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3 years ago

What system will break the chewed food down to form your human cell can use

Physics

1 answer:

PolarNik [594]3 years ago

4 0

Your stomach. afterwards it is dissolved by the stomach acid and the nutrients travel to your cells

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Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.650 µC charge and flies due west at a speed of

Ganezh [65]

Answer : $F=2808\times 10^{-11}\ N$

Explanation :

It is given that,

Charge, $q=0.650\ \mu C=0.65\times 10^{-6}\ C$

Velocity of Aircraft, $v=540\ m/s$ (in west)

Magnetic field, $B=8 \times 10^{-5}\ T$ ( in north )

Using the relation as :

$F=q(v\times B)$

Magnetic force is ,

$F=0.65\times 10^{-6}\ C\times 540\ m/s\times 8 \times 10^{-5}\ T$

$F=2808\times 10^{-11}\ N$

Using Right hand thumb rule, the direction of force is in the plane perpendicular to the velocity and the magnetic field i.e. $-\hat{k}$ .

4 0

4 years ago

Describe a structural adaption that a snake has. What challenge does it overcome? PLZ HELP ASAP

PolarNik [594]

It has a flexible vertibrae which allows it to move faster and fit into places . Plus it also allows it to get its prey easier

6 0

3 years ago

How does the angle of launch affect the kinetic energy of a rubber band?

Lady_Fox [76]

Answer:

The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.

3 0

3 years ago

while test flying the new f-22 raptor fighter jet, test pilot bob put the plane in a horizontal loop while flying at the speed o

sveticcg [70]

Given

v = 343 m/s

ac = 5g

ac = 5*9.8 m/s^2

ac = 49 m/s^2

where,

v: velocity

ac = centripetal aceleration

Procedure

We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration ac; centripetal means “toward the center” or “center seeking”.

Formula

$\begin{gathered} a_c=\frac{v^2}{r} \\ r=\frac{v^2}{a_c} \\ r=\frac{(343m/s)^2}{49m/s^2} \\ r=2401\text{ m} \end{gathered}$

The minimum radius not to exceed the centripetal acceleration is 2401 m.

8 0

2 years ago

If position vector r = bt^2i + ct^3j, where b and c are positive constants, when does the velocity vector make an angle of 450 w

vazorg [7]

$\vec r(t)=bt^2\,\vec\imath+ct^3\,\vec\jmath$

The velocity at time $t$ is

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=2bt\,\vec\imath+3ct^2\,\vec\jmath$

Take two vectors that point in the positive $x$ and positive $y$ directions, such as $\vec\imath$ and $\vec\jmath$ . The dot products of the velocity vector with $\vec\imath$ and $\vec\jmath$ are

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\imath=2bt=\sqrt{4b^2t^2+9c^2t^4}\cos\theta$

and

$\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\jmath=3ct^2=\sqrt{4b^2t^2+9c^2t^4}\cos\theta$

We want the angles between these vectors to be 45º, for which we have $\cos45^\circ=\frac1{\sqrt2}$ . So

$\begin{cases}2\sqrt2\,bt=\sqrt{4b^2t^2+9c^2t^4}\\3\sqrt2\,ct^2=\sqrt{4b^2t^2+9c^2t^4}\end{cases}\implies3\sqrt2\,ct^2-2\sqrt2\,bt=0$

$\implies t(3ct-2b)=0$

$\implies t=0\text{ or }t=\dfrac{2b}{3c}$

When $t=0$ , the velocity vector is equal to the zero vector, which technically has no direction/doesn't make an angle with any other vector. So the only time this happens is for

$\boxed{t=\dfrac{2b}{3c}}$

7 0

3 years ago