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2 years ago

15

The use of air bags in cars reduces the force of impact by a factor of 110.(The resulting force is only as great.) What can be s

aid about how the airbag changed the duration of the collision?

2 answers:

Strike441 [17]2 years ago

5 0

The variation of momentum (= the impulse) of the car during the impact is
$\Delta p = F \Delta t$
$\Delta p$ does not change whether the car has an airbag or not, because
$\Delta p = m\Delta v$
and 1) the mass of the car is always the same 2) the change in velocity of the car is always the same,

so if $\Delta p$ is constant and F is reduced by a factor 110, then $\Delta t$ (the duration of the collision) must be increased by a factor 110 with the airbag.

topjm [15]2 years ago

4 0

the answer is C. it increases by a factor of 110, i got 100 on the test

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2 years ago

This sphere is now connected by a long, thin conducting wire to another sphere of radius R2 that is several meters from the firs

alekssr [168]

Here is the full question

A metal sphere with Radius R₁ has a charge Q₁. Take the electric potential to be zero at an infinite distance from the sphere

a) What are the electric field and electric potential at the surface of the sphere?

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b) what is the total charge on each sphere?

Assume that the amount of charge on the wire is much less than the charge on each sphere.

Answer:

a) The electric field (E) at the surface is the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1^2}$

The electric potential (V) at the surface of the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1}$

= $ER_1$

b)

The total charge of the first sphere $q_1 = \frac{Q_1R_1}{R_1+R_2}$

The total charge of the second sphere $q_2= \frac{Q_1R_2}{R_1+R_2}$

Explanation:

Given that;

the radius of the sphere = R

The radius of the first sphere = $R_1$

The radius of the second sphere = $R_2$

Charge on the first sphere = $Q_1$

a) The electric field (E) at the surface is the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1^2}$

The electric potential (V) at the surface of the first sphere = $\frac{1}{4 \pi \epsilon _0}\frac{Q_1}{R_1}$

= $ER_1$

b) From the question. before the part b question; we learnt that the first sphere is now connected to another sphere;

Now that the two sphere are joined . Charges flows from one to another until their potentials are equal.

As Such; We use $q_1 \ and \ q_2$ to represent their charges respectively

The potential on the surface of the first sphere;

$V_1 = \frac{1}{4 \pi \epsilon _0}\frac{q_1}{R_1}$

The potential on the surface of the second sphere;

$V_2 = \frac{1}{4 \pi \epsilon _0}\frac{q_2}{R_2}$

$V_1=V_2$

∴

$\frac{1}{4 \pi \epsilon _0}\frac{q_1}{R_1}= \frac{1}{4 \pi \epsilon _0}\frac{q_2}{R_2}$

Thus, we can say :

$\frac{q_1}{q_2}= \frac{R_1}{R_2}$

and $Q_1 = q_1 + q_2$

As such ;

The total charge of the first sphere $q_1 = \frac{Q_1R_1}{R_1+R_2}$

The total charge of the second sphere $q_2= \frac{Q_1R_2}{R_1+R_2}$

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2 years ago

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Answer:

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You replace the values of the parameters in the equation (2):

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2 years ago

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svet-max [94.6K]

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2 years ago

Read 2 more answers

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rodikova [14]

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$N_{x}=W*Sin(\alpha)=mg*Sin(\alpha)\\\\N_{y}=W*Cos(\alpha)=mg*Cos(\alpha)$

Now, using the given information, we have:

$mass=m=1150Kg\\\alpha=8.70\°\\g=9.81\frac{m}{s^{2}}$

Calculating, we have:

$N_{x}=mg*Sin(\alpha)$

$N_{x}=1150Kg*9.81\frac{m}{s^{2}}*Sin(8.70\°)\\\\N_{x}=11281.5\frac{Kg.m}{s^{2} }*Sin(8.70\°)=1706.45\frac{Kg.m}{s^{2} }=1706.45.23N$

Hence, we have that the x-component of the normal force is equal to <u>1706.45 N.</u>

Have a nice day!

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2 years ago