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2 years ago

11

Most persons are of the belief that the old time hard or solid iron cars are much safer in collisions than the new model softer

cars. But how can newer, softer cars be safer in a collision, considering momentum?

1 answer:

insens350 [35]2 years ago

5 0

Softer cars will crumple better in a collision- they absorb more of the impact force and therefore are safer

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On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole

ohaa [14]

The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
$F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\$
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. $\alpha$ is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
$g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
g_0=g'\\
g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\
ag_0=g_0-w^2r\\
w^2r=g_0(a-1)
$
Our final equation is:
$g=g_0-g_0(a-1)cos^2(\alpha)$
Colatitude is:
$\alpha_c=90^\circ-\alpha$
The answer is:
$g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)$

5 0

3 years ago

Find the wavelength λ of the 80.0-khz wave emitted by the bat. express your answer in millimeters.

vfiekz [6]

Answer:

4.29 millimeters

Explanation:

Bats emit ultrasound waves: in air, ultrasound waves travel at a speed of

$v=343 m/s$

The frequency of the waves emitted by this bat is:

$f=80.0 kHz = 80,000 Hz$

Therefore we can find the wavelength of the wave emitted by the bat by using the relationship between speed, frequency and wavelength:

$\lambda=\frac{v}{f}=\frac{343 m/s}{80,000 Hz}=4.29\cdot 10^{-3} m=4.29 mm$

4 0

3 years ago

What experimental evidence led to the development of this atomic model from the one before it?

Marina86 [1]

Answer:

A few of the positive particles aimed at a gold foil seemed to bounce back

Explanation:

7 0

2 years ago

Read 2 more answers

Two electric charges A and B were placed facing each other at a distance of separation "r". The common electrostatic force betwe

lutik1710 [3]

Answer:

From the formula of force:

$F = \frac{kAB}{ {r}^{2} } \\$

since AB and k are constants:

$F \: \alpha \: \frac{1}{ {r}^{2} } \\ \\ F = \frac{x}{ {r}^{2} }$

x is a constant of proportionality

• when force is 4N, separation distance is 1

$4 = \frac{x}{1} \\ x = 4$

therefore, equation becomes

$F = \frac{4}{ {r}^{2} } \\$

when r is doubled, r becomes 2. find F:

$F = \frac{4}{ {2}^{2} } \\ \\ F = \frac{4}{4} \\ \\ { \underline{force \: is \: 1N}}$

5 0

3 years ago

Help me with this???

Vikentia [17]

Yo sup??

Average velocity=total distance covered/total time taken

total distance covered=4 + 8=12 miles

total time taken=6 hours

Therefore

average velocity=12/6

=2 miles/hour

Hope this helps

8 0

3 years ago