J$

5 0

4 years ago

A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a

EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is: