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3 years ago

explain why a high-speed collision between two cars would cause more damage than a low-speed collision between the same two cars

1 answer:

8 0

The force equation can easily prove this. F=ma. This states that the force on an object is equal to mass times acceleration. If the mass stays the same and the velocity of the cars increases than that means there is a larger force. This is because in both cases the cars are stopping in almost an instant and the times of the crashes are theoretically identical. Acceleration is the change in velocity over time. If the velocity is higher with the same amount of time than that means there is a higher acceleration. If you plug a higher acceleration into the force equation then you wind up with a higher force and in turn a more damaging collision.

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A satellite is in a circular orbit around the Earth at an altitude of 3.18x10 m. Find the period and th orbital speed of the sat

Leona [35]

Answer:

112.17 m/s

56.427 years

Explanation:

h = 3.18 x 10^10 m

R = 6.4 x 10^6 m

r = R + h = 3.18064 x 10^10 m

M = 6 x 10^24 kg

The formula for the orbital velocity is given by

$v = \sqrt{\frac{G M }{r}}$

$v = \sqrt{\frac{6.67 \times 10^{-11}\times 6\times 10^{24} }{3.18064\times 10^{10}}}$

v = 112.17 m/s

Orbital period, T = 2 x 3.14 x 3.18064 x 10^10 / 112.17

T = 0.178 x 10^10 s

T = 56.427 years

8 0

3 years ago

How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight

Natalka [10]

By using Ohm's law, we can find what should be the resistance of the wire, R:
$R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega$

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
$r=35 mm=0.35 \cdot 10^{-3} m$
So the area is
$A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2$

And by using the resistivity of the Aluminum, $\rho=2.65 \cdot 10^{-8} \Omega m$ , we can use the relationship between resistance R and resistivity:
$R= \frac{\rho L}{A}$
to find L, the length of the wire:
$L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m$

4 0

2 years ago

Which of these common substances is a heterogeneous mixture?

LenaWriter [7]

Answer:

(C) If the composition of a mixture appears uniform no matter where you sample it, is homogeneous; sand on a beach *IS HETEROGENEOUS* because when you look at it up close, you can identify different types of particles, such as sand, shells, and organic matter.

Explanation:

(A) Pure Water is a collection of solely H2O molecules therefore Pure Water is classified as a *Compound*.

(B) Table Salt is NOT a heterogeneous mixture because the particles of salt can't be separated, and it is a *Pure Substance*.

(D) Maple Syrup is a homogeneous mixture because the solutes are fully dissolved and not easily identified. In other words, Maple Syrup is uniform throughout.

-Hope this helps!

4 0

2 years ago

The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1

telo118 [61]

<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life $h$ of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: $h_{B-13}=5(10)^{-10}s=0.0000000005s$

beryllium-15: $h_{B-15}=2(10)^{-7}s=0.0000002s$

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

$h_{B-15}=X.h_{B-13}$

Where $X$ is the amount we want to find:

$X=\frac{h_{B-15}}{h_{B-13}}$

$X=\frac{2(10)^{-7}s}{5(10)^{-10}s}$

Finally:

$X=400$

Therefore:

The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.

8 0

2 years ago

For each of the problems below, you will need to draw a graph to find the solution.

Gelneren [198K]

Answer:

Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. Use standard gravity, a = 9.80665 m/s2, for equations involving the Earth's gravitational force as the acceleration rate of an object.

Explanation:

3 0

2 years ago