Answer quick please!!

An aircraft has a liftoff speed of 33 m/s. What is the minimum constant acceleration an

JulsSmile [24]

Answer: The minimum acceleration for the air plane is 2.269m/s2.

Explanation: To solve such problem the equation of motion are applicable.

The initial velocity is 0 since the airplane was initially standing. We are going to use this equation

V^2=U^2+2as

33^2=0+2a (240)

a= 2.269m/s2

5 0

3 years ago

On the way to school, Jed traveled 100m north, 300m east, 100 north, 100m east, and 100m north. A.) Find the total distance trav

Oduvanchick [21]

Answer:

Total distance = 700 m

Displacement = 500 m

Explanation:

Notice that Jed travelled a total of 3 x 100 m = 300 m in the North direction, and 300 m + 100 m = 400 m in the East direction. Therefore the total distance he travelled is: 300 + 400 = 700 m.

But the actual displacement is given by the Pythagorean theorem as the hypotenuse of a right angle triangle of legs 300 m and 400 m:

displacement = $\sqrt{300^2+400^2} =\sqrt{250000} =500\,m$

5 0

3 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

2 years ago

In Trial III, a different, looser, spring is used; its force constant is 23.1 N/m. The suspended mass is the same as the one in

cricket20 [7]

Answer:

T=0.827s

Explanation:

The period of a spring can be calculated with the equation

$T=2\pi w$

But we know as well that w is given by,

$w=\sqrt{\frac{k}{m}}$

Replacing,

$w=\frac{2\pi}{T}= \sqrt{\frac{k}{m}}\\T=2\pi\sqrt{\frac{k}{m}}\\T=2\pi\sqrt{\frac{0.4}{23.1}}$

So we have that

$T=0.827s$

7 0

2 years ago

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa

Sonja [21]

Answer:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

$c_{pA} > c_{pB}$

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

$T_{iA}= T_{iB}$

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

$Q = m c_p \Delta T$

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

$Q_A = Q_B$

And if we replace the formula for sensible heat we got:

$m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B$

And if we replace for the change of the temperature we got:

$m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})$

And since $T_{iA}= T_{iB}= T$ we have this:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

4 0

3 years ago